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Mathematics LibreTexts

4-7. Exponential and Logarithmic Models

Exponential and Logarithmic Models
In this section, you will:
  • Model exponential growth and decay.
  • Use Newton’s Law of Cooling.
  • Use logistic-growth models.
  • Choose an appropriate model for data.
  • Express an exponential model in base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi></mrow></annotation-xml></semantics></math>.
<figure class="small" id="CNX_Precalc_Figure_04_07_001" style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 1; word-spacing: 0px; -webkit-text-stroke-width: 0px;"> <figcaption>A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus (credit: Georgia Tech Research Institute)</figcaption> Inside a nuclear research reactor.</figure>

We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling.

Modeling Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math>

where <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> is equal to the value at time zero, <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>e</mi></annotation-xml></semantics></math> is Euler’s constant, and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>k</mi></annotation-xml></semantics></math> is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math> where <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> is the starting value, and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>e</mi></annotation-xml></semantics></math> is Euler’s constant. Now <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>k</mi></annotation-xml></semantics></math> is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in [link] and [link]. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis.

<figure class="small" id="CNX_Precalc_Figure_04_07_002"> <figcaption>A graph showing exponential growth. The equation is <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>2</mn><msup><mi>e</mi><mrow><mn>3</mn><mi>x</mi></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math></figcaption> Graph of y=2e^(3x) with the labeled points (-1/3, 2/e), (0, 2), and (1/3, 2e) and with the asymptote at y=0.</figure> <figure class="small" id="CNX_Precalc_Figure_04_07_003"> <figcaption>A graph showing exponential decay. The equation is <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>3</mn><msup><mi>e</mi><mrow><mo>−</mo><mn>2</mn><mi>x</mi></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math></figcaption> Graph of y=3e^(-2x) with the labeled points (-1/2, 3e), (0, 3), and (1/2, 3/e) and with the asymptote at y=0.</figure>

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4.01134972</mn><mtext> </mtext><mo>×</mo><mtext> </mtext><msup><mrow><mn>10</mn></mrow><mrow><mn>13</mn></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math> So, we could describe this number as having order of magnitude <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mn>10</mn></mrow><mrow><mn>13</mn></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math>

Characteristics of the Exponential Function, y = A0ekt

An exponential function with the form <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math> has the following characteristics:

  • one-to-one function
  • horizontal asymptote: <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>0</mn></mrow></annotation-xml></semantics></math>
  • domain: <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mo>–</mo><mi>∞</mi><mo>,</mo><mo> </mo><mi>∞</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
  • range: <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo stretchy="false">(</mo><mn>0</mn><mo>,</mo><mi>∞</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
  • x intercept: none
  • y-intercept: <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><msub><mi>A</mi><mn>0</mn></msub></mrow><mo>)</mo></mrow></mrow></annotation-xml></semantics></math>
  • increasing if <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>></mo><mn>0</mn></mrow></annotation-xml></semantics></math> (see [link])
  • decreasing if <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo><</mo><mn>0</mn></mrow></annotation-xml></semantics></math> (see [link])
<figure class="medium" id="CNX_Precalc_Figure_04_07_004"> <figcaption>An exponential function models exponential growth when <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>></mo><mn>0</mn></mrow></annotation-xml></semantics></math> and exponential decay when <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo><</mo><mn>0.</mn></mrow></annotation-xml></semantics></math></figcaption> Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k>0 and with the labeled points (1/k, (A_0)e), (0, A_0), and (-1/k, (A_0)/e). The second graph is of when k<0 and with the labeled points (-1/k, (A_0)e), (0, A_0), and (1/k, (A_0)/e).</figure>
Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> we use the fact that <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> is the amount at time zero, so <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub><mo>=</mo><mn>10.</mn></mrow></annotation-xml></semantics></math> To find <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>,</mo></mrow></annotation-xml></semantics></math> use the fact that after one hour <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo><mrow><mi>t</mi><mo>=</mo><mn>1</mn></mrow><mo>)</mo></mrow></mrow></annotation-xml></semantics></math> the population doubles from <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>10</mn></mrow></annotation-xml></semantics></math> to <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>20.</mn></mrow></annotation-xml></semantics></math> The formula is derived as follows

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mn>20</mn><mo>=</mo><mn>10</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> e k⋅1    2= e k Divide by 10 ln2=k Take the natural logarithm

so <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>=</mo><mi>ln</mi><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics></math> Thus the equation we want to graph is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>10</mn><msup/></mrow></annotation-xml></semantics></math> e (ln2)t =10 ( e ln2 ) t =10· 2 t . The graph is shown in [link].

<figure class="small" id="CNX_Precalc_Figure_04_07_005"> <figcaption>The graph of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>10</mn><msup><mi>e</mi><mrow><mrow><mo>(</mo><mrow><mi>ln</mi><mn>2</mn></mrow><mo>)</mo></mrow><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math></figcaption> A graph starting at ten on the y-axis and rising rapidly to the right.</figure>
Analysis

The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mn>10</mn></mrow><mn>4</mn></msup><mo>.</mo></mrow></annotation-xml></semantics></math>The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mrow><mn>10</mn></mrow><mn>7</mn></msup><mo>,</mo></mrow></annotation-xml></semantics></math> so we could say that the population has increased by three orders of magnitude in ten hours.

Half-Life

We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>A</mi><mn>0</mn></msub><mo>=</mo><msub><mi>A</mi><mi>o</mi></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math>

We find that the half-life depends only on the constant <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>k</mi></annotation-xml></semantics></math> and not on the starting quantity <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub><mo>.</mo></mrow></annotation-xml></semantics></math>

The formula is derived as follows

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 1 2 A 0 = A o e kt                1 2 = e kt Divide by  A 0 .      ln( 1 2 )=kt Take the natural log. −ln(2)=ktApply laws of logarithms.  − ln(2) k =t Divide by k.

Since <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>,</mo></mrow></annotation-xml></semantics></math> the time, is positive, <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>k</mi></annotation-xml></semantics></math> must, as expected, be negative. This gives us the half-life formula

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>=</mo><mo>−</mo><mfrac><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo></mrow><mi>k</mi></mfrac></mrow></annotation-xml></semantics></math>

Given the half-life, find the decay rate.

  1. Write <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><msub><mi>A</mi><mi>o</mi></msub><msup><mi>e</mi><mrow><mi>k</mi><mi>t</mi></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math>
  2. Replace <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>A</mi></annotation-xml></semantics></math> by <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> and replace <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>t</mi></annotation-xml></semantics></math> by the given half-life.
  3. Solve to find <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>.</mo></mrow></annotation-xml></semantics></math> Express <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>k</mi></annotation-xml></semantics></math> as an exact value (do not round).

Note: It is also possible to find the decay rate using <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>=</mo><mo>−</mo><mfrac><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo></mrow><mi>t</mi></mfrac><mo>.</mo></mrow></annotation-xml></semantics></math>

Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

This formula is derived as follows.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>        </mtext><mi>A</mi><mo>=</mo><msub/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> A 0 e kt The continuous growth formula.   0.5 A 0 = A 0 e k⋅5730 Substitute the half-life for t and 0.5 A 0  for f(t).      0.5= e 5730k Divide by  A 0 . ln(0.5)=5730k Take the natural log of both sides.          k= ln(0.5) 5730Divide by the coefficient of k.         A= A 0 e ( ln(0.5) 5730 )t Substitute for r in the continuous growth formula.

 

The function that describes this continuous decay is <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mrow><mo>(</mo><mrow><mfrac><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>0.5</mn><mo stretchy="false">)</mo></mrow><mrow><mn>5730</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mi>t</mi></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math> We observe that the coefficient of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>,</mo></mrow></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>ln</mi><mo stretchy="false">(</mo><mn>0.5</mn><mo stretchy="false">)</mo></mrow><mrow><mn>5730</mn></mrow></mfrac><mo>≈</mo><mo>−</mo><mn>1.2097</mn></mrow></annotation-xml></semantics></math> is negative, as expected in the case of exponential decay.

The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mo>−</mo><mn>0.0000000087</mn><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math>

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>t</mi></annotation-xml></semantics></math> years is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>≈</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e ( ln(0.5) 5730 )t

where

 

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>A</mi></annotation-xml></semantics></math> is the amount of carbon-14 remaining
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></annotation-xml></semantics></math> is the amount of carbon-14 when the plant or animal began decaying.

 

This formula is derived as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>        </mtext><mi>A</mi><mo>=</mo><msub/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> A 0 e kt The continuous growth formula.  0.5 A 0 = A 0 e k⋅5730 Substitute the half-life for t and 0.5 A 0  for f(t).      0.5= e 5730k Divide by  A 0 . ln(0.5)=5730k Take the natural log of both sides.          k= ln(0.5) 5730Divide by the coefficient of k.         A= A 0 e ( ln(0.5) 5730 )t Substitute for r in the continuous growth formula.

To find the age of an object, we solve this equation for <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>:</mo></mrow></annotation-xml></semantics></math>

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>=</mo><mfrac><mrow><mi>ln</mi><mrow><mo>(</mo><mrow><mfrac><mi>A</mi><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></mfrac></mrow><mo>)</mo></mrow></mrow><mrow><mo>−</mo><mn>0.000121</mn></mrow></mfrac></mrow></annotation-xml></semantics></math>

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>r</mi></annotation-xml></semantics></math> be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>≈</mo><msub><mi>A</mi><mn>0</mn></msub><msup><mi>e</mi><mrow><mo>−</mo><mn>0.000121</mn><mi>t</mi></mrow></msup></mrow></annotation-xml></semantics></math> we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>=</mo><mfrac><mi>A</mi><mrow><msub><mi>A</mi><mn>0</mn></msub></mrow></mfrac><mo>≈</mo><msup><mi>e</mi><mrow><mo>−</mo><mn>0.000121</mn><mi>t</mi></mrow></msup><mo>.</mo></mrow></annotation-xml></semantics></math> We solve this equation for <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>,</mo></mrow></annotation-xml></semantics></math> to get

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>=</mo><mfrac><mrow><mi>ln</mi><mrow><mo>(</mo><mi>r</mi><mo>)</mo></mrow></mrow><mrow><mo>−</mo><mn>0.000121</mn></mrow></mfrac></mrow></annotation-xml></semantics></math>

Given the percentage of carbon-14 in an object, determine its age.

  1. Express the given percentage of carbon-14 as an equivalent decimal,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>.</mo></mrow></annotation-xml></semantics></math>
  2. Substitute for k in the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> ln( r ) −0.000121  and solve for the age,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>.</mo></mrow></annotation-xml></semantics></math>
Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>20</mn><mi>%</mi><mo>=</mo><mn>0.20</mn><mtext>  </mtext></mrow></annotation-xml></semantics></math>for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>k</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>in the equation and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>t</mi><mo>:</mo></mrow></annotation-xml></semantics></math>

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>t</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> ln(r) −0.000121 Use the general form of the equation.    = ln(0.20) −0.000121 Substitute for r.    ≈13301Round to the nearest year.

The bone fragment is about 13,301 years old.

Analysis

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>13,301 years</mtext><mo>±</mo><mtext>1% or 13,301 years</mtext><mo>±</mo><mtext>133 years</mtext><mtext>.</mtext></mrow></annotation-xml></semantics></math>

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

less than 230 years, 229.3157 to be exact

Calculating Doubling Time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time.

Given the basic exponential growth equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kt , doubling time can be found by solving for when the original quantity has doubled, that is, by solving<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><msub/></mrow></annotation-xml></semantics></math> A 0 = A 0 e kt .

The formula is derived as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>2</mn><msub/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> A 0 = A 0 e kt        2= e kt Divide by  A 0 .   ln2=kt Take the natural logarithm.         t= ln2 k Divide by the coefficient of t.

Thus the doubling time is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> ln2 k
Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

The formula is derived as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> ln2 k The doubling time formula. 2= ln2 k Use a doubling time of two years. k= ln2 2 Multiply by k and divide by 2. A =A 0 e ln2 2 t Substitute k into the continuous growth formula.

The function is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e ln2 2 t .

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e ln2 3 t

Using Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the genericexponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> e kt + T s

This formula is derived as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>A</mi><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> b ct + T s T(t)=A e ln( b ct ) + T s Laws of logarithms. T(t)=A e ctlnb + T s Laws of logarithms. T(t)=A e kt + T sRename the constant c ln b, calling it k.
Newton’s Law of Cooling

The temperature of an object,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo>,</mo></mrow></annotation-xml></semantics></math> in surrounding air with temperature<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> T s  will behave according to the formula

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>A</mi><msup/></mrow></annotation-xml></semantics></math> e kt + T s
where
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is time
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the difference between the initial temperature of the object and the surroundings
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is a constant, the continuous rate of cooling of the object

 

Given a set of conditions, apply Newton’s Law of Cooling.

  1. Set<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> T s  equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature).
  2. Substitute the given values into the continuous growth formula<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>A</mi><msup/></mrow></annotation-xml></semantics></math> e k t + T s  to find the parameters<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>.</mo></mrow></annotation-xml></semantics></math>
  3. Substitute in the desired time to find the temperature or the desired temperature to find the time.
Using Newton’s Law of Cooling

A cheesecake is taken out of the oven with an ideal internal temperature of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>165°F,</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>and is placed into a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>35°F</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>refrigerator. After 10 minutes, the cheesecake has cooled to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>150°F</mtext><mtext>.</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>If we must wait until the cheesecake has cooled to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>70°F</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>before we eat it, how long will we have to wait?

Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>A</mi><msup/></mrow></annotation-xml></semantics></math> e kt +35

We know the initial temperature was 165, so<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mn>6</mn><mn>5</mn><mo>.</mo></mrow></annotation-xml></semantics></math>

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>165</mn><mo>=</mo><mi>A</mi><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> e k0 +35 Substitute (0,165).      A=130 Solve for A.

We were given another data point,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo stretchy="false">(</mo><mn>1</mn><mn>0</mn><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mn>5</mn><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics></math> which we can use to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>               </mtext><mn>150</mn><mo>=</mo><mn>130</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> e k10 +35 Substitute (10, 150).                115=130 e k10 Subtract 35.                 115 130 = e 10kDivide by 130.          ln( 115 130 )=10k Take the natural log of both sides.                     k= ln( 115 130 ) 10 =−0.0123Divide by the coefficient of k.

This gives us the equation for the cooling of the cheesecake:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mn>3</mn><mn>0</mn><msup/></mrow></annotation-xml></semantics></math> e –0.0123t +35.

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mn>70</mn><mo>=</mo><mn>130</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> e −0.0123t +35 Substitute in 70 for T(t).               35=130 e −0.0123t Subtract 35.            35 130 = e −0.0123tDivide by 130.   ln( 35 130 )=−0.0123t Take the natural log of both sides                    t= ln( 35 130 ) −0.0123 ≈106.68Divide by the coefficient of t.

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>70°F</mtext><mtext>.</mtext></mrow></annotation-xml></semantics></math>

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

6.026 hours

Using Logistic Growth Models

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext fontstyle="italic">a, b,</mtext></mrow></annotation-xml></semantics></math> and <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext fontstyle="italic">c,</mtext></mrow></annotation-xml></semantics></math> the logistic growth of a population over time<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is represented by the model

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> c 1+a e −bx

The graph in [link] shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

<figure class="small" id="CNX_Precalc_Figure_04_07_006">Graph of f(x)=c/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c/(1+a)). The point of maximum growth is (ln(a)/b, c/2).</figure>
Logistic Growth

The logistic growth model is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> c 1+a e −bx

where

 

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>c</mi></mfrac></mrow></annotation-xml></semantics></math> 1+a  is the initial value
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the carrying capacity, or limiting value
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>b</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is a constant determined by the rate of growth.

 

Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>=</mo><mn>0.6030.</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

We substitute the given data into the logistic growth model

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> c 1+a e −bx

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mn>1000.</mn></mrow></annotation-xml></semantics></math>  To find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>,</mo></mrow></annotation-xml></semantics></math>we use the formula that the number of cases at time<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> c 1+a =1, from which it follows that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mn>999.</mn><mtext/></mrow></annotation-xml></semantics></math>This model predicts that, after ten days, the number of people who have had the flu is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 1000 1+999 e −0.6030x ≈293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mo>=</mo><mn>1000.</mn></mrow></annotation-xml></semantics></math>

Analysis

Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph in [link] gives a good picture of how this model fits the data.

<figure class="medium" id="CNX_Precalc_Figure_04_07_007"> <figcaption>The graph of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 1000 1+999 e −0.6030x</figcaption> Graph of f(x)=1000/(1+999e^(-0.5030x)) with the y-axis labeled as “Cases” and the x-axis labeled as “Days”. There was 1 case on day 0, 20 on day 5, 294 on day 10, and 1000 on day 21.</figure>

Using the model in [link], estimate the number of cases of flu on day 15.

895 cases on day 15

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed in [link]? Find the model, and use a graph to check your choice.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> 1 2 3 4 5 6 7 8 9
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>y</mi></annotation-xml></semantics></math> 0 1.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394

First, plot the data on a graph as in [link]. For the purpose of graphing, round the data to two significant digits.

<figure class="small" id="CNX_Precalc_Figure_04_07_008">Graph of the previous table’s values.</figure>

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><mi>ln</mi><mo stretchy="false">(</mo><mi>b</mi><mi>x</mi><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Plugging in the first point,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 1,0 ), gives<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>0</mn><mo>=</mo><mi>a</mi><mi>ln</mi><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>We reject the case that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>(if it were, all outputs would be 0), so we know<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>ln</mi><mo stretchy="false">(</mo><mi>b</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>Thus<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x ). Next we can use the point<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 9,4.394 ) to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>:</mo></mrow></annotation-xml></semantics></math>

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><mi>ln</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 4.394=aln(9)            a= 4.394 ln(9)

Because<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 4.394 ln( 9 ) ≈2,an appropriate model for the data is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x ).

To check the accuracy of the model, we graph the function together with the given points as in [link].

<figure class="small" id="CNX_Precalc_Figure_04_07_009"> <figcaption>The graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><mi>ln</mi><mi>x</mi><mo>.</mo></mrow></annotation-xml></semantics></math></figcaption> Graph of previous table’s values showing that it fits the function y=2ln(x) with an asymptote at x=0.</figure>

We can conclude that the model is a good fit to the data.

Compare [link] to the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ) shown in [link].

<figure class="small" id="CNX_Precalc_Figure_04_07_009a"> <figcaption>The graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 )</figcaption> Graph of previous table’s values showing that it fits the function y=2ln(x) with an asymptote at x=0.</figure>

The graphs appear to be identical when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>></mo><mn>0.</mn></mrow></annotation-xml></semantics></math>  A quick check confirms this conclusion:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 )=2ln( x ) for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>></mo><mn>0.</mn></mrow></annotation-xml></semantics></math>

However, if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo><</mo><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics></math> the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ) includes a “extra” branch, as shown in [link]. This occurs because, while<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x ) cannot have negative values in the domain (as such values would force the argument to be negative), the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x 2 ) can have negative domain values.

<figure class="small" id="CNX_Precalc_Figure_04_07_010">Graph of y=ln(x^2).</figure>

Does a linear, exponential, or logarithmic model best fit the data in [link]? Find the model.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> 1 2 3 4 5 6 7 8 9
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>y</mi></annotation-xml></semantics></math> 3.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034

Exponential.<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2</mn><msup/></mrow></annotation-xml></semantics></math> e 0.5x .

Expressing an Exponential Model in Base e

While powers and logarithms of any base can be used in modeling, the two most common bases are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>10</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>In science and mathematics, the base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is often preferred. We can use laws of exponents and laws of logarithms to change any base to base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

Given a model with the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> b x , change it to the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kx .

  1. Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> b x  as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> e ln( b x ) .
  2. Use the power rule of logarithms to rewrite y as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> e xln( b ) =a e ln( b )x .
  3. Note that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>a</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0  and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>=</mo><mi>ln</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> b ) in the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kx .
Changing to base e

Change the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>2.5</mn><msup/></mrow></annotation-xml></semantics></math> (3.1) x  so that this same function is written in the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kx .

The formula is derived as follows

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>y</mi><mo>=</mo><mn>2.5</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> (3.1) x    =2.5 e ln( 3.1 x ) Insert exponential and its inverse.    =2.5 e xln3.1 Laws of logs.    =2.5 e ( ln3.1 ) xCommutative law of multiplication

Change the function<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mn>3</mn><msup/></mrow></annotation-xml></semantics></math> (0.5) x  to one having<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>as the base.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>y</mi><mo>=</mo><mn>3</mn><msup/></mrow></annotation-xml></semantics></math> e ( ln0.5 )x

Key Equations

Half-life formula If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kt , <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo><</mo><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics></math> the half-life is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics></math> ln(2) k .
Carbon-14 dating <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> ln( A A 0 ) −0.000121 .
 
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msub><mi>A</mi></msub></mrow></annotation-xml></semantics></math> 0  <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the amount of carbon-14 when the plant or animal died
 
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the amount of carbon-14 remaining today
 
is the age of the fossil in years
Doubling time formula If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kt , <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>k</mi><mo>></mo><mn>0</mn><mo>,</mo></mrow></annotation-xml></semantics></math> the doubling time is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> ln2 k
Newton’s Law of Cooling <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mi>A</mi><msup/></mrow></annotation-xml></semantics></math> e kt + T s , where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> T s  is the ambient temperature,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mi>T</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>−</mo><msub/></mrow></annotation-xml></semantics></math> T s , and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the continuous rate of cooling.

Key Concepts

  • The basic exponential function is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> b x . If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>></mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics></math>we have exponential growth; if<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>0</mn><mo><</mo><mi>b</mi><mo><</mo><mn>1</mn><mo>,</mo></mrow></annotation-xml></semantics></math> we have exponential decay.
  • We can also write this formula in terms of continuous growth as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kx , where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> A 0  is the starting value. If<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> A 0  is positive, then we have exponential growth when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>></mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>and exponential decay when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo><</mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>See [link].
  • In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See [link].
  • We can find the age,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>,</mo></mrow></annotation-xml></semantics></math> of an organic artifact by measuring the amount,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>,</mo></mrow></annotation-xml></semantics></math> of carbon-14 remaining in the artifact and using the formula<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> ln( k ) −0.000121  to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>t</mi><mo>.</mo><mtext>  </mtext></mrow></annotation-xml></semantics></math>See [link].
  • Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See [link].
  • We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See [link].
  • We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See [link].
  • We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See [link].
  • Any exponential function with the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mi>a</mi><msup/></mrow></annotation-xml></semantics></math> b x  can be rewritten as an equivalent exponential function with the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> A 0 e kx  where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>k</mi><mo>=</mo><mi>ln</mi><mi>b</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>See [link].

Section Exercises

Verbal

With what kind of exponential model would half-life be associated? What role does half-life play in these models?

Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.

With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?

Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied.

What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain.

An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 10 2  times, or 2 orders of magnitude greater, than the mass of Earth.

Numeric

The temperature of an object in degrees Fahrenheit after t minutes is represented by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>68</mn><msup/></mrow></annotation-xml></semantics></math> e −0.0174t +72. To the nearest degree, what is the temperature of the object after one and a half hours?

For the following exercises, use the logistic growth model<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 150 1+8 e −2x .

Find and interpret<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Round to the nearest tenth.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mn>0</mn><mo stretchy="false">)</mo><mo>≈</mo><mn>16.7</mn><mo>;</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>The amount initially present is about 16.7 units.

Find and interpret<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mn>4</mn><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics></math>Round to the nearest tenth.

Find the carrying capacity.

150

Graph the model.

Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
–2 0.694
–1 0.833
0 1
1 1.2
2 1.44
3 1.728
4 2.074
5 2.488

exponential;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><msup/></mrow></annotation-xml></semantics></math> 1.2 x

Rewrite<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1.68</mn><msup/></mrow></annotation-xml></semantics></math> ( 0.65 ) x  as an exponential equation with base<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>e</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>to five significant digits.

Technology

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

 

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
1 2
2 4.079
3 5.296
4 6.159
5 6.828
6 7.375
7 7.838
8 8.238
9 8.592
10 8.908

 

logarithmic

Graph of the question’s table.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
1 2.4
2 2.88
3 3.456
4 4.147
5 4.977
6 5.972
7 7.166
8 8.6
9 10.32
10 12.383
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
4 9.429
5 9.972
6 10.415
7 10.79
8 11.115
9 11.401
10 11.657
11 11.889
12 12.101
13 12.295

logarithmic

Graph of the question’s table.
<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mi>x</mi></annotation-xml></semantics></math> <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></annotation-xml></semantics></math>
1.25 5.75
2.25 8.75
3.56 12.68
4.2 14.6
5.65 18.95
6.75 22.25
7.25 23.75
8.6 27.8
9.25 29.75
10.5 33.5

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>t</mi><mtext>  </mtext></mrow></annotation-xml></semantics></math>years is modeled by the equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>P</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t )= 1000 1+9 e −0.6t .

Graph the function.

Graph of P(t)=1000/(1+9e^(-0.6t))

What is the initial population of fish?

To the nearest tenth, what is the doubling time for the fish population?

about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1.4</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>years

To the nearest whole number, what will the fish population be after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>years?

To the nearest tenth, how long will it take for the population to reach<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>900</mn><mo>?</mo></mrow></annotation-xml></semantics></math>

about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>7.3</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>years

What is the carrying capacity for the fish population? Justify your answer using the graph of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

Extensions

A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>4</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>half-lives;<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>8.18</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes

The formula for an increasing population is given by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><msub/></mrow></annotation-xml></semantics></math> P 0 e rt  where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics></math> P 0  is the initial population and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>r</mi><mo>></mo><mn>0.</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>Derive a general formula for the time t it takes for the population to increase by a factor of M.

Recall the formula for calculating the magnitude of an earthquake,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 3 log( S S 0 ). Show each step for solving this equation algebraically for the seismic moment<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>S</mi><mo>.</mo></mrow></annotation-xml></semantics></math>

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>          </mtext><mi>M</mi><mo>=</mo><mfrac/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> 2 3 log( S S 0 ) log( S S 0 )= 3 2 M          S S 0 = 10 3M 2            S= S 0 10 3M 2

What is the y-intercept of the logistic growth model<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> c 1+a e −rx ? Show the steps for calculation. What does this point tell us about the population?

Prove that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> b x = e xln( b )  for positive<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1.</mn></mrow></annotation-xml></semantics></math>

Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>y</mi><mo>=</mo><msup/></mrow></annotation-xml></semantics></math> b x  for some non-negative real number<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mo>≠</mo><mn>1.</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>Then,

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>ln</mi><mo stretchy="false">(</mo><mi>y</mi><mo stretchy="false">)</mo><mo>=</mo><mi>ln</mi><mo stretchy="false">(</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics></math> b x ) ln(y)=xln(b) e ln(y) = e xln(b)       y= e xln(b)

Real-World Applications

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.

To the nearest hour, what is the half-life of the drug?

Write an exponential model representing the amount of the drug remaining in the patient’s system after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mo>=</mo><mn>125</mn><msup/></mrow></annotation-xml></semantics></math> e ( −0.3567t ) ;A≈43 mg

Using the model found in the previous exercise, find<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> 10 ) and interpret the result. Round to the nearest hundredth.

For the following exercises, use this scenario: A tumor is injected with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>0.5</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>grams of Iodine-125, which has a decay rate of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1.15%</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>per day.

To the nearest day, how long will it take for half of the Iodine-125 to decay?

about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>60</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>days

Write an exponential model representing the amount of Iodine-125 remaining in the tumor after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram.

A scientist begins with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>250</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>grams of a radioactive substance. After<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>250</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes, the sample has decayed to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>32</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>250</mn><msup/></mrow></annotation-xml></semantics></math> e (−0.00914t) ; half-life: about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>76</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes

The half-life of Radium-226 is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1590</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.

The half-life of Erbium-165 is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>10</mtext><mtext>.4</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>r</mi><mo>≈</mo><mo>−</mo><mn>0.0667</mn><mo>,</mo></mrow></annotation-xml></semantics></math> So the hourly decay rate is about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>6.67</mn><mi>%</mi></mrow></annotation-xml></semantics></math>

A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>573</mtext><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>years.)

A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1350</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> 3 </mtext></mrow></annotation-xml></semantics></math>hours?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1350</mn><msup/></mrow></annotation-xml></semantics></math> e (0.03466t) ; after 3 hours:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mo stretchy="false">(</mo><mn>180</mn><mo stretchy="false">)</mo><mo>≈</mo><mn>691</mn><mo>,</mo><mn>200</mn></mrow></annotation-xml></semantics></math>

For the following exercises, use this scenario: A biologist recorded a count of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mn>360</mn><mtext>  </mtext></mrow></annotation-xml></semantics></math>bacteria present in a culture after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mn>5</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>1000</mn><mtext>  </mtext></mrow></annotation-xml></semantics></math>bacteria present after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext>  </mtext><mn>20</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes.

To the nearest whole number, what was the initial population in the culture?

Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>256</mn><msup/></mrow></annotation-xml></semantics></math> e (0.068110t) ; doubling time: about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>10</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes

For the following exercises, use this scenario: A pot of boiling soup with an internal temperature of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>100°</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>Fahrenheit was taken off the stove to cool in a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>69° F</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>room. After fifteen minutes, the internal temperature of the soup was<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>95° F</mtext><mtext>.</mtext></mrow></annotation-xml></semantics></math>

Use Newton’s Law of Cooling to write a formula that models this situation.

To the nearest minute, how long will it take the soup to cool to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>80° F?</mtext></mrow></annotation-xml></semantics></math>

about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>88</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes

To the nearest degree, what will the temperature be after<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>2</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>and a half hours?

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>165°F</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math> and is allowed to cool in a<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>75°F</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>room. After half an hour, the internal temperature of the turkey is<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>145°F</mtext><mtext>.</mtext></mrow></annotation-xml></semantics></math>

Write a formula that models this situation.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>T</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>90</mn><msup/></mrow></annotation-xml></semantics></math> e (−0.008377t) +75, where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is in minutes.

To the nearest degree, what will the temperature be after 50 minutes?

To the nearest minute, how long will it take the turkey to cool to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>110° F?</mtext></mrow></annotation-xml></semantics></math>

about<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>113</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

Number line to show log(x) is between -1 and 0.
Number line to show log(x) is between 1 and 2.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>log</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> x )=1.5; x≈31.623

Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 10 −10   W m 2 ,Vacuum:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 10 −4 W m 2 ,Jet:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics></math> 10 2   W m 2

Recall the formula for calculating the magnitude of an earthquake,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>M</mi><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> 2 3 log( S S 0 ). One earthquake has magnitude<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>3</mtext><mo>.</mo><mtext>9</mtext><mtext> </mtext></mrow></annotation-xml></semantics></math>on the MMS scale. If a second earthquake has<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mtext>75</mtext><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

MMS magnitude:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>5.82</mn></mrow></annotation-xml></semantics></math>

For the following exercises, use this scenario: The equation<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>N</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t )= 500 1+49 e −0.7t  models the number of people in a town who have heard a rumor after t days.

How many people started the rumor?

To the nearest whole number, how many people will have heard the rumor after 3 days?

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>N</mi><mo stretchy="false">(</mo><mn>3</mn><mo stretchy="false">)</mo><mo>≈</mo><mn>71</mn></mrow></annotation-xml></semantics></math>

As<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>t</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>increases without bound, what value does<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>N</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t ) approach? Interpret your answer.

For the following exercise, choose the correct answer choice.

A doctor and injects a patient with<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>13</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>milligrams of radioactive dye that decays exponentially. After<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>12</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>minutes, there are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>4.75</mn><mtext> </mtext></mrow></annotation-xml></semantics></math>milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?

  1. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t )=13 ( 0.0805 ) t
  2. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t )=13 e 0.9195t
  3. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo>=</mo><mn>13</mn><msup/></mrow></annotation-xml></semantics></math> e (−0.0839t)
  4. <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>f</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics></math> t )= 4.75 1+13 e −0.83925t

C

Glossary

carrying capacity
in a logistic model, the limiting value of the output
doubling time
the time it takes for a quantity to double
half-life
the length of time it takes for a substance to exponentially decay to half of its original quantity
logistic growth model
a function of the form<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac/></mrow></annotation-xml></semantics></math> c 1+a e −bx  where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics></math> c 1+a  is the initial value,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>c</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is the carrying capacity, or limiting value, and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>b</mi><mtext> </mtext></mrow></annotation-xml></semantics></math>is a constant determined by the rate of growth
Newton’s Law of Cooling
the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature
order of magnitude
the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal