Evaluating Exponential Functions

Learning Outcomes

Identify and evaluate exponential functions

A function of the form $f(x)=b^{x}, b>0, b \neq 1$ is called an exponential function. In an exponential function the base is a positive constant other than $1$ , and the exponent is a variable expression. For example, if $f(x)=3^{x}$ , then $f(2)=3^{2}=9, f(0)=3^{0}=1$ and $f(-2)=3^{-2}=\frac{1}{3^{2}}=\frac{1}{9}$ .

Exponential Growth function

A function that models exponential growth grows by a rate proportional to the current amount. For any real number x and any positive real numbers a and b such that $b\ne 1$ , an exponential growth function has the form

f (x) = a b^{x}

$f\left(x\right)=a{b}^{x}$

where

a is the initial or starting value of the function.
b is the growth factor or growth multiplier per unit x

To evaluate an exponential function of the form $f\left(x\right)=a \cdot {b}^{x}$ , we simply substitute x with the given value, and calculate the resulting power. For example:

Let $f(x)=7 \cdot 2^{x}$

Then $2$ is the base of the exponential function. Since $f(x)=7 \cdot 2^{0} = 7 \cdot 1= 7$ , the constant $a=7$ in this example is the initial value of the function when the variable $x$ is $0$ .

If $x$ is $1$ , then
$f(1)=7 \cdot 2^{1}$
$= 7 \cdot 2$
$= 14$ .

If $x$ is $3$ , then
$f(3)=7 \cdot 2^{3}$
$= 7 \cdot 8$
$= 56$ .

To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. Consider the following example.

Example

Let $f\left(x\right)=5{\left(3\right)}^{x+1}$ . Evaluate $f\left(2\right)$ without using a calculator.

Show Solution

Follow the order of operations. Be sure to pay attention to the parentheses.

\begin{matrix} f (x) & = 5 {(3)}^{x + 1} \\ f (2) & = 5 {(3)}^{2 + 1} & Substitute x = 2. \\ = 5 {(3)}^{3} & Add the exponents . \\ = 5 (27) & Simplify the power . \\ = 135 & Multiply . \end{matrix}

$\begin{array}{c}f\left(x\right)\hfill & =5{\left(3\right)}^{x+1}\hfill & \hfill \\ f\left(2\right)\hfill & =5{\left(3\right)}^{2+1}\hfill & \text{Substitute }x=2.\hfill \\ \hfill & =5{\left(3\right)}^{3}\hfill & \text{Add the exponents}.\hfill \\ \hfill & =5\left(27\right)\hfill & \text{Simplify the power}\text{.}\hfill \\ \hfill & =135\hfill & \text{Multiply}\text{.}\hfill \end{array}$

In the following video, we present more examples of evaluating an exponential function at several different values.

A General Note: The Number $e$

The letter e represents the irrational number

{(1 + \frac{1}{n})}^{n}

${\left(1+\frac{1}{n}\right)}^{n}$

as n increases without bound

The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, $e\approx 2.718282$ . The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.

The Continuous Growth/Decay Formula

For all real numbers r, t, and all positive numbers a, continuous growth or decay is represented by the formula

A (t) = a e^{r t}

$A\left(t\right)=a{e}^{rt}$

where

a is the initial value,
r is the continuous growth or decay rate per unit time,
and t is the elapsed time.

If r > $0$ , then the formula represents continuous growth. If r < $0$ , then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

A (t) = P e^{r t}

$A\left(t\right)=P{e}^{rt}$

where

P is the principal or the initial amount invested,
r is the growth or interest rate per unit time,
and t is the period or term of the investment.

Example

The population of a town, in thousands, $t$ years after $2013$ is modeled by the function $A(t)=13.2 \cdot e^{0.07t}$ . Find the population of the town in the year…

2013
2020

Show Answer

The year $2013$ is the starting year. Since $t$ represents the number of years after $2013$ , in this case $t = 2013 - 2013 = 0$ . The population in $2013$ is found by evaluating $A (0)$ .
- $A(0) = 13.2 \cdot e^{0.07-0}$
  $= 13.2 \cdot e^{0}$
  $= 13.2 \cdot 1$
  $= 13.2$
- Population is given in thousands, so the population in $2013$ was $13,200$ .
The year $2020$ corresponds to $t = 2020 - 2013 = 7$ . The population in $2020$ is found by evaluating $A (7)$ .
- $A(7)=13.2 \cdot e^{0.07 \cdot 7}$
  $\approx 13.2 \cdot 2.718282^{0.49}$
  $13.2 \cdot 1.63231627$
  $=21.54657477$
- Since the population is given in thousands, the population in $2020$ was approximately $21,547$ . Since the population is a whole number, we round the solution to the nearest unit.

Note that constants $\pi$ and $e$ are built-in keys on many calculators. They are carried to greater accuracy, and so your answer using the $e$ button on a calculator will be slightly different. In this example, such a calculator will produce the result $A(7)=21.5465741$ .

In the following video, we show an example of calculating the remaining amount of a radioactive substance after it decays for a length of time.

Module 5: Continuous Random Variables