Section4.5Exponential Models

SubsectionFitting an Exponential Function through Two Points

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

Example4.62

Find an exponential function 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{equation*} \begin{aligned}[t] 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{aligned} \end{equation*}

This is a system of equations in the two unknowns, $a$ and $b\text{,}$ but it is not a linear system. We can solve the system by the method of elimination, but we will divide one of the equations by the other.

\begin{equation*} \begin{aligned}[t] \frac{ab^5}{ab^2}\amp= \frac{121.5}{4.5}\\ b^3 \amp = 27 \end{aligned} \end{equation*}

Note that by dividing the two equations, we eliminated $a\text{,}$ and we can now solve for $b\text{.}$

\begin{equation*} \begin{aligned}[t] b^3 \amp = 27 \\ b \amp = \sqrt[3]{27} = 3 \end{aligned} \end{equation*}

Next we substitute $b = 3$ into either of the two equations and solve for $a\text{.}$

\begin{equation*} \begin{aligned}[t] a(3)^2 \amp = 4.5 \\ a \amp= \frac{4.5}{9} \\ \amp = 0.5 \end{aligned} \end{equation*}

Thus, $a = 0.5$ and $b = 3\text{,}$ so the function is $f (x) = 0.5(3^ x )\text{.}$

Caution4.63

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 4.62, 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{equation*} \begin{aligned}[t] L(x) \amp = -73.5 + 39x \\ P(x) \amp = 0.372x^{3.6} \\ E(x) \amp = 0.5(3^x ) \\ \end{aligned} \end{equation*}

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 $a\text{.}$

Checkpoint4.64

Use the ratio method to find an exponential function whose graph includes the points $(1, 20)$ and $(3, 125)\text{.}$

$f (x) = 8(2.5)^x$

We can use the ratio method to find an exponential growth or decay model if we know two function values.

Example4.65

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?

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{equation*} \begin{aligned}[t] ab^7 \amp = 720 \\ ab^{10} \amp = 1620 \end{aligned} \end{equation*}

We divide the second equation by the first to find

\begin{equation*} \begin{aligned}[t] \frac{ab^{10}}{ab^7} \amp= \frac{1620}{720} \\ b^3 \amp= 2.25 \end{aligned} \end{equation*}

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{equation*} \begin{aligned}[t] a(1.31)^7 \amp = 720 \\ a \amp = \frac{720}{1.317} \\ \amp = 108.75 \end{aligned} \end{equation*}

Thus, $C(t) = 108.75(1.31)^t$ , and the annual inflation rate was $31$%.

Checkpoint4.66

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 $6\text{.}$ (Source: National Earthquake Information Center, U.S. Geological Survey)

1. Find a formula for the number of earthquakes, $N(m)\text{,}$ in terms of their magnitude.
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. How many earthquakes of magnitude $8$ occurred?
1. $N(m) = 558,526,329(0.106)^m$

2. $59,212,751\text{;}$ $~9$

SubsectionDoubling Time

Instead of giving the rate of growth of a population, we can specify its rate of growth by giving the time it takes for the population to double.

Example4.67

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 $P(t) = 74(1.02)^t\text{,}$ 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{equation*} \begin{aligned}[t] 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}\\ \amp\approx 35 \text{ years} \end{aligned} \end{equation*}

It will take the population about $35$ years to double.

2. Twice 148 million is 296 million, so we solve the equation

\begin{equation*} \begin{aligned}[t] 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}\\ \amp\approx 35 \text{ years} \end{aligned} \end{equation*}

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.

In Example 4.67, it took the population $35$ years to double. Notice that the calculations in parts (a) and (b) are identical after the first step. In fact, we can start at any point, and it will take the population $35$ years to double. We say that $35$ years is the doubling time for this population. In the Homework problems, you will show that any increasing exponential function has a constant doubling time.

Checkpoint4.68

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. 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.
1. $P(t) = 26.9(1.032)^t$ million

2. $22$ years

3. $P(0) = 26.9\text{;}$ $~P(22)\approx 53.8\text{,}$ $~P(44)\approx 107.6\text{,}$ $~P(66)\approx 215.1$

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*} 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.

Example4.69

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 $1.006\text{.}$ For exponential growth, $b = 1 + r\text{,}$ so $r = 0.006\text{,}$ or $0.6$%.
Checkpoint4.70

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

$1.9\%$

SubsectionHalf-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.

Example4.71

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{equation*} \begin{aligned}[t] 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} \\ \amp= 2.2 \end{aligned} \end{equation*}
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.

 $t$ $0$ $2.2$ $4.4$ $6.6$ $8.8$ $Q(t)$ $200$ $100$ $50$ $25$ $12.5$
Checkpoint4.72

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.
2. What is the half-life of alcohol in the body?
1. $A(t) = A_0(0.85)^t$

2. $4.3$ hours

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 $0.5\text{.}$ 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 Example 4.71.

Half-Life

If $H$ is the half-life for an exponential function $Q(t)\text{,}$ then

\begin{equation*} 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.

Example4.73

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.

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 $0.5\text{,}$ 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 $12.32$%.
Checkpoint4.74

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?

$2.28\%$

SubsectionAnnuities and Amortization

An annuity is a sequence of equal payments or deposits made at equal time intervals. A retirement fund is an example of an annuity. For ordinary annuities, payments are made at the end of each compounding period. The future value of an annuity is the sum of all the payments plus all the interest earned.

Future Value of an Annuity

If you make $n$ payments per year for $t$ years into an annuity that pays interest rate $r$ compounded $n$ times per year, the future value, $FV\text{,}$ of the annuity is

\begin{equation*} FV =\dfrac{P\left[\left(1+\dfrac{r}{n}\right)^{nt}-1\right]}{\dfrac{r}{n}} \end{equation*}

where each payment is $P$ dollars.

Example4.75

Greta plans to contribute $$200$ a month to a retirement fund that pays $5$% interest compounded monthly. 1. What is the future value of Greta's retirement fund after $15$ years? 2. For how many years must she contribute in order to accumulate$100,000?
Solution
1. We evaluate the formula for $FV$ when $P = 200\text{,}$ $r = 0.05\text{,}$ $n = 12\text{,}$ and $t = 15\text{.}$ Substituting these values into the formula, we find
\begin{equation*} \begin{aligned}[t] FV \amp=\frac{200\left[\left(1+\frac{0.05}{12}\right)^{12(15)}-1\right]}{\frac{0.05}{12}} \\ \amp = \frac{200[(1.00416)^{180}-1]}{0.0041\overline{6}} \\ \amp = 53,457.79 \end{aligned} \end{equation*}
In $15$ years, Greta's retirement fund will be worth $53,457.79\text{.}$ 2. We would like to find the value of $t$ when $P = 200\text{,}$ $r = 0.05\text{,}$ $n = 12\text{,}$ and $FV = 100,000\text{,}$ so we must solve the equation \begin{equation*} \begin{aligned}[t] 100,000\amp=\frac{200\left[\left(1+\frac{0.05}{12}\right)^{12t}-1\right]}{\frac{0.05}{12}}\\ \amp \blert{\text{Isolate the expression in brackets.}} \\ \frac{1}{200}\left(\frac{0.05}{12}\right)100,000\amp = \left(1+\frac{0.05}{12}\right)^{12t}-1\\ \amp \blert{\text{Simplify. Add 1 to both sides.}} \\ 2.08\overline{3} + 1 \amp = (1.0041\overline{6})^{12t}\\ \amp \blert{\text{Take the log of both sides.}}\\ \log 3.08\overline{3} \amp = 12t\log(1.0041\overline{6}) ~~~~~~~~ \blert{\text{Solve for } t.}\\ t\amp = \frac{\log 3.08\overline{3}}{12\log(1.0041\overline{6})}\\ \amp\approx 22.6 \end{aligned} \end{equation*} Greta must contribute for over $22$ years in order to accumulate$100,000\text{.}$
Checkpoint4.76

Rufus is saving for a new car. He puts $$2500$ a year into an account that pays $4$% interest compounded annually. How many years will it take him to accumulate$$20,000\text{?}$ (Round up to the next whole year.)

$8$ years

In Example 4.75, we knew the monthly deposits into the annuity and calculated how much the sum of all the deposits (plus interest) would be in the future. Now imagine that you have just retired and you want to begin drawing monthly payments from your retirement fund. The total amount accumulated in your fund is now its present value, and that amount must cover your future withdrawal payments.

Present Value of an Annuity

If you wish to receive $n$ payments per year for $t$ years from a fund that earns interest rate $r$ compounded $n$ times per year, the present value, $PV\text{,}$ of the annuity must be

\begin{equation*} PV =\frac{P\left[1-\left(1+\dfrac{r}{n}\right)^{-nt}\right]}{\dfrac{r}{n}} \end{equation*}

where each payment is $P$ dollars.

Example4.77

Candace Welthy is setting up a college fund for her nephew Delbert that will provide $400$ a month for the next $5$ years. If the interest rate is $4$% compounded monthly, how much money should she deposit now to cover the fund? Solution We would like to find the present value of an annuity in which $P = 400\text{,}$ $r = 0.04\text{,}$ $n = 12\text{,}$ and $t = 5\text{.}$ Substituting these values into the formula gives \begin{equation*} \begin{aligned}[t] PV \amp = \frac{400\left[1-\left(1+\frac{0.04}{12}\right)^{-(12)(5)}\right]}{\frac{0.04}{12}}\\ \amp= \frac{400[1 - (1.00\overline{3})]^{-60}}{0.00\overline{3}}\\ \amp= 21,719.63 \end{aligned} \end{equation*} Delbert's Aunt Welthy should deposit$21,719.63\text{.}$

Payments on a loan, such as a home mortgage, are also an annuity, but in this case the monthly payments do not collect interest; instead, we must pay interest on the present value of the loan. Repaying a loan (plus interest) by making a sequence of equal payments is called amortizing the loan.

44

Francine plans to make monthly payments into an account to save up for a cruise vacation. She wants to save $$25,000$ for the trip. How many$$200$ payments will she need if the account pays $3\%$ interest compounded monthly? What if the rate is $4\%\text{?}$

For Problems 45 and 46, use the formula for present value of an annuity.

47

Moore's law predicts that the number of transistors per computer chip will continue to grow exponentially, with a doubling time of $18$ months.

1. Write a formula for Moore's law, with $t$ in years and $M_0 = 2200$ in $1970\text{.}$

2. From $1970$ to $1999\text{,}$ the number of transistors per chip was actually modeled approximately by $N(t) = 2200(1.356)^t\text{.}$ How does this function compare with your answer to part (a)?

3. Complete the table showing the number of transistors per chip in recent years, the number predicted by Moore's law, and the number predicted by $N(t)\text{.}$

 Name of chip Year Moore'slaw $N(t)$ Actualnumber Pentium IV $2000$ $42,000,000$ Pentium M (Banias) $2003$ $77,000,000$ Pentium M (Dothan) $2004$ $140,000,000$
4. What is the doubling time for $N(t)\text{?}$

48

If the population of a particular animal is very small, inbreeding will cause a loss of genetic diversity. In a population of $N$ individuals, the percent of the species' original genetic variation that remains after $t$ generations is given by

\begin{equation*} V=V_0\left(1-\dfrac{1}{2N} \right)^t \end{equation*}

(Source: Chapman and Reiss, 1992)

1. Assuming $V_0 = 100\text{,}$ graph $V$ as a function of $t$ for three different values of $N\text{:}$ $N = 1000\text{,}$ $100\text{,}$ and $10\text{.}$

2. Fill in the table to compare the values of $V$ after $5\text{,}$ $50\text{,}$ and $100$ generations.

 Population size Number of generations $5$ $50$ $100$ $1000$ $100$ $10$
3. Studies of the cheetah have revealed variation at only $3.2\%$ of its genes. (Other species show variation at $10\%$ to $43\%$ of their genes.) The population of cheetah may be less than $5000\text{.}$ Assuming the population can be maintained at its current level, how many generations will it take before the cheetah's genetic variation is reduced to $1\%\text{?}$