# 15.1: Applications of exponential functions

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Before giving specific applications, we note that the exponential function $$y=c\cdot b^x$$ is uniquely determined by providing any two of its function values.

## Example $$\PageIndex{1}$$

Let $$f(x)=c\cdot b^x$$. Determine the constant $$c$$ and base $$b$$ under the given conditions.

1. $$f(0)=5$$, $$f(1)=20$$
2. $$f(0)=3$$, $$f(4)=48$$
3. $$f(2)=160$$, $$f(7)=5$$
4. $$f(-2)=55$$, $$f(1)=7$$

Solution

1. Applying $$f(0)=5$$ to $$f(x)=c\cdot b^x$$, we get

$5=f(0)=c\cdot b^0=c\cdot 1 =c \nonumber$

Indeed, in general, we always have $$f(0)=c$$ for any exponential function. The base $$b$$ is then determined by substituting the second equation $$f(1)=20$$.

$20=f(1)=c\cdot b^1=5 \cdot b \quad\stackrel{(\div 5)}\implies \quad b=4 \nonumber$

Therefore, $$f(x)=5\cdot 4^x$$. Note that in the last implication, we used that the base must be positive.

1. As before, we get $$3=f(0)=c\cdot b^0=c$$, and

$48=f(4)=c\cdot b^4=3\cdot b^4 \quad\stackrel{(\div 3)}\implies \quad 16=b^4 \quad\stackrel{(\text{exponentiate by }\frac 1 4)}\implies \quad b=16^{\frac 1 4}=2 \nonumber$

Therefore, $$f(x)=3\cdot 2^x$$.

1. When $$f(0)$$ is not given, it is easiest to solve for $$b$$ first. We can see this as follows. Since $$160=f(2)=c\cdot b^2$$ and $$5=f(7)=c\cdot b^7$$, the quotient of these equations eliminates $$c$$.

\begin{aligned} \dfrac{160}{5}&=\dfrac{c\cdot b^2}{c \cdot b^7}=\dfrac{1}{b^5}\\ \implies 32&=b^{-5} \\ \\ \Longrightarrow b&=32^{-\frac 1 5} \quad (\text{exponentiate by }(-\frac {1} 5))\\ &=\dfrac{1}{32^{\frac 1 5}}\\ &=\dfrac 1 2 \end{aligned} \nonumber

Then $$c$$ is determined by any of the original equations.

$160=f(2)=c\cdot b^2=c\cdot \Big(\frac 1 2\Big)^2=c \cdot \frac 1 4 \quad \implies \quad c=4\cdot 160=640 \nonumber$

Therefore, $$f(x)=640\cdot \Big(\dfrac 1 2\Big)^x$$.

1. This solution is similar to part (c).

\begin{aligned} \dfrac{55}{7}&=\dfrac{f(-2)}{f(1)}=\dfrac{c\cdot b^{-2}}{c \cdot b^1}=\dfrac{1}{b^3}\\ \implies b^3&=\dfrac{7}{55} \\ \implies b&=\Big(\dfrac {7}{55}\Big)^{\frac 1 3}\approx 0.503 \end{aligned} \nonumber

$\begin{gathered} 55=f(-2)=c\cdot b^{-2}=c\cdot \Big(\Big(\frac {7}{55}\Big)^{\frac 1 3}\Big)^{-2}=c\cdot \Big(\frac {7}{55}\Big)^{\frac {-2} 3} \\ \implies c=\frac{55}{\Big(\dfrac {7}{55}\Big)^{\frac {-2} 3}}=55\cdot \Big(\frac {7^{\frac {2} 3}}{55^{\frac {2} 3}}\Big)=55^{\frac 1 3}\cdot 7^{\frac 2 3}=\sqrt[3]{55\cdot 7^2}=\sqrt[3]{2695} \approx 13.916\end{gathered}$

therefore, $$f(x)=\sqrt[3]{2695}\cdot \Big(\sqrt[3]{\dfrac 7 {55}}\,\,\Big)^x$$.

Many situations are modeled by exponential functions.

## Example $$\PageIndex{2}$$

The mass of a bacteria sample is $$2\cdot 1.02^t$$ grams after $$t$$ hours.

1. What is the mass of the bacteria sample after $$4$$ hours?
2. When will the mass reach $$10$$ grams?

Solution

1. The formula for the mass $$y$$ in grams after $$t$$ hours is $$y(t)=2\cdot 1.02^t$$. Therefore, after $$4$$ hours, the mass in grams is

$y(4)=2\cdot 1.02^4 \approx 2.16 \nonumber$

1. We are seeking the number of hours $$t$$ for which $$y=10$$ grams. Therefore, we have to solve:

$10=2\cdot 1.02^t \quad\quad \stackrel{\text{(\div 2)}}\implies \quad\quad 5=1.02^t \nonumber$

We need to solve for the variable in the exponent. In general, to solve for a variable in the exponent requires an application of a logarithm on both sides of the equation.

$5=1.02^t \quad\quad \stackrel{\text{(apply \log)}}\implies \quad\quad \log(5)=\log(1.02^t) \nonumber$

Recall one of the main properties that we need to solve for $$t$$:

$\label{EQU:log(xt)} \log(x^t)=t\cdot \log(x) \nonumber$

Using this fact, we can now solve for $$t$$:

\begin{aligned} \log(5)&=\log(1.02^t) \\ \implies \log(5)&=t\cdot \log(1.02) \\ \implies \dfrac{\log(5)}{\log(1.02)}&=t \quad \text{(divide by \log(1.02))} \end{aligned} \nonumber

Using the $$\boxed{log}$$ key on the calculator, we may approximate this as

$$t=\dfrac{\log (5)}{\log (1.02)} \approx 81.3$$

After approximately $$81.3$$ hours, the mass will be $$10$$ grams.

Note that the formula $$2\cdot 1.02^t$$ comes from the following calculations. Suppose that after each hour there is an increase of 2 percent of the mass and that the initial mass of the bacteria is 2 grams. Then after the first hour we see that the mass in grams is

$2+2\cdot(.02)=2(1+.02)=2(1.02) \nonumber$

and that after the second hour the mass is

$2(1.02)+2(1.02)(.02)=2(1.02)(1+.02)=2(1.02)^2 \nonumber$

and that after three hours the mass is

$2(1.02)^2+2(1.02)^2(.02)=2(1.02)^2(1+.02)=2(1.02)^3 \nonumber$

Continuing in this manner we can see that, at least for whole numbers of hours $$t$$, the weight is given by

$2\cdot 1.02^t \nonumber$

The same sort of idea can be used in all of the following applications.

## Example $$\PageIndex{3}$$

The population size of a country was $$12.7$$ million in the year $$2000$$, and $$14.3$$ million in the year $$2010$$.

1. Assuming an exponential growth for the population size, find the formula for the population depending on the year $$t$$.
2. What will the population size be in the year $$2015$$, assuming the formula holds until then?
3. When will the population reach $$18$$ million?

Solution

1. The growth is assumed to be exponential, so that $$y(t)=c\cdot b^t$$ describes the population size depending on the year $$t$$, where we set $$t=0$$ corresponding to the year $$2000$$. Then the example describes $$y(0)=c$$ as $$c=12.7$$, which we assume in units of millions. To find the base $$b$$, we substitute the data of $$t=10$$ and $$y(t)=14.3$$ into $$y(t)=c\cdot b^t$$.

\begin{aligned} 14.3&=12.7\cdot b^{10} \\ \implies \dfrac{14.3}{12.7}&=b^{10}\\ \implies \Big(\dfrac{14.3}{12.7}\Big)^{\frac 1 {10}}&=(b^{10})^{\frac 1 {10}}=b \\ \implies b&=\Big(\dfrac{14.3}{12.7}\Big)^{\frac 1 {10}}\approx 1.012 \end{aligned} \nonumber

The formula for the population size is $$y(t)\approx 12.7\cdot 1.012^t$$.

1. We calculate the population size in the year $$2015$$ by setting $$t=15$$:

$y(15)=12.7\cdot 1.012^{15}\approx 15.2 \nonumber$

1. We seek $$t$$ so that $$y(t)=18$$. We solve for $$t$$ using the logarithm.

\begin{aligned} 18&=12.7\cdot 1.012^{t} \\ \implies \dfrac{18}{12.7}&=1.012^{t} \\ \implies \log\Big(\dfrac{18}{12.7}\Big)&=\log(1.012^t) \\ \implies \log\Big(\dfrac{18}{12.7}\Big)&=t\cdot \log(1.012)\\ \implies t&=\dfrac{\log\Big(\frac{18}{12.7}\Big)}{\log(1.012)}\approx 29.2 \end{aligned} \nonumber

The population will reach $$18$$ million in the year $$2029$$.

In many instances the exponential function $$f(x)=c\cdot b^x$$ is given via a rate of growth $$r$$.

## Definition: Rate of Growth

An exponential function with a rate of growth $$r$$ is a function $$f(x)=c\cdot b^x$$ with base

$\boxed{b=1+r} \nonumber$

For an example of why this is a reasonable definition see the note following Example $$\PageIndex{2}$$.

## Note

We want to point out that some authors use a different convention than the one given in definition Rate of Growth above. Indeed, sometimes a function with rate of growth $$r$$ is defined as an exponential function with base $$b=e^r$$, whereas for us it has base $$b=1+r$$. Since $$e^r$$ can be expanded as $$e^r=1+r+\dfrac{r^2}{2}+\dots$$, we see that the two versions only differ by an error of order $$2$$.

## Example $$\PageIndex{4}$$

The number of PCs that are sold in the U.S. in the year $$2011$$ is approximately $$350$$ million with a rate of growth of $$3.6\%$$ per year. Assuming the rate stays constant over the next years, how many PCs will be sold in the year $$2015$$?

Solution

Since the rate of growth is $$r=3.6\%=0.036$$, we obtain a base of $$b=1+r=1.036$$, giving the number of PCs to be modeled by $$c(1.036)^t$$. If we set $$t=0$$ for the year $$2011$$, we find that $$c=350$$, so the number of sales is given by $$y(t)=350\cdot 1.036^t$$. Since the year $$2015$$ corresponds to $$t=4$$, we can calculate the number of sales in the year $$2015$$ as

$y(4)=350\cdot 1.036^4\approx 403 \nonumber$

Approximately $$403$$ million PCs will be sold in the year $$2015$$.

## Example $$\PageIndex{5}$$

The size of an ant colony is decreasing at a rate of $$1\%$$ per month. How long will it take until the colony has reached $$80\%$$ of its original size?

Solution

Since $$r=-1\%=-0.01$$, we obtain the base $$b=1+r=1-0.01=0.99$$. We have a colony size of $$y(t)=c\cdot 0.99^t$$ after $$t$$ months, where $$c$$ is the original size. We need to find $$t$$ so that the size is at $$80\%$$ of its original size $$c$$, that is, $$y(t)=80\%\cdot c=0.8\cdot c$$.

\begin{aligned} 0.8\cdot c&=c\cdot 0.99^{t} \\ \implies 0.8&=0.99^{t} \quad \text{(\div c)} \\ \implies \log(0.8)&=\log(0.99^t) \\ \implies \log(0.8)&=t\cdot \log(0.99) \\ \implies t&=\dfrac{\log(0.8)}{\log(0.99)}\approx 22.2 \end{aligned} \nonumber

After approximately $$22.2$$ months, the ant colony has decreased to $$80\%$$ of its original size.

## Example $$\PageIndex{6}$$

1. The population size of a country is increasing at a rate of $$4\%$$ per year. How long does it take until the country has doubled its population size?
2. The number of new flu cases is decreasing at a rate of $$5\%$$ per week. After how much time will the number of new flu cases reach a quarter of its current level?

Solution

1. The rate of change is $$r=4\%=0.04$$, so that population size is an exponential function with base $$b=1+0.04=1.04$$. Therefore, $$f(x)=c\cdot 1.04^x$$ denotes the population size, with $$c$$ being the initial population in the year corresponding to $$x=0$$. In order for the population size to double, $$f(x)$$ has to reach twice its initial size, that is:

\begin{aligned} f(x)&=2c \\ \implies c\cdot 1.04^x&=2c \\ \implies 1.04^x&=2 \quad (\div c)\\ \implies \log(1.04^x)&=\log(2)\\ \implies x\log(1.04)&=\log(2) \\ \implies x&=\dfrac{\log(2)}{\log(1.04)}\approx 17.7 \end{aligned} \nonumber

It will take about $$17.7$$ years until the population size has doubled.

1. Since the number of new flu cases is decreasing, the rate of growth is negative, $$r=-5\%=-0.05$$ per week, so that we have an exponential function with base $$b=1+r=1+(-0.05)=0.95$$. To reach a quarter of its initial number of flu cases, we set $$f(x)=c\cdot 0.95^x$$ equal to $$\dfrac 1 4 c$$.

\begin{aligned} c\cdot 0.95^x&=\dfrac 1 4 c \\ \implies 0.95^x&=\cfrac 1 4 \quad (\div c)\\ \implies x\log(0.95)&=\log(\cfrac 1 4) \\ \implies x&=\cfrac{\log(\frac 1 4)}{\log(0.95)}\approx 27.0 \end{aligned} \nonumber

It will therefore take about $$27$$ weeks until the number of new flu cases has decreased to a quarter of its current level.

This page titled 15.1: Applications of exponential functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.