7.4: Exponential Growth
The next growth we will examine is exponential growth. Linear growth occurs by adding the same amount in each unit of time. Exponential growth happens when an initial population increases by the same percentage or factor over equal time increments or generations. This is known as relative growth and is usually expressed as percentage.
Suppose that every year, only 10% of the fish in a lake have surviving offspring. If there were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in the lake last year, there would now be 1100 fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.
Suppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each year. We start with the initial population, 1000 fish, \(P=1000\). How do we calculate the population after 1 year? The new population, \(A\), will be the initial population plus an additional 10%. Symbolically:
\(A=P+0.10P\)
Notice this could be condensed to a shorter form by factoring:
\(A=P+0.10P=P(1+0.10)=P(1.10)\)
While 10% is the growth rate , 1.10 is the growth multiplier . Notice that 1.10 can be thought of as “the original 100% plus an additional 10%”
To calculate the fish population after 1 year,
\(A=P(1.10)=1000(1.10)=1100\)
We could then calculate the population in the next years by using the population from the previous year:
\(A=1100(1.10)=1210\)
\(A=1210(1.10)=1331\)
Notice that in the first year, the population grew by 100 fish, in the second year, the population grew by 110 fish, and in the third year the population grew by 121 fish.
While there is a constant percentage growth, the actual increase in number of fish is increasing each year.
Graphing these values we see that this growth doesn’t quite appear linear.
To get a better picture of how this percentage-based growth affects things, we need an explicit form, so we can quickly calculate values further out in the future.
Like we did for the linear model, we will start building from the recursive equation:
For year 1: \(A=1000(1.10)\)
For year 2: \(A=1000(1.10)(1.10))=1000(1.10)^{2}\)
For year 3: \(A=1000(1.10)(1.10)(1.10)=1000(1.10)^{3}\)
For year 4: \(A=1000(1.10)(1.10)(1.10)(1.10)=1000(1.10)^{4}\)
Observing a pattern, we can generalize the explicit form to be:
\(A=1000(1.10)^{n}\)
From this, we can quickly calculate the number of fish in 10, 20, or 30 years:
\(A=1000(1.10)^{10}=2594\)
\(A=1000(1.10)^{20}=6727\)
\(A=1000(1.10)^{30}=17449\)
Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly by 100 fish each year, the population would have only reached 4000 in 30 years compared to almost 18000 with this percent-based growth, called exponential growth.
In exponential growth, the population grows proportional to the size of the population, so as the population gets larger, the same percent growth will yield a larger numeric growth.
If a quantity starts at size \(P\) and grows by \(R\%\) (written as a decimal, \(r\)) every time period, then the amount \(A\) after \(n\) years can be determined using the following formula:
\(A=P(1+r)^{n}\)
We call \(r\) the growth rate .
The term \((1+r)\) is called the growth multiplier , or common ratio.
Between 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand people. If this growth rate was to continue, what would the population of Olympia be in 2014?
Solution
As we did before, we first need to define what year will correspond to \(n = 0\). Since we know the population in 2008, it would make sense to have 2008 correspond to \(n = 0\), then year 2014 would then be \(n = 6\). We have \(P = 245,000\).
We know the growth rate is 3%, giving \(r = 0.03\).
Using the explicit form:
\(A=245,000(1+0.03)^{6}=245,000(1.19405)=292,542.25\)
The model predicts that in 2014, Olympia would have a population of about 293 thousand people.
To evaluate expressions like \((1.03)^6\), it will be easier to use a calculator than multiply 1.03 by itself six times. Most scientific calculators have a button for exponents. It is typically either labeled like:
[^], \([y^x]\), or \([x^y]\).
To evaluate \(1.03^6\) we’d type 1.03 [^] 6, or 1.03 \([y^x]\) 6. Try it out - you should get an answer around 1.1940523.
India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, what will India’s population grow to by 2020?
- Answer
-
Using \(n = 0\) corresponding with 2008,
\(A=1.14(1+0.0134)^{12}=\) about 1.337 billion people in 2020
A friend is using the equation \(A=4600(1.072)^{n}\) to predict the annual tuition at a local college. She says the formula is based on the number of years after 2010. What does this equation tell us?
Solution
In the equation, \(P=4600\), which is the starting value of the tuition when \(n = 0\). This tells us that the tuition in 2010 was $4,600.
The growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the tuition is expected to grow by 7.2% each year.
Putting this together, we could say that the tuition in 2010 was $4,600, and is expected to grow by 7.2% each year.
In the previous example, we had to calculate the 10 th root of a number. This is different than taking the basic square root, \(\sqrt{ }\). Many scientific calculators have a button for general roots. It is typically labeled like:
[\(\sqrt[n]{ }\)], [\(\sqrt[x]{ }\)], or [\(\sqrt[y]{x}\)]
To evaluate the 3rd root of 8, for example, we’d either type 3 [\(\sqrt[x]{ }\)] 8, or 8 [\(\sqrt[x]{ }\)] 3, depending on the calculator. Try it on yours to see which to use – you should get an answer of 2.
If your calculator does not have a general root button, all is not lost. You can instead use the property of exponents which states that \(\sqrt[n]{a}=a^{1 / n}\). So, to compute the 3 rd root of 8, you could use your calculator’s exponent key to evaluate \((2)^{2}-1=\). To do this, type:
8 \([y^x]\) ( 1 \([\div]\) 3 )
The parentheses tell the calculator to divide 1/3 before doing the exponent.
In 1990, the residential energy use in the US was responsible for 962 million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric tons[1]. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?
Solution
Similar to before, we will correspond \(n = 0\) with 1990, as that is the year for the first piece of data we have. That will make \(P=962\) (million metric tons of \(\mathrm{CO}_{2}\)). In this problem, we are not given the growth rate, but instead are given that \(A=1182\), the amount in 2000 (after 10 years).
When \(n = 10\), the explicit equation looks like:
\(A=P(1+r)^{10}\)
We know the value for \(P\), so we can put that into the equation:
\(A= 962(1+r)^{10}\)
We also know that \(A=1182\), so substituting that in, we get
\(1182= 962(1+r)^{10}\)
We can now solve this equation for the growth rate, \(r\). Start by dividing by 962.
\(\begin{array}{ll} \frac{1182}{962}=(1+r)^{10} & \text{Take the 10th root of both sides} \\ \sqrt[10]{\frac{1182}{962}}=1+r & \text{Subtract 1 from both sides} \\ r=\sqrt[10]{\frac{1182}{962}}-1=0.0208=2.08 \% \end{array}\)
So if the emissions are growing exponentially, they are growing by about 2.08% per year. We can now predict the emissions in 2050, 60 years after 1990, so \(n = 60\),
\(A=962(1+0.0208)^{60}=3308.4\) million metric tons of \(\mathrm{CO}_{2}\) in 2050.
As a note on rounding, notice that if we had rounded the growth rate to 2.1%, our calculation for the emissions in 2050 would have been 3347. Rounding to 2% would have changed our result to 3156. A very small difference in the growth rates gets magnified greatly in exponential growth. For this reason, it is recommended to round the growth rate as little as possible.
If you need to round, keep at least three significant digits - numbers after any leading zeros. So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027 could be reasonably rounded to 0.00103.
Looking back at example 3, for the sake of comparison, what would the carbon emissions be in 2050 if emissions grow linearly at the same rate?
Solution
Again we let \(n = 0\) correspond to the year 1990 when the emissions were 962 million metric tons. After 10 years, the emissions increased to 1182 million metric tons. Since we are looking at linear growth, find the slope, then write the equation of the line. This is a good review of section 7.2.
Calculate the slope, \(m\), using the two points (0,962) and (10,1182). We get \(m=22\). Thus, the linear model is:
\(y=22x+962\)
This tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050 (60 years after 1990),
\(y=22(60)+962=2282\) million metric tons.
You will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.
So how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:
1) Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?
2) Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.
The number of users on a social networking site was 45 thousand in February when they officially went public, and grew to 60 thousand by October. If the site is growing exponentially, and growth continues at the same rate, how many users should they expect two years after they went public?
- Answer
-
Here we will measure n in months rather than years, with \(n=0\) corresponding to February when the site went public. This gives \(P=45\) thousand. October is 8 months later, so \(A=60\) when \(n=8\).
\(A=P(1+r)^{n}\)
\(60=45(1+r)^{8}\)
\(\frac{60}{45}=(1+r)^{8}\)
\(\sqrt[8]{\frac{60}{45}}=1+r\)
\(r=\sqrt[8]{\frac{60}{45}}-1=0.0366\), or 3.66%
The general explicit equation is \(A=45(1.0366)^{n}\).
Predicting 24 months (2 years) after they went public:
\(A=45(1.0366)^{24}=106.63\) thousand users.
Doubling Time Model
A water tank up on the San Francisco Peaks is contaminated with a colony of 80,000 E. coli bacteria. The population doubles every five days. We want to find a model for the population of bacteria present after \(t\) days. The amount of time it takes the population to double is five days, so this is our time unit. After \(t\) days have passed, then \(t/5\) is the number of time units that have passed. Starting with the initial amount of 80,000 bacteria, our doubling model becomes:
\[A = 80,000(2)^{\frac{t}{5}} \nonumber\]
Using this model, how large is the colony in two weeks’ time? We have to be careful that the units on the times are the same; 2 weeks = 14 days.
Solution
\[A = 80,000(2)^{\frac{14}{5}} = 557,152 \nonumber\]
The colony is now 557,152 bacteria.
If \(D\) is the doubling time of a quantity (the amount of time it takes the quantity to double) and \(P\) is the initial amount of the quantity then the amount of the quantity present after \(t\) units of time is
\[A = P(2)^{\frac{t}{D}}\]
The doubling time of a population of flies is eight days. If there are initially 100 flies, how many flies will there be in 17 days?
Solution
To solve this problem, use the doubling time model with \(D=8\) and \(P = 100\). So the doubling time model for this problem is:
\[A = 100(2)^{t/8} \nonumber\]
When \(t = 17\, days\),
\[A = 100(2)^{\frac{17}{8}} = 436 \nonumber\]
There are 436 flies after 17 days.
Note: The population of flies follows an exponential growth model.
Sometimes we want to solve for the length of time it takes for a certain population to grow given their doubling time. To solve for the exponent, we use the log button on the calculator.
Suppose that a bacteria population doubles every six hours. If the initial population is 4000 individuals, how many hours would it take the population to increase to 25,000?
Solution
We know that \(P = 4000\), \(D = 6\), and \(A = 25000\), so the doubling time model for this problem is:
\[25000 = 4000(2)^{\frac{t}{6}}\nonumber\]
We need to solve for \(t\)
\[\dfrac{25,000}{4000} = \dfrac{\cancel{4000}(2)^{\frac{t}{6}}}{\cancel{4000}}\nonumber\]
\[6.25 = (2)^{\frac{t}{6}}\nonumber\]
Now, take the log of both sides of the equation
\[\text{log}(6.25) = \text{log}(2)^{\frac{t}{6}}\nonumber\]
Bring the exponent out front using rules of logarithms
\[\text{log}(6.25) = (\dfrac{t}{6}) \text{log}(2) \nonumber\]
Divide both sides of the equation by \(\text{log}(2)\)
\[\dfrac{\text{log}(6.25)}{\text{log}(2)} = \dfrac{t}{6}\nonumber\]
Simplify using your calculator
\[\begin{align*} 2.6439 &= \dfrac{t}{6} \\ t &= 15.8631 \end{align*}\]
The population would increase to 25,000 bacteria in approximately 15.9 hours.
A bird population on a certain island has an annual growth rate of 2.5% per year. The population doubling time is given as 28 years. Approximate the number of years it will take the population to double. If the initial population is 20 birds, use it to find the bird population of the island in 17 years.
Solution
With the bird population doubling in 28 years, we use the doubling time model to find the population is 17 years.
\[A = 20(2)^{\frac{t}{28}} \nonumber\]
When \(t = 17\) years
\[A = 20(2)^{\frac{17}{28}} = 30.46 \nonumber\]
There will be 30 birds on the island in 17 years.
Exponential Decay and Half-Life Model
The half-life of a material is the time it takes for a quantity of material to be cut in half. This term is commonly used when describing radioactive metals like uranium or plutonium. For example, the half-life of carbon-14 is 5730 years.
If a substance has a half-life, this means that half of the substance will be gone in a unit of time. In other words, the amount decreases by 50% per unit of time.
If \(H\) is the half-life of a quantity (the amount of time it takes the quantity be cut in half) and \(P\) is the initial amount of the quantity then the amount of the quantity present after \(t\) units of time is
\[A = P(\dfrac{1}{2})^{\frac{t}{H}}\]
Let’s say a substance has a half-life of eight days. If there are 40 grams present now, how much is left after three days?
Solution
We want to find a model for the quantity of the substance that remains after t days. The amount of time it takes the quantity to be reduced by half is eight days, so this is our time unit. After t days have passed, then t8 is the number of time units that have passed. Starting with the initial amount of 40, our half-life model becomes:
\[A = 40(\dfrac{1}{2})^{\frac{t}{8}} \nonumber\]
With \(t=3\)
\[A = 40(\dfrac{1}{2})^{\frac{3}{8}} = 30.8 \nonumber\]
There are 30.8 grams of the substance remaining after three days.
Lead-209 is a radioactive isotope. It has a half-life of 3.3 hours. Suppose that 40 milligrams of this isotope is created in an experiment, how much is left after 14 hours?
Solution
Use the half-life model to solve this problem.
\(P = 40\) and \(H = 3.3\), so the half-life model for this problem is:
\[A = 40(\dfrac{1}{2})^{\frac{t}{3.3}} \nonumber\]
With \(t=14\) hours,
\[A = 40(\dfrac{1}{2})^{\frac{14}{3.3}} = 2.1 \nonumber\]
There are 2.1 milligrams of Lead-209 remaining after 14 hours.
Note: The milligrams of Lead-209 remaining follows a decreasing exponential growth model.
Radioactive carbon-14 is used to determine the age of artifacts because it concentrates in organisms only when they are alive. It has a half-life of 5730 years. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls.
Solution
In this problem, we want to estimate the age of the scrolls, so we need to find \(t\). In 1947, 76% of the carbon-14 remained. This means that the amount, \(A\), remaining at time \(t\), divided by the original amount of carbon-14, \(P\), is equal to 76%.
Thus, \(\dfrac{A}{P} = 0.76\). Use this fact to solve for \(t\).
\[\begin{align*} A &= P(\dfrac{1}{2})^{\frac{t}{5730}} \\ \dfrac{A}{P} &= (\dfrac{1}{2})^{\frac{t}{5730}} \\ 0.76 &= (\dfrac{1}{2})^{\frac{t}{5730}} \end{align*}\]
Now, take the log of both sides of the equation. And let's rewrite the fraction as a decimal.
\[\text{log}(0.76) = \text{log}(0.5)^{\frac{t}{5730}} \nonumber\]
Bring the exponent out front using rules of logarithms
\[\text{log}(0.76) = (\dfrac{t}{5730}) \text{log}(0.5) \nonumber\]
Divide both sides of the equation by \(\text{log}(0.5)\)
\[\dfrac{\text{log}(0.76)}{\text{log}(0.5)} = \dfrac{t}{5730}\nonumber\]
Simplify using your calculator
\[\begin{align*} 0.3959 &= \dfrac{t}{5730} \\ t &= 2268.6713 \end{align*}\]
The Dead Sea Scrolls are over 2268 years old.
The population of wild elephants is decreasing by 7% per year which gives the half -life for this population to be approximately 10 years. If there are currently 7000 elephants left in the wild, and the population continues to decrease at this rate, how many elephants will remain in the wild in 25 years?
- Answer
-
\(P = 7000\) and \(H = 10\) years, so the half-life model for this problem is:
\[A = 7000(\dfrac{1}{2})^{\frac{t}{10}} \nonumber\]
When \(t=25\),
\[A = 7000(\dfrac{1}{2})^{\frac{25}{10}} = 1237.4 \nonumber\]
There will be approximately 1237 wild elephants left in 25 years.
Figure \(\PageIndex{3}\): Elephant Population over a 70 Year Span.
Note: The population of elephants follows a decreasing exponential growth model.
|
Rules of Exponents |
Rules of Logarithm for the Common Logarithm (Base 10) |
|---|---|
|
Definition of an Exponent \(a^{n} = a\cdot a\cdot a\cdot a\cdot ..... \cdot a\) (n a's multiplied together) |
Definition of a Logarithm \(10^{y} = x \text{ if and only if } \text{log}x = y\) |
|
Zero Rule \(a^{0} = 1\) |
|
|
Product Rule \(a^{m} \cdot a^{n}= a^{m+n}\) |
Product Rule \(\text{log}(xy) =\text{log }(x) + \text{log }(y)\) |
|
Quotient Rule \(\dfrac{a^{m}}{a^{n}}= a^{m-n}\) |
Quotient Rule \(\text{log}(\dfrac{x}{y}) =\text{log }(x) - \text{log }(y)\) |
|
Power Rule \((a^{n})^{m}= a^{n\cdot m}\) |
Power Rule \(\text{log }x^{r} =r \text{log }(x) (x > 0)\) |
|
Distributive Rules \((ab)^{n}= a^{n} \cdot a^{n}, (\dfrac{a}{b})^{n} = \dfrac{a^{n}}{b^{n}} \) |
\(\text{log}10^{x} =x \text{log}(10) = x\) |
|
Negative Exponent Rules \( a^{-n} = \dfrac{1}{a^{n}}, (\dfrac{a}{b})^{-n} = (\dfrac{b}{a})^{n} \) |
\(10^{\text{log}x} = x (x > 0)\) |
[1] www.eia.doe.gov/oiaf/1605/ggrpt/carbon.html