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Intermediate Algebra: Functions and Graphs

Katherine Yoshiwara

Section 7.5 Exponential Models

Using what we have learned about exponential functions and logarithms, we can now explore some exponential models. In this secction we’ll see how to fit an exponential function to data and how to use doubling time and half-life.

Subsection 7.5.1 Fitting an Exponential Function through Two Points

To write a formula for an exponential function \(f(x)=ab^x\text{,}\) we need to know the initial value, \(a\), and the growth or decay factor, \(b\text{.}\) We can find these two parameters if we know any two function values.

Example 7.5.1.

Find an exponential function \(f(x)=ab^x\) that has the values \(f(2) = 4.5\) and \(f(5) = 121.5\text{.}\)

Solution.

We would like to find values of \(a\) and \(b\) so that the given function values satisfy \(f(x) = ab^x\) . By substituting the function values into the formula, we can write two equations.
\begin{align*} f(2) \amp= 4.5~~~ \text{ means }~~~x = 2, y = 4.5, \amp\amp\text{so}~~~~~ab^2 = 4.5\\ f(5) \amp= 121.5~~~ \text{ means }~~~x = 5, y = 121.5, \amp\amp\text{so}~~~~~ab^5 = 121.5 \end{align*}
We now have a system of equations in the two unknowns \(a\) and \(b\text{:}\)
\begin{gather*} ab^2 = 4.5\\ ab^5 = 121.5 \end{gather*}
but it is not a linear system. We can solve the system by the method of elimination, but instead of adding the equations, we will divide one of the equations by the other.
\begin{align*} \frac{ab^5}{ab^2}\amp= \frac{121.5}{4.5}\\ b^3 \amp = 27 \end{align*}
Note that by dividing the two equations, we eliminated \(a\text{,}\) and we can now solve for \(b\text{.}\)
\begin{align*} b^3 \amp = 27 \\ b \amp = \sqrt[3]{27} = 3 \end{align*}
Next we substitute \(b = 3\) into either of the two equations and solve for \(a\text{.}\)
\begin{align*} a(3)^2 \amp = 4.5 \\ a \amp= \frac{4.5}{9} = 0.5 \end{align*}
Thus, \(a = 0.5\) and \(b = 3\text{,}\) and the function is \(f(x) = 0.5(3^x)\text{.}\)

Caution 7.5.2.

Knowing only two points on the graph of \(f\) is not enough to tell us what kind of function \(f\) is. Through the two points in Example 7.5.1, we can also fit a linear function or a power function.
You can check that the three functions below all satisfy \(f(2) = 4.5\) and \(f(5) = 121.5\text{.}\) The graphs of the functions are shown at right.
\begin{align*} L(x) \amp = -73.5 + 39x \\ P(x) \amp = 0.372x^{3.6} \\ E(x) \amp = 0.5(3^x ) \end{align*}
three functions through the same two points
However, if we already know that we are looking for an exponential function, we can follow the steps below to find its formula. This method is sometimes called the ratio method. (Of course, if one of the known function values is the initial value, we can find \(b\) without resorting to the ratio method.)

To find an exponential function \(f(x)=ab^x\) through two points:.

  1. Use the coordinates of the points to write two equations in \(a\) and \(b\text{.}\)
  2. Divide one equation by the other to eliminate \(a\text{.}\)
  3. Solve for \(b\text{.}\)
  4. Substitute \(b\) into either equation and solve for \(a\text{.}\)

Checkpoint 7.5.3. Practice 1.

Use the ratio method to find an exponential function whose graph includes the points \((1, 20)\) and \((3, 125)\text{.}\)
Answer.
\(f (x) = 8(2.5)^x\)

Checkpoint 7.5.4. QuickCheck 1.

You have written a system of equations to fit an exponential function through two points. What is the next step?
  1. Calculate the slope.
  2. Subtract one equation from the other.
  3. Divide one equation by the other.
  4. Take the log of both sides.
We can use the ratio method to find an exponential growth or decay model if we know two function values.

Example 7.5.5.

The unit of currency in Ghana is the cedi, denoted by ¢. Beginning in 1986, the cedi underwent a period of exponential inflation. In 1993, one U.S. dollar was worth ¢720, and in 1996, the dollar was worth about ¢1620. Find a formula for the number of cedi to the dollar as a function of time since 1986. What was the annual inflation rate during the decade from 1986 to 1996? \(~\alert{\text{[TK]}}\)

Solution.

We want to find a function \(C(t) = ab^t\) for the number of cedi to the dollar, where \(t = 0\) in 1986. We have two function values, \(C(7) = 720\text{,}\) and \(C(10) = 1620\text{,}\) and with these values we can write two equations.
\begin{align*} ab^7 \amp = 720 \\ ab^{10} \amp = 1620 \end{align*}
We divide the second equation by the first to find
\begin{align*} \frac{ab^{10}}{ab^7} \amp= \frac{1620}{720} \\ b^3 \amp= 2.25 \end{align*}
Now we can solve this last equation for \(b\) to get \(b = \sqrt[3]{2.25}\approx 1.31\text{.}\) Finally, we substitute \(b = 1.31\) into the first equation to find \(a\text{.}\)
\begin{align*} a(1.31)^7 \amp = 720 \\ a \amp = \frac{720}{1.317}\\ \amp = 108.75 \end{align*}
Thus, \(C(t) = 108.75(1.31)^t\text{.}\) Finally, recall that \(b=1+r\) to see that the annual inflation rate was 31%.

Checkpoint 7.5.6. Practice 2.

The number of earthquakes that occur worldwide is a decreasing exponential function of their magnitude on the Richter scale. Between 2000 and 2005, there were 7480 earthquakes of magnitude 5 and 793 earthquakes of magnitude 6. (Source: National Earthquake Information Center, U.S. Geological Survey)
  1. Find a formula for the number of earthquakes that occurred between 2000 and 2005, \(N(m)\text{,}\) in terms of their magnitude, \(m\text{.}\) (Notice that in this problem \(N(m)\) is not a function of time; \(N\) is a function of Richter magnitude \(m\text{.}\)) Round your value for \(a\) to the nearest million, because greater accuracy is not realistic.
  2. It is difficult to keep an accurate count of small earthquakes. Use your formula to estimate the number of magnitude 1 earthquakes that occurred between 2000 and 2005.
  3. How many earthquakes of magnitude 8 occurred over those five years?
Answer.
  1. \(\displaystyle N(m) = 559,000,000(0.106)^m\)
  2. \(\displaystyle 59,254,000\)
  3. \(\displaystyle 9\)

Subsection 7.5.2 Doubling Time

Instead of giving the rate of growth of a population, we often specify its rate of growth by giving the time it takes for the population to double. Let us see how those two descriptions are connected.

Example 7.5.7.

In 2005, the population of Egypt was 74 million and was growing by 2% per year.
  1. If it continues to grow at the same rate, how long will it take the population of Egypt to double?
  2. How long will it take the population to double again?
  3. Illustrate the results on a graph.

Solution.

  1. The population of Egypt is growing according to the formula
    \begin{equation*} P(t) = 74(1.02)^t \end{equation*}
    where \(t\) is in years and \(P(t)\) is in millions. We would like to know when the population will reach 148 million (twice 74 million), so we solve the equation
    \begin{align*} 74(1.02)^t \amp= 148\amp\amp \blert{\text{Divide both sides by 74.}}\\ 1.02^t \amp= 2\amp\amp \blert{\text{Take the log of both sides.}}\\ t \log {(1.02)} \amp= \log {(2)}\amp\amp \blert{\text{Divide both sides by log (1.02).}}\\ t \amp= \frac{\log {(2)}}{\log {(1.02)}} \approx 35 \text{ years} \end{align*}
    It will take the population about 35 years to double.
  2. Twice 148 million is 296 million, so we solve the equation
    \begin{align*} 148(1.02)^t \amp= 296\amp\amp \blert{\text{Divide both sides by 148.}}\\ 1.02^t \amp= 2\amp\amp \blert{\text{Take the log of both sides.}}\\ t \log {(1.02)} \amp= \log {(2)}\amp\amp \blert{\text{Divide both sides by log (1.02).}}\\ t \amp= \frac{\log {(2})}{\log {(1.02)}} \approx 35 \text{ years} \end{align*}
    It will take the population about 35 years to double again.
  3. A graph of \(P(t) = 74(1.02)^t\) is shown below. Note that the population doubles every 35 years.
    graph showing doubling time
In the Example above, notice that the calculations in parts (a) and (b) are identical after the first step, and give the same result, 35 years. In fact, we can start at any time, and it will take the population 35 years to double. We say that 35 years is the doubling time for this population. Every increasing exponential function has a constant doubling time. And if a function has a constant doubling time, it must be exponential.

Checkpoint 7.5.8. Practice 3.

In 2005, the population of Uganda was 26.9 million people and was growing by 3.2% per year.
  1. Write a formula for the population of Uganda as a function of years since 2005.
  2. To the nearest year,how long will it take the population of Uganda to double?
  3. Use your formula from part (a) to verify the doubling time for three doubling periods.
Answer.
  1. \(P(t) = 26.9(1.032)^t\) million
  2. 22 years
  3. \(P(0) = 26.9\text{;}\) \(P(22)\approx 53.8\text{,}\) so \(P(22)\approx 2\cdot P(0)\text{;}\)
    \(P(44)\approx 107.6\text{,}\) so \(P(44)\approx 2\cdot P(22)\text{;}\)
    \(P(66)\approx 215.1\text{,}\) so \(P(66)\approx 2\cdot P(44)\)

Checkpoint 7.5.9. QuickCheck 2.

Which of these statements are true?
  1. The doubling time of a population depends on its initial value.
  2. An increasing exponential function has a constant doubling time.
  3. The doubling time is twice the percent growth rate.
  4. The doubling time is half the percent growth rate.
If we know the doubling time for a population, we can immediately write down its growth law. Because the population of Egypt doubles in 35 years, we can write
\begin{equation*} P(t) = 74 \cdot 2^{t/35} \end{equation*}
In this form, the growth factor for the population is \(2^{1/35}\text{,}\) and you can check that, to five decimal places, \(2^{1/35} = 1.02\text{.}\)

Doubling Time.

If \(D\) is the doubling time for an exponential function \(P(t)\text{,}\) then
\begin{equation*} \blert{P(t) = P_0~ 2^{t/D}} \end{equation*}
So, from knowing the doubling time, we can easily find the growth rate of a population.

Example 7.5.10.

At its current rate of growth, the population of the United States will double in 115.87 years.
  1. Write a formula for the population of the United States as a function of time.
  2. What is the annual percent growth rate of the population?

Solution.

  1. The current population of the United States is not given, so we represent it by \(P_0\text{.}\) With \(t\) expressed in years, the formula is then
    \begin{equation*} P(t) = P_0 ~2^{t/115.87} \end{equation*}
  2. We write \(2^{t/115.87}\) in the form \(\left(2^{1/115.87}\right)^t\) to see that the growth factor is \(b = 2^{1/115.87}\text{,}\) or 1.006. For exponential growth, \(b = 1 + r\text{,}\) so \(r = 0.006\text{,}\) or 0.6%.

Checkpoint 7.5.11. Practice 4.

At its current rate of growth, the population of Mexico will double in 36.8 years. What is its annual percent rate of growth?
Answer.
1.9%

Subsection 7.5.3 Half-Life

The half-life of a decreasing exponential function is the time it takes for the output to decrease to half its original value. For example, the half-life of a radioactive isotope is the time it takes for half of the substance to decay. The half-life of a drug is the time it takes for half of the drug to be eliminated from the body. Like the doubling time, the half-life is constant for a particular function; no matter where you start, it takes the same amount of time to reach half that value.

Example 7.5.12.

If you take \(200\) mg of ibuprofen to relieve sore muscles, the amount of the drug left in your body after \(t\) hours is \(Q(t) = 200(0.73)^t\text{.}\)
  1. What is the half-life of ibuprofen?
  2. When will 50 mg of ibuprofen remain in your body?
  3. Use the half-life to sketch a graph of \(Q(t)\text{.}\)

Solution.

  1. To find the half-life, we calculate the time elapsed when only half the original amount, or 100 mg, is left.
    \begin{align*} 200(0.73)^t \amp = 100\amp\amp \blert{\text{Divide both sides by 200.}}\\ 0.73^t \amp = 0.5\amp\amp \blert{\text{Take the log of both sides.}}\\ t\log {(0.73)} \amp = \log 0.5\amp\amp \blert{\text{Divide both sides by log (0.73).}}\\ t \amp= \frac{\log {(0.5)}}{\log {(0.73)}} = 2.2 \end{align*}
    The half-life is 2.2 hours.
  2. After 2.2 hours, 100 mg of ibuprofen is left in the body. After another 2.2 hours, half of that amount, or 50 mg, is left. Thus, 50 mg remain after 4.4 hours.
  3. We locate multiples of 2.2 hours on the horizontal axis. After each interval of 2.2 hours, the amount of ibuprofen is reduced to half its previous value. The graph is shown below.
    graph showing half-life
    \(t\) \(0\) \(2.2\) \(4.4\) \(6.6\) \(8.8\)
    \(Q(t)\) \(200\) \(100\) \(50\) \(25\) \(12.5\)

Checkpoint 7.5.13. Practice 5.

Alcohol is eliminated from the body at a rate of 15% per hour.
  1. Write a decay formula for the amount of alcohol remaining in the body, using \(A_0\) for the initial amount of alcohol.
  2. What is the half-life of alcohol in the body?
Answer.
  1. \(\displaystyle A(t) = A_0(0.85)^t\)
  2. 4.3 hours

Checkpoint 7.5.14. QuickCheck 3.

The half-life of DDT is 15 years. This means that:
  1. 30 pounds of DDT dissolve in one year.
  2. 100 pounds of DDT dissolve in 30 years.
  3. After 30 years, 100 pounds of DDT is reduced to 25 pounds.
  4. Each half-pound of DDT takes 15 years to dissolve.
Just as we can write an exponential growth law in terms of its doubling time, we can use the half-life to write a formula for exponential decay. For example, the half-life of ibuprofen is 2.2 hours, so every 2.2 hours the amount remaining is reduced by a factor of 0.5. After \(t\) hours a 200-mg dose will be reduced to
\begin{equation*} Q(t) = 200(0.5)^{t/2.2} \end{equation*}
Once again, you can check that this formula is equivalent to the decay function given in the previous Example.

Half-Life.

If \(H\) is the half-life for an exponential function \(Q(t)\text{,}\) then
\begin{equation*} \blert{Q(t)=Q_0 ~(0.5)^{t/H}} \end{equation*}
Radioactive isotopes are molecules that decay into more stable molecules, emitting radiation in the process. Although radiation in large doses is harmful to living things, radioactive isotopes are useful as tracers in medicine and industry, and as treatment against cancer. The decay laws for radioactive isotopes are often given in terms of their half-lives.

Example 7.5.15.

Cobalt-60 is used in cold pasteurization to sterilize certain types of food. Gamma rays emitted by the isotope during radioactive decay kill any bacteria present without damaging the food. The half-life of cobalt-60 is 5.27 years. \(~\alert{\text{[TK]}}\)
  1. Write a decay law for cobalt-60.
  2. What is the annual decay rate for cobalt-60?

Solution.

  1. We let \(Q(t)\) denote the amount of cobalt-60 left after \(t\) years, and let \(Q_0\) denote the initial amount. Every 5.27 years, \(Q(t)\) is reduced by a factor of 0.5, so
    \begin{equation*} Q(t) = Q_0 (0.5)^{t/5.27} \end{equation*}
  2. We rewrite the decay law in the form \(Q(t) = Q_0 (1 - r )^t\) as follows:
    \begin{equation*} Q(t) = Q_0 (0.5)^{t/5.27}=Q_0 \left((0.5)^{1/5.27}\right)^t = Q_0 (0.8768)^t \end{equation*}
    Thus, \(1 - r = 0.8768\text{,}\) so \(r = 0.1232\text{,}\) or 12.32%.

Checkpoint 7.5.16. Practice 6.

Cesium-137, with a half-life of 30 years, is one of the most dangerous by-products of nuclear fission. What is the annual decay rate for cesium-137?
Answer.
2.28%

Checkpoint 7.5.17. QuickCheck 4.

Decide whether each statement is true or false.
  1. Every increasing exponential function has a constant doubling time.
  2. If the doubling time of a population is 5 years, then its growth factor is given by \(2^{1/5}\text{.}\)
  3. The half-life of a substance is half the time it takes for all of the substance to decay.
  4. We can sketch the graph of an exponential decay function if we know its half-life and initial value.

Subsection 7.5.4 Compound Interest

In Section 7.1.5 we considered a formula for interest compounded annually. Many accounts compound interest more frequently than once a year. If the interest is compounded \(n\) times per year, then in \(t\) years there will be \(nt\) compounding periods, and in each period the account earns interest at a rate of \(\dfrac{r}{n}\text{.}\) The amount accumulated is given by a generalization of our earlier formula.

Compound Interest.

The amount \(A(t)\) accumulated (principal plus interest) in an account earning interest compounded \(n\) times annually is
\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt} \end{equation*}
where \(P\) is the principal invested, \(r\) is the interest rate, and \(t\) is the time period, in years.

Example 7.5.18.

Nigel deposited $1000 in an account that pays 4% interest. Calculate the amount in his account after 5 years if the interest is compounded
  1. quarterly
  2. monthly

Solution.

  1. “Quarterly” means 4 times a year, so we use the compound interest formula with \(~P=1000, ~r=0.04, ~n=4, ~\text{and}~ t=5\text{.}\)
    \begin{align*} A(5) \amp = 1000 \left(1 + \dfrac{0.04}{4}\right)^{4(5)}\\ \amp = 1000(1.01)^{20} = 1220.19 \end{align*}
    If interest is compounded quarterly, the balance in the account after 5 years is $1220.19.
  2. There are 12 months in a year, so we use the compound interest formula with \(~P=1000, ~r=0.04, ~n=12, ~\text{and}~ t=5\text{.}\)
    \begin{align*} A(5) \amp = 1000 \left(1 + \dfrac{0.04}{2}\right)^{12(5)}\\ \amp = 1000(1.0\overline{3})^{60} = 1221.00 \end{align*}
    If interest is compounded quarterly, the balance in the account after 5 years is $1221.
In Example 7.1.18 above, you can see that the larger the value of \(n\text{,}\) the greater the value of \(A\text{.}\) More frequent compounding periods result in a higher account balance.

Checkpoint 7.5.19. Practice 7.

Calculate the amount in Nigel’s account after 5 years if the interest is compounded weekly. (See Example 7.1.18. There are 52 weeks in a year.)
Answer.
$1221.31

Exercises 7.5.5 Problem Set 7.5

Warm Up

Exercise Group.
For Problems 1-4, solve. Round your answers to hundredths.
1.
\(4.6(x-3)^{1.8}+12=18\)
2.
\(2x^2-4.7x=3.8\)
3.
\(28(1.65)^{-0.3t} = 2.53\)
4.
\(5(10)^x=30\)
Exercise Group.
For Problems 5–8, find an equation for the line with the given properties.
5.
slope \(= \dfrac{-2}{3}\text{,}\) \(y\)-intercept is \((0,-1)\)
6.
slope \(=-2\text{,}\) passes through \((-1,2)\)
7.
passes through \((0,4)\) and \((2,3)\)
8.
passes through \((-3,0)\) and \((0,-5)\)

Skills Practice

Exercise Group.
For Problems 9 and 10, find an exponential function that has the given values.
9.
\(P(0)=8,~P(5)=0.25\)
10.
\(f(2)=9, ~f(3)=12\)
Exercise Group.
For Problems 11 and 12, find a formula for the exponential function.
11.
mystery exponential growth
12.
mystery exponential decay
Exercise Group.
For Problems 13 and 14,
  1. fit a linear function to the points,
  2. fit an exponential function to the points,
  3. graph both functions in the same window.
13.
\((0,2.6)\text{,}\) \((1,1.3)\)
14.
\((-2,0.75),~(4,6)\)

Applications

Exercise Group.
For Problems 15-18, write a growth or decay formula for the exponential function. Then find the percent growth or decay rate.
15.
A population starts with 2000 and has a doubling time of 5 years.
16.
You have 10 grams of a radioactive isotope whose half-life is 42 years.
17.
A certain medication has a half-life of 18 hours in the body. You are given an initial dose of \(D_0\) mg.
18.
The doubling time of a certain financial investment is 8 years. You invest an amount \(M_0\text{.}\)
19.
In 1798, the English political economist Thomas R. Malthus claimed that human populations, unchecked by environmental or social constraints, double every 25 years, regardless of the initial population size.
  1. Write a growth law for human populations under these conditions.
  2. What is the growth rate in unconstrained conditions?
20.
David Sifry observed in 2005 that over the previous two years, the number of Weblogs, or blogs, was doubling every 5 months. (Source: www.sifry.com/alerts/archives)
  1. Write a formula for the number of blogs \(t\) years after January 2005, assuming it continues to grow at the same rate.
  2. What is the growth rate for the number of blogs?
21.
Radioactive potassium-42, which is used by cardiologists as a tracer, decays at a rate of 5.4% per hour.
  1. Find the half-life of potassium-42.
  2. How long will it take for three-fourths of the sample to decay? For seven-eighths of the sample?
  3. Suppose you start with 400 milligrams of potassium-42. Using your answers to (a) and (b), make a rough sketch of the decay function.
22.
Caffeine leaves the body at a rate of 15.6% each hour. Your first cup of coffee in the morning has 100 mg of caffeine.
  1. How long will it take before you have 50 mg of that caffeine in your body?
  2. How long will it take before you have 25 mg of that caffeine in your body?
  3. Using your answers to (a) and (b), make a rough sketch of the decay function.
23.
Dichloro-diphenyl-trichloroethane (DDT) is a pesticide that was used in the middle decades of the twentieth century to control malaria. After 1945, it was also widely used on crops in the United States, and as much as one ton might be sprayed on a single cotton field. However, after the toxic effects of DDT on the environment began to appear, the chemical was banned in 1972.
  1. A common estimate for the half-life of DDT in the soil is 15 years. Write a decay law for DDT in the soil.
  2. In 1970, many soil samples in the United States contained about 0.5 mg of DDT per kg of soil. The NOAA (National Oceanic and Atmospheric Administration) safe level for DDT in the soil is 0.008 mg/kg. When will DDT content in the soil be reduced to a safe level?
24.
In 1986, the inflation rate in Bolivia was 8000% annually. The unit of currency in Bolivia is the boliviano.
  1. Write a formula for the price of an item as a function of time. Let \(P_0\) be its initial price.
  2. How long did it take for prices to double? Give both an exact value and a decimal approximation rounded to two decimal places.
  3. Suppose \(P_0 = 5\) bolivianos. Graph your function in the window \(\text{Xmin} = 0\text{,}\) \(\text{Xmax} = 0.94\text{,}\) \(\text{Ymin} = 0\text{,}\) \(\text{Ymax} = 100\text{.}\)
  4. Use intersect to verify that the price of the item doubles from 5 to 10 bolivianos, from 10 to 20, and from 20 to 40 in equal periods of time.
Exercise Group.
In Problems 25 and 26,
  1. Write a decay law for the isotope.
  2. Use the decay law to answer the question. (Round to the nearest ten years.)
25.
In living organisms, carbon-14 occurs with a fixed ratio to nonradioactive carbon-12. After a plant or animal dies, the carbon-14 decays into stable carbon with a halflife of 5730 years. When samples from the Shroud of Turin were analyzed in 1988, they were found to have 91.2% of their original carbon-14. How old were those samples in 1988?
26.
Rubidium-strontium radioactive dating is used in geologic studies to measure the age of minerals. Rubidium-87 decays into strontium-87 with a half-life of 48.8 billion years. Several meteors were found to have 93.7% of their original rubidium. How old are the meteors?
Exercise Group.
For Problems 27-30, use the following formula for compound interest. If \(P\) dollars is invested at an annual interest rate \(r\) (expressed as a decimal) compounded \(n\) times yearly, the amount \(A\) after \(t\) years is given by
\begin{equation*} A=P \left(1+\dfrac{r}{n}\right)^{nt} \end{equation*}
27.
What rate of interest is required so that $1000 will yield $1900 after 5 years if the interest rate is compounded monthly?
28.
What rate of interest is required so that $400 will yield $600 after 3 years if the interest rate is compounded quarterly?
29.
How long will it take a sum of money to triple if it is invested at 10% compounded daily?
30.
How long will it take a sum of money to increase by a factor of 5 if it is invested at 10% compounded quarterly?
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