Section 5.1 Exponential Functions: Practical Meaning

Objectives

After completing this section, you should be able to do the following.

State the definition of an exponential function and identify its initial value and growth factor.
Convert between the growth factor and the percent growth (or decay) rate of an exponential function.
Apply exponential functions to model real-life growth and decay processes, such as compound interest, population growth, and radioactive decay.

Linear functions $y = f(t) = mt + b\text{,}$ which you studied in Chapter 2, change by a constant amount $m$ per unit change in $t\text{.}$ Exponential functions change by a constant factor.

Example 5.1.1.

You deposit $\$1500$ into a savings account that pays $5\%$ interest annually. Let $B = B(t)$ be your balance after $t$ years.

What are $B(0)\text{,}$ $B(1)\text{,}$ $B(2)\text{,}$ and $B(3)\text{?}$
Find a formula for the function $B(t)\text{.}$
What is your balance after $10$ years?

Solution.

Your initial deposit was made at time $t = 0\text{,}$ so $B(0)=1500$ is the amount of the initial deposit.

After $1$ year, the bank adds $5\%$ interest on your initial deposit $B(0)$ to your balance. Since 5\% of $B(0)$ can be calculated by computing $0.05 \cdot B(0)\text{,}$ your balance after $1$ year is

\begin{equation*} B(1) = B(0) + 0.05 \cdot B(0) = B(0) \left( 1 + 0.05 \right) = B(0) \cdot 1.05. \end{equation*}

Since $B(0) = 1500\text{,}$

\begin{equation*} B(1) = B(0) \cdot 1.05 = 1500 \cdot 1.05 = 1575. \end{equation*}

This means that during the first year, your savings account earned $\$75$ in interest ($5\%$ of $1500$).

At the end of the second year, the bank will add $5\%$ of $1575$ to your account—the interest from the first year is now earning interest, too! Equivalently, the bank will multiply $B(1)$ by $1.05\text{.}$ Hence:

\begin{equation*} B(2) = B(1) \cdot 1.05 = 1575 \cdot 1.05 = 1653.75. \end{equation*}

Notice that $1653.75 - 1575 = 78.75\text{.}$ So during the second year your account earned $\$78.75$ in interest, which is $5\%$ of $1575\text{.}$

The amount of money by which your balance increases each year is not constant; but the factor by which it increases is always $1.05\text{,}$ which corresponds to a $5\%$ increase. To see a clear pattern, observe that since $B(1) = B(0) \cdot 1.05\text{:}$

\begin{equation*} B(2) = B(1) \cdot 1.05 = (B(0) \cdot 1.05) \cdot 1.05 = B(0) \cdot 1.05^2. \end{equation*}

Similarly:

\begin{equation*} B(3) = B(2) \cdot 1.05 = (B(0) \cdot 1.05^2) \cdot 1.05 = B(0) \cdot 1.05^3. \end{equation*}

Substituting $B(0) = 1500\text{,}$ we get $B(3) = 1500 \cdot 1.05^3 \approx 1736.44\text{.}$ Your balance after $3$ years is $\$1736.44\text{.}$
Each year your current balance is multiplied by the factor $1.05\text{.}$ After $t$ years, your initial balance $B(0)$ will be multiplied $t$ times by $1.05\text{:}$

\begin{equation*} B(t) = B(0) \cdot 1.05^t = 1500 \cdot 1.05^t. \end{equation*}

So $B(t) = 1500 \cdot 1.05^t$ gives your balance after $t$ years.
Using the latter formula, we calculate $B(10) = 1500 \cdot 1.05^{10} \approx 2443.34\text{.}$ Your balance after $10$ years is $\$2443.34\text{.}$

A function of the form $B(t) = 1500 \cdot 1.05^t\text{,}$ where the independent variable $t$ is in the exponent, is called an exponential function.

Definition 5.1.2. Exponential Function.

Let $A$ and $b$ be given constants with $A \gt 0\text{,}$ $b \gt 0\text{,}$ and $b \neq 1\text{.}$ Then the function:

\begin{equation*} f(t) = A \cdot b^t \end{equation*}

is called an exponential function with the base $b$ (also called the growth factor) and initial value $A\text{.}$

The function $B(t) = 1500 \cdot 1.05^t$ of Example 5.1.1 is an exponential function with the base—or equivalently growth factor—$b = 1.05$ and initial value $A = 1500\text{.}$

Notice that for any exponential function $f(t) = A \cdot b^t\text{,}$

\begin{equation*} f(0) = A \cdot b^0 = A \end{equation*}

as $b^0 = 1$ for any $b\text{.}$ This means $A$ is always the value of an exponential function at $0\text{,}$ hence the name “initial value”.

Exponential functions are used very often to model real-life processes of exponential growth and exponential decay.

Subsection Exponential Growth

If the base (growth factor) $b$ is greater than $1\text{,}$ the exponential function $f(t) = A \cdot b^t$ is increasing. For every unit change in $t\text{,}$ the current value of the function is multiplied by the factor $b\text{,}$ so the value increases.

This is the case for the function $B(t) = 1500 \cdot 1.05^t$ in Example 5.1.1 since $b = 1.05 \gt 1\text{;}$ the growth of your savings account balance is exponential.

Whenever $b$ is greater than $1\text{,}$ then for some positive number $r$ we can write

\begin{equation*} b = 1 + r. \end{equation*}

The constant $r\text{,}$ when expressed as a percentage, is called the percent growth rate. For $B(t) = 1500 \cdot 1.05^t\text{,}$ since $1.05 = 1 + 0.05\text{,}$ the percent growth rate is $r = 0.05\text{,}$ or $r = 5\%\text{:}$ the percentage of the current balance that is added to your account every year.

Exponential functions are often used to model population growth for populations: human populations, animal populations, or populations of bacteria or insects in laboratory experiments.

Bacteria, which are one-cell organisms, reproduce by each cell dividing into two daughter cells with a frequency that depends on the kind of bacteria. Such a population grows slowly at first when the population consists of a small number of bacteria, and faster and faster with a larger and larger number of bacteria ready to divide. It is reasonable to expect that the population will grow not by a constant number of bacteria per unit of time but by a constant percentage of its current size.

In other words, we expect the population to grow exponentially.

Example 5.1.3.

A population of Escherichia coli (E. coli) bacteria in nutrient-rich laboratory conditions grows by 3.53 percent per minute

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6015860/, accessed: 6/26/20

. Let $P = f(t)$ be the number of bacteria $t$ minutes after the experiment began and suppose that at $t = 0\text{,}$ there are $2000$ bacteria.

What is the percent growth rate? What is the growth factor? Write a formula for $f(t)\text{.}$
How many bacteria are there at $t = 20\text{?}$ At $t = 40\text{?}$ At $t = 60\text{?}$

Solution.

Since the population increases by a constant percent, $f(t)$ is an exponential function: $f(t) = A \cdot b^t\text{.}$ The initial amount is $A=2000\text{.}$ Every minute the population increases by $3.53\%$ of its current size, so the percent growth rate is $r = 3.53\%$ or equivalently $r = 0.0353\text{.}$ The growth factor is $b = 1 + r = 1 + 0.0353 = 1.0353\text{.}$ Hence:

\begin{equation*} P = f(t) = 2000 \cdot 1.0353^t. \end{equation*}
We use the formula obtained in (a) and calculate:

\begin{align*} f(20) \amp = 2000 \cdot 1.0353^{20} \approx 4002.7\\ f(40) \amp = 2000 \cdot 1.0353^{40} \approx 8010.8\\ f(60) \amp = 2000 \cdot 1.0353^{60} \approx 16032.6 \end{align*}

You may notice that the E. coli population approximately doubles during the first $20$ minutes, then doubles again during the next $20$ minutes, and again after $20$ more minutes. This is not a coincidence. Processes of exponential growth have what we call a doubling time; that is, the time needed for the quantity to double. We will cover this in depth in Section 5.3.

Subsection Exponential Decay

If the base $b$ of an exponential function $f(t) = A \cdot b^t$ is less than $1\text{,}$ multiplying by $b$ decreases the value; hence the exponential function is decreasing. Commonly, $b$ is called the “growth factor” whether $b \gt 1$ or $b \lt 1\text{,}$ even though for $b \lt 1$ there is decay rather than growth. The base $b$ can still be written as:

\begin{equation*} b = 1 + r \end{equation*}

for some number $r\text{.}$ However, when $b \lt 1\text{,}$ the number $r$ is negative. We still call $r$ the “percent growth rate”. This terminology is counterintuitive, although one can argue that a negative growth rate means the quantity is exhibiting the opposite of growth—that is, it is exhibiting decay.

Many real-life processes are modeled by decaying exponential functions, including elimination of a drug from the body and processes of radioactive decay.

Example 5.1.4.

The common antianxiety medication Diazepam is eliminated from the body at a daily rate of $26\%\text{.}$ Suppose that a patient takes a one-time dose of $10$ mg of Diazepam. Let $D = D(t)$ be the amount of Diazepam left in his bloodstream $t$ days after the dose. Find a formula for $D(t)\text{.}$

Solution.

The initial amount is $D(0) = 10\text{.}$ During the first day, $26\%$ of the initial dose is eliminated. Hence:

\begin{equation*} D(1) = D(0) - 0.26 \cdot D(0) = D(0) (1 - 0.26) = D(0) (1 + (-0.26)). \end{equation*}

During the second day, $26\%$ of $D(1)$ is eliminated:

\begin{equation*} D(2) = D(1) - 0.26 \cdot D(1) = D(1) (1 + (-0.26)) = D(0) (1 + (-0.26))^2 \end{equation*}

and so on. Each next day we multiply the dose from the day before by $1 + (-0.26)\text{.}$ The amount left after $t$ days is:

\begin{equation*} D(t) = D(0) (1 + (-0.26))^t. \end{equation*}

Hence, $D(t)$ is an exponential function. The growth factor is

\begin{equation*} b = 1 + (-0.26) = 0.74, \end{equation*}

the daily percent growth rate is $r = -0.26\text{,}$ and the initial amount is $A = D(0) = 10\text{.}$ The final version of the formula is:

\begin{equation*} D(t) = 10 \cdot 0.74^t. \end{equation*}

Every day $26\%$ of the amount from the day before gets eliminated, so $74\%$ of the amount from the day before stays in the body.

The growth factor $0.74$ is less than $1$ and the daily percent growth rate of $-26\%$ is negative since the amount is decaying. Often we say that the daily percent decay rate is $26\%\text{.}$

Notice that after $2$ days, the patient still has $D(2) = 10 \cdot 0.74^2 \approx 5.48$ mg in his system—more than a half of the initial dose. Such a slow elimination rate causes a medication buildup if a daily dose is taken.

For the sake of convenience, let’s summarize the relationship between the growth factor $b$ and the percent growth rate $r\text{.}$

Result 5.1.5.

Let an exponential function $f(t) = A \cdot b^t$ be given (increasing or decreasing). Then:

$b$ is called the “growth factor”.
The constant $r$ such that $b = 1 + r$ is called the “percent growth rate” or simply the “growth rate”.
Given $r\text{,}$ we calculate $b$ by taking $b = 1 + r\text{.}$
Given $b\text{,}$ we calculate $r$ by taking $r = b - 1\text{.}$

Another common real-life application of exponential functions is in the study of radioactivity. All radioactive isotopes decay and they decay exponentially. Some decay extremely slowly and some decay very quickly. Carbon-14, a radioactive isotope of Carbon used in carbon dating, decays so slowly that it takes thousands of years for half of an initial amount to decay. Iodine-131 used in the treatment of thyroid cancer, decays in a matter of days.

Example 5.1.6.

The daily percent decay rate of Iodine-131 is $8.3\%\text{.}$ Let $I = I(t)$ be the amount of Iodine-131, in $\mu$g, left after $t$ days if the initial amount is 10$\mu$g.

Find a formula for $I(t)\text{.}$ What is the growth factor? What is the percent growth rate?
Estimate how long it will take for half of initial amount to decay.

Solution.

The daily percent decay rate is $8.3\%\text{.}$ Hence, $r = -0.083$ and the growth factor is $b = 1 + (-0.083) = 0.917\text{.}$ The initial amount is 10$\mu$g. Hence:

\begin{equation*} I(t) = 10 \cdot 0.917^t. \end{equation*}
We are looking for $t$ such that $I(t) = 5\text{.}$ At this point, we don’t have tools other than trial and error, so let’s calculate a few values. Since $I(10) = 10 \cdot 0.917^{10} \approx 4.2\text{,}$ then $10$ days is a bit too long. And, $I(7) = 10 \cdot 0.917^{7} \approx 5.45$ so $7$ days is too short. At $8$ days, we have $I(8) = 10 \cdot 0.917^{8} \approx 4.9998$ which is very close. It will take about $8$ days for half of the initial amount to decay.

Note: From now on, we will often use the more common notation $I(t) = 10(0.917)^t$ as well as the notation $I(t) = 10 \cdot 0.917^t\text{.}$

Example 5.1.7.

The following exponential functions describe populations of four towns $t$ years after January 1, 2000.

$\displaystyle P(t) = 700(1.15)^t$
$\displaystyle A(t) = 10000(0.89)^t$
$\displaystyle Q(t) = 12000(0.92)^t$
$\displaystyle S(t) = 1500(1.021)^t$

For each town, identify the initial population at $t = 0\text{,}$ the growth factor, and the annual percent growth rate. Also determine if the population of each town is increasing or decreasing.

Solution.

The initial population is $700$ people, since $P(0) = 700(1.15)^0 = 700\text{,}$ and the growth factor is $1.15\text{.}$ The percent growth rate is then $r = 1.15 - 1 = 0.15 = 15\%\text{.}$ Since the growth factor of $1.15$ is greater than $1$ and the growth rate is positive, the population of the town is increasing.
The initial population is $10000$ people and the growth factor is $0.89\text{,}$ so the percent growth rate is $r = 0.89 - 1 = -0.11 = -11\%\text{.}$ In other words, the decay rate is $11\%$ which implies the population of the town is decreasing.
The initial population is $12000$ people and the growth factor is $0.92\text{,}$ therefore the percent growth rate is $r = 0.92 - 1 = -0.08 = -8\%\text{.}$ In other words, the decay rate is $8\%\text{.}$ The population of the town is decreasing.
The initial population is $1500$ people and the growth factor is $1.021\text{,}$ meaning the percent growth rate is $2.1\%\text{.}$ The growth factor of $1.021$ is greater than $1$ and the growth rate is positive so the population of the town is increasing.

Example 5.1.8.

Let $V(t)$ be the value, in dollars, of an antique lamp $t$ years after its purchase. The lamp was purchased for $\$5000$ and its value $V(t)$ increases by $\$400$ each year.

Find a formula for the function $V(t)\text{.}$ What kind of function is it?
What is the rate of increase of $V(t)$ in dollars per year?
What is the value of the lamp $10$ years after its purchase?

Solution.

The value $V(t)$ increases by a fixed amount of dollars each year. Hence, $V(t)$ is a linear function:

\begin{equation*} V(t) = 5000 + 400t \end{equation*}
The slope $m = 400$ represents the constant rate of increase of the value $V(t)$ in dollars per year.
The value after $10$ years is $V(10) = 5000 + 400 \cdot 10 = 9000$ dollars.

Example 5.1.9.

Let $V(t)$ be the value, in dollars, of an antique lamp, $t$ years after its purchase. The lamp was purchased for $\$5000$ and its value $V(t)$ increases by $12\%$ each year.

Find a formula for the function $V(t)\text{.}$ What kind of function is it?
What is the growth factor and the annual percent growth rate of $V(t)\text{?}$
What is the value of the lamp $10$ years after its purchase?

Solution.

This time the value of the lamp is increasing not by a constant amount each year but by a constant percentage of the current value each year. An increase of $12\%$ means adding $12\%$ of $V(t)$ to itself; that is, multiplying $V(t)$ by a constant factor $1.12$ each year. Hence, the function $V(t)$ is exponential and equal to:

\begin{equation*} V(t) = 5000 (1.12)^t \end{equation*}
The growth factor is $1.12\text{;}$ the annual percent growth rate is $0.12$ or $12\%\text{.}$
The value after $10$ years is $V(10) = 5000 (1.12)^{10} \approx 15529.24$ dollars.

Example 5.1.10.

Is a given function $Q(t)$ exponential? If yes, rewrite $Q(t)$ in the form $Q(t) = A \cdot b^t\text{.}$ Identify $A$ and $b\text{.}$

$\displaystyle Q(t) = 50 (2)^{\frac{t}{6}}$
$\displaystyle Q(t) = 550 (t^2)^{\frac{1}{3}}$
$\displaystyle Q(t) = (\sqrt{2})^{2t}$
$\displaystyle Q(t) = 12000 \left( \frac{1}{8} \right)^{\frac{t}{3}}$

Solution.

By utilizing appropriate exponent rules, we have $Q(t) = 50 (2)^{\frac{t}{6}} = 50 \left( 2^{\frac{1}{6}} \right)^t = 50 (\sqrt[6]{2})^t\text{.}$ Hence, $Q(t)$ is exponential with $A = 50$ and $b = \sqrt[6]{2}\text{.}$ Since $\sqrt[6]{2} \approx 1.1225\text{,}$ in an applied problem we could write $Q(t) = 50 (1.1225)^t\text{.}$
We can simplify $Q(t)$ as $Q(t) = 550 t^{\frac{2}{3}}\text{,}$ so $Q(t)$ is a power function and therefore not an exponential function. Recall that the independent variable is the base in a power function whereas it is the exponent in an exponential function.
$Q(t) = ((\sqrt{2})^2)^t = 2^t\text{.}$ The function $Q(t)$ is exponential with $A = 1$ and $b = 2\text{.}$
Simplifying we get: $Q(t) = 12000 \left( \left( \frac{1}{8} \right)^{1/3} \right)^t = 12000 \left( \frac{1}{2} \right)^t\text{.}$ The function $Q(t)$ is exponential with $A = 12000$ and $b = \dfrac{1}{2}\text{.}$

Exercises Exercises

Exponential Growth or Decay.

For each of the following, decide if a given exponential function represents a process of exponential growth or decay. For each function identify the initial value and the growth factor.

1.

$\displaystyle f(t)=120(1.7)^t$

Solution.

exponential growth; initial value: $120\text{;}$ growth factor: $1.7$

2.

$\displaystyle g(t)=1500(0.95)^t$

Solution.

exponential decay; initial value: $1500\text{;}$ growth factor: $0.95$

3.

$\displaystyle h(t)=60(0.83)^t$

Solution.

exponential decay; initial value: $60\text{;}$ growth factor: $0.83$

4.

$\displaystyle m(t)=130(1.02)^t$

Solution.

exponential growth; initial value: $130\text{;}$ growth factor: $1.02$

5.

You deposit $\$2000$ dollars into a savings account that pays $3.5\%$ annually. Let $B=B(t)$ be your balance $t$ years later.

(a)

Find a formula for $B(t)$ in the form $B(t)=A\cdot b^t\text{.}$

Solution.

$B(t)=2000(1.035)^t\text{;}$ exponential function

(b)

Find your balance after $7$ years.

Solution.

$\$2544.56$

(c)

Find the growth factor and the percent growth rate of the function $B(t)\text{.}$

Solution.

growth factor: $1.035\text{;}$ growth rate: $3.5\%$

6.

Let $V(t)$ be the value, in dollars, of an antique desk $t$ years after its purchase. The desk was purchased for $\$7000$ and its value $V(t)$ increases by $\$500$ per year.

(a)

Find a formula for the function $V(t)\text{.}$ What kind of function is it?

Solution.

$V(t)=500t+7000\text{;}$ linear function

(b)

What is the rate of increase of $V(t)$ in dollars per year?

Solution.

$500$ dollars per year

(c)

What is the value of the desk after $12$ years?

Solution.

$\$13000$

7.

Let $V(t)$ be the value, in dollars, of an antique desk, $t$ years after its purchase. The desk was purchased for $\$7000$ and its value $V(t)$ increases by $9.5\%$ per year.

(a)

Find a formula for the function $V(t)\text{.}$ What kind of function is it?

Solution.

$V(t)=7000(1.095)^t\text{;}$ exponential function

(b)

What is the growth factor and the percent growth rate of $V(t)\text{?}$

Solution.

growth factor: $1.095\text{;}$ growth rate: $9.5\%$

(c)

What is the value of the desk after $12$ years?

Solution.

$\$20800.20$

8.

Let $P(t)$ be the population of a town $t$ years after the year $1990\text{.}$ The population $P(t)$ was $12000$ people in $1990\text{,}$ that is, at $t=0\text{,}$ and it has been increasing by $750$ people each year.

(a)

Write a formula for the function $P(t)\text{.}$ What kind of function is it?

Solution.

$P(t)=750t+12000\text{;}$ linear function

(b)

What is the rate of increase of the population in people per year?

Solution.

$750$ people per year

9.

(a)

Write a formula for the function $P(t)\text{.}$ What kind of function is it?

Solution.

$P(t)=12000(1.107)^t\text{;}$ exponential function

(b)

What is the growth factor and the annual percent growth rate of the population?

Solution.

growth factor: $1.107\text{;}$ growth rate: $10.7\%$

10.

A biologist studies the effects of three different nutrients, $a\text{,}$ $b\text{,}$ and $c\text{,}$ on the growth of a particular kind of bacterium. $t$ hours after the experiment began, the number of bacteria in the culture fed nutrient $a$ is $A(t)\text{,}$ the number of bacteria in the culture fed nutrient $b$ is $B(t)\text{,}$ the number of bacteria in the culture fed nutrient $c$ is $C(t)\text{.}$ The biologist observes that the functions $A(t)\text{,}$ $B(t)$ and $C(t)$ are given by the following formulas:

\begin{equation*} A(t)=500(1.09)^t, \quad B(t)=500(1.15)^t, \quad C(t)=500(0.75)^t. \end{equation*}

(a)

Which of the nutrients stimulates growth of the bacteria the most? What is the hourly percent growth rate of the culture fed that nutrient?

Solution.

nutrient b; $15\%$

(b)

One of the nutrients proves toxic to the bacterium. Which one is it? What is the percent growth rate of the culture fed that nutrient?

Solution.

nutrient c; $-25\%$

11.

Following a dose of $40$ mg, a medication leaves a patient’s body at an hourly percent rate of $3.7\%\text{.}$

(a)

Write a formula for the amount $M(t)$ of the medication, in mg, left in the body $t$ hours after the dose. What is the growth factor of $M(t)\text{?}$

Solution.

$M(t)=40(0.963)^t\text{;}$ growth factor: $0.963$

(b)

How much of the medication remains in the body after $10$ hours?

Solution.

approximately $27.44$ mg

12.

Cesium-137, a radioactive isotope of Cesium, decays very slowly. Let $25$ mg of Cesium-137 be present initially and let $C(t)$ be the amount, in mg, remaining after $t$ years. Then:

\begin{equation*} C(t)=25(0.9772)^t \end{equation*}

(a)

What is the growth factor of Cesium-137?

Solution.

$0.9772$

(b)

What is the percent growth rate?

Solution.

$-2.28\%$

(c)

How much Cesium-137 remains after $30$ years?

Solution.

approximately $12.52$ mg

Simplifying Exponential Functions.

For each of the following, decide whether a given function is exponential. If yes, rewrite the function in the form $y=A\cdot b^t\text{.}$ Identify the initial value, the growth factor, and decide if the function is increasing or decreasing. Round off your answers to four decimal places.

13.

$\displaystyle y=30(0.8)^{2t}$

Solution.

$y=30(0.64)^t\text{;}$ initial value: $30\text{;}$ growth factor: $0.64\text{;}$ decreasing

14.

$\displaystyle y=\sqrt{160(1.09)^{2t}}$

Solution.

$y=12.6491(1.09)^t\text{;}$ initial value: $12.6491\text{;}$ growth factor: $1.09\text{;}$ increasing

15.

$\displaystyle y=30\sqrt{t^4}$

Solution.

Not an exponential function.

16.

$\displaystyle y=70(2^{\frac{t}{5}})$

Solution.

$y=70(1.1487)^t\text{;}$ initial value: $70\text{;}$ growth factor: $1.1487\text{;}$ increasing

17.

$\displaystyle y=5(2^{t-1})$

Solution.

$y=2.5(2)^t\text{;}$ initial value: $2.5\text{;}$ growth factor: $2\text{;}$ increasing

18.

$\displaystyle y=60\left(\frac{1}{2}\right)^{\frac{t}{3}}$

Solution.

$y=60(0.7937)^t\text{;}$ initial value: $60\text{;}$ growth factor: $0.7937\text{;}$ decreasing

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