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Mathematics LibreTexts

4.6: Exponential and Logarithmic Models

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    While we have explored some basic applications of exponential and logarithmic functions, in this section we explore some important applications in more depth.

    Radioactive Decay

    In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes change over time. One of the common terms associated with radioactive decay is half-life.

    Definition: Half Life

    The half-life of a radioactive isotope is the time it takes for half the substance to decay.

    Given the basic exponential growth/decay equation \(h(t)=ab^{t}\), half-life can be found by solving for when half the original amount remains; by solving

    \[\frac{1}{2} a=a(b)^{t}\]

    or more simply \(\frac{1}{2} =b^{t}\). Notice how the initial amount is irrelevant when solving for half-life.

    Example \(\PageIndex{1}\)

    Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?

    Solution

    We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant a. Then the decay of Bismuth-210 can be described by the equation \(Q(d)=a\eqref{GrindEQ__0_87_}^{d}\).

    To find the half-life, we want to determine when the remaining quantity is half the original: \(\frac{1}{2} a\). Solving,

    \(\frac{1}{2} a=a\eqref{GrindEQ__0_87_}^{d}\) Divide by a,

    \(\frac{1}{2} =0.87^{d}\) Take the log of both sides

    \(\log \left(\frac{1}{2} \right)=\log \left(0.87^{d} \right)\) Use the exponent property of logs

    \(\log \left(\frac{1}{2} \right)=d\log \left(0.87\right)\) Divide to solve for d

    \(d=\frac{\log \left(\frac{1}{2} \right)}{\log \left(0.87\right)} \approx 4.977\) days

    This tells us that the half-life of Bismuth-210 is approximately 5 days.

    Example \(\PageIndex{2}\):

    Cesium-137 has a half-life of about 30 years. If you begin with 200mg of cesium-137, how much will remain after 30 years? 60 years? 90 years?

    Solution

    Since the half-life is 30 years, after 30 years, half the original amount, 100mg, will remain.

    After 60 years, another 30 years have passed, so during that second 30 years, another half of the substance will decay, leaving 50mg.

    After 90 years, another 30 years have passed, so another half of the substance will decay, leaving 25mg.

    Example \(\PageIndex{3}\):

    Cesium-137 has a half-life of about 30 years. Find the annual decay rate.

    Solution

    Since we are looking for an annual decay rate, we will use an equation of the form \(Q(t)=a(1+r)^{t}\). We know that after 30 years, half the original amount will remain. Using this information

    \(\frac{1}{2} a=a(1+r)^{30}\) Dividing by a

    \(\frac{1}{2} =(1+r)^{30}\) Taking the 30\({}^{th}\) root of both sides

    \(\sqrt[{30}]{\frac{1}{2} } =1+r\) Subtracting one from both sides, \[r=\sqrt[{30}]{\frac{1}{2} } -1\approx -0.02284\]

    This tells us cesium-137 is decaying at an annual rate of 2.284% per year.

    Exercise \(\PageIndex{1}\)

    Chlorine-36 is eliminated from the body with a biological half-life of 10 days3. Find the daily decay rate.

    Answer

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    Example \(\PageIndex{4}\)

    Carbon-14 is a radioactive isotope that is present in organic materials, and is commonly used for dating historical artifacts. Carbon-14 has a half-life of 5730 years. If a bone fragment is found that contains 20% of its original carbon-14, how old is the bone?

    Solution

    To find how old the bone is, we first will need to find an equation for the decay of the carbon-14. We could either use a continuous or annual decay formula, but opt to use the continuous decay formula since it is more common in scientific texts. The half life tells us that after 5730 years, half the original substance remains. Solving for the rate,

    \(\frac{1}{2} a=ae^{r5730}\) Dividing by a

    \(\frac{1}{2} =e^{r5730}\) Taking the natural log of both sides

    \(\ln \left(\frac{1}{2} \right)=\ln \left(e^{r5730} \right)\) Use the inverse property of logs on the right side

    \(\ln \left(\frac{1}{2} \right)=5730r\) Divide by 5730 \[r=\frac{\ln \left(\frac{1}{2} \right)}{5730} \approx -0.000121\]

    Now we know the decay will follow the equation \(Q(t)=ae^{-0.000121t}\). To find how old the bone fragment is that contains 20% of the original amount, we solve for t so that Q(t) = 0.20a.

    \[0.20a=ae^{-0.000121t}\] \[0.20=e^{-0.000121t}\] \[\ln (0.20)=\ln \left(e^{-0.000121t} \right)\] \[\ln (0.20)=-0.000121t\] \(t=\frac{\ln \eqref{GrindEQ__0_20_}}{-0.000121} \approx 13301\) years

    The bone fragment is about 13,300 years old.

    Exercise \(\PageIndex{1}\)

    In Example 2, we learned that Cesium-137 has a half-life of about 30 years. If you begin with 200mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

    Answer

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    Doubling Time

    For decaying quantities, we asked how long it takes for half the substance to decay. For growing quantities we might ask how long it takes for the quantity to double.

    Definition: Doubling Time

    The doubling time of a growing quantity is the time it takes for the quantity to double.

    Given the basic exponential growth equation \(h(t)=ab^{t}\), doubling time can be found by solving for when the original quantity has doubled; by solving \(2a=a(b)^{x}\), or more simply \(2=b^{x}\). Like with decay, the initial amount is irrelevant when solving for doubling time.

    Example \(\PageIndex{5}\):

    Cancer cells sometimes increase exponentially. If a cancerous growth contained 300 cells last month and 360 cells this month, how long will it take for the number of cancer cells to double?

    Defining t to be time in months, with t = 0 corresponding to this month, we are given two pieces of data: this month, (0, 360), and last month, (-1, 300).

    From this data, we can find an equation for the growth. Using the form \(C(t)=ab^{t}\), we know immediately a = 360, giving \(C(t)=360b^{t}\). Substituting in (-1, 300), \[\begin{array}{l} {300=360b^{-1} } \\ {300=\frac{360}{b} } \\ {b=\frac{360}{300} =1.2} \end{array}\]

    This gives us the equation \(C(t)=360\eqref{GrindEQ__1_2_}^{t}\)

    To find the doubling time, we look for the time when we will have twice the original amount, so when C(t) = 2a. \[2a=a(1.2)^{t}\] \[2=(1.2)^{t}\] \[\log \left(2\right)=\log \left(1.2^{t} \right)\] \[\log \left(2\right)=t\log \left(1.2\right)\] \(t=\frac{\log \left(2\right)}{\log \left(1.2\right)} \approx 3.802\) months for the number of cancer cells to double.

    Example \(\PageIndex{6}\)

    Use of a new social networking website has been growing exponentially, with the number of new members doubling every 5 months. If the site currently has 120,000 users and this trend continues, how many users will the site have in 1 year?

    Solution

    We can use the doubling time to find a function that models the number of site users, and then use that equation to answer the question. While we could use an arbitrary a as we have before for the initial amount, in this case, we know the initial amount was 120,000.

    If we use a continuous growth equation, it would look like\(N(t)=120e^{rt}\), measured in thousands of users after t months. Based on the doubling time, there would be 240 thousand users after 5 months. This allows us to solve for the continuous growth rate: \[240=120e^{r5}\] \[2=e^{r5}\] \[\ln 2=5r\] \[r=\frac{\ln 2}{5} \approx 0.1386\]

    Now that we have an equation, \(N(t)=120e^{0.1386t}\), we can predict the number of users after 12 months:

    \(N\eqref{GrindEQ__12_}=120e^{0.1386\eqref{GrindEQ__12_}} =633.140\) thousand users.

    So after 1 year, we would expect the site to have around 633,140 users.

    Exercise \(\PageIndex{3}\)

    If tuition at a college is increasing by 6.6% each year, how many years will it take for tuition to double?

    Answer

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    Newton’s Law of Cooling

    When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off towards the surrounding air temperature. This "leveling off" will correspond to a horizontal asymptote in the graph of the temperature function. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function.

    Definition: Newton’s Law of Cooling

    Newton’s Law of CoolingThe temperature of an object, T, in surrounding air with temperature T\({}_{s}\) will behave according to the formula

    \[T(t)=ae^{kt} +T_{s}\]

    Where

    • \(t\) is time
    • \(a\) is a constant determined by the initial temperature of the object
    • \(k\) is a constant, the continuous rate of cooling of the object

    While an equation of the form \(T(t)=ab^{t} +T_{s}\) could be used, the continuous growth form is more common.

    Example \(\PageIndex{7}\):

    A cheesecake is taken out of the oven with an ideal internal temperature of 165 degrees Fahrenheit, and is placed into a 35 degree refrigerator. After 10 minutes, the cheesecake has cooled to 150 degrees. If you must wait until the cheesecake has cooled to 70 degrees before you eat it, how long will you have to wait?

    Solution

    Since the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially towards 35, following the equation \[T(t)=ae^{kt} +35\]

    We know the initial temperature was 165, so \(T(0)=165\). Substituting in these values, \[\begin{array}{l} {165=ae^{k0} +35} \\ {165=a+35} \\ {a=130} \end{array}\]

    We were given another pair of data, \(T\eqref{GrindEQ__10_}=150\), which we can use to solve for k \[150=130e^{k10} +35\] \[\begin{array}{l} {115=130e^{k10} } \\ {\frac{115}{130} =e^{10k} } \\ {\ln \left(\frac{115}{130} \right)=10k} \\ {k=\frac{\ln \left(\frac{115}{130} \right)}{10} =-0.0123} \end{array}\]

    Together this gives us the equation for cooling: \(T(t)=130e^{-0.0123t} +35\).

    Now we can solve for the time it will take for the temperature to cool to 70 degrees. \[70=130e^{-0.0123t} +35\] \[35=130e^{-0.0123t}\] \[\frac{35}{130} =e^{-0.0123t}\] \[\ln \left(\frac{35}{130} \right)=-0.0123t\] \[t=\frac{\ln \left(\frac{35}{130} \right)}{-0.0123} \approx 106.68\]

    It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool. Of course, if you like your cheesecake served chilled, you’d have to wait a bit longer.

    Exercise \(\PageIndex{4}\)

    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?

    Answer

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    Logarithmic Scales

    Logarithmic Functions:Logarithmic Scales

    For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly.

    Planet Distance (millions of km)
    Mercury 58
    Venus 108
    Earth 150
    Mars 228
    Jupiter 779
    Saturn 1430
    Uranus 2880
    Neptune 4500

    Placed on a linear scale – one with equally spaced values – these values get bunched up.

    However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent.4

    Planet Distance (millions of km) log(distance)
    Mercury 58 1.76
    Venus 108 2.03
    Earth 150 2.18
    Mars 228 2.36
    Jupiter 779 2.89
    Saturn 1430 3.16
    Uranus 2880 3.46
    Neptune 4500 3.65

    \[321\]

    Section 4.6 Exponential and Logarithmic Models

    Sometimes, as shown above, the scale on a logarithmic number line will show the log values, but more commonly the original values are listed as powers of 10, as shown below. \[321\]

    Example \(\PageIndex{8}\):

    Estimate the value of point P on the log scale above

    The point P appears to be half way between -2 and -1 in log value, so if V is the value of this point,

    \(\log (V)\approx -1.5\) Rewriting in exponential form, \[V\approx 10^{-1.5} =0.0316\]

    Example \(\PageIndex{9}\):

    Place the number 6000 on a logarithmic scale.

    Since \(\log (6000)\approx 3.8\), this point would belong on the log scale about here:

    Exercise \(\PageIndex{5}\)

    Plot the data in the table below on a logarithmic scale5

    Answer

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