4.3.1: Evaluating Exponential Expressions

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Learning Objectives

By the end of this section, you will be able to:

Graph basic exponential equations
Solve exponential equations
Use exponential models in applications

Be Prepared

Before you get started, take this readiness quiz.

1. Simplify $\dfrac{x^{3}}{x^{2}}$.

2. Evaluate:

a. $2^{0}$

b. $\left(\dfrac{1}{3}\right)^{0}$

3. Evaluate:

a. $2^{−1}$

b. $\left(\dfrac{1}{3}\right)^{-1}$

Basic Exponential Expressions

The expressions we have studied so far do not give us a model for many naturally occurring phenomena. From the growth of populations and the spread of viruses to radioactive decay and compounding interest, the models are very different from what we have studied so far. These models involve exponential expressions.

An exponential expression is an expression of the form $a^{x}$ where $a>0$ and $a≠1$.

Definition $\PageIndex{1}$

An exponential expression, where $a>0$ and $a≠1$, is an expression of the form

$a^{x}$,

or an expression containing expressions of that form.

Notice that in this expression, the variable is the exponent. In our expressions so far, the variables were the base.

This figure shows three expressions: y equals negative 3x plus 4, which is marked as linear; y equals 2x squared plus 5x minus 3, which is marked as quadratic; and y equals 6 to the x power, which is marked exponential. For the expressions marked linear and quadratic, x is the base. For the expressions marked exponential, x is the exponent for the base 6.

Our definition says $a≠1$. If we let $a=1$, then $a^{x}$ becomes $1^{x}$. But we know $1^{x}=1$ for all real numbers.

Our definition also says $a>0$. If we let a base be negative, say $−4$, then $(−4)^{x}$ is not a real number when $x=\dfrac{1}{2}$.

In fact, $(−4)^{x}$ would not be a real number any time $x$ is a fraction with an even denominator. So our definition requires $a>0$.

We will assume that the base in an exponential expression is positive.

We notice $a^{0}=1$ for any $a$.

Also, $a^{1}=a$ and $a^{-1}=\dfrac{1}{a}$ for any $a$.

The expression $a^x$, $a>1$ makes sense for any value of $x$ (we know that it is true for $x$ rational and will accept this as fact for other values of $x$).

It is clear that $a^x>0$ for rational values of $x$ and is in fact true for all values of $x$.

Also, $\left(\dfrac{a}{b}\right)^{-x}=\left(\dfrac{b}{a}\right)^x$, so for example, $\left(2\right)^{-x}=\left(\dfrac{1}{2}\right)^x$.

Note $\PageIndex{2}$

Natural Base $e$

In applications, it happens that different bases are convenient for writing relevant expressions. There is one number which is written "$e$" . We will not delve into the details of this number except to say that this number is irrational and

$e \approx 2.718281827.$

Most calculators will include this particular base on a button often labeled $e^x$.

Notice that for each $x$, $2^{x}<e^{x}<3^{x}$ since $2<e<3$. So you may get a rough idea of the values for particular values of $x$ without using a calculator!

Note

One-to-One Property of Exponential Equations

For $a>0$ and $a≠1$,

if $a^{x}=a^{y}$, then $x=y$.

Using Exponential Models in Applications

Exponential equations model many situations. If you own a bank account, you have experienced the use of an exponential equation. There are two formulas that are used to determine the balance in the account when interest is earned. If a principal, $P$, is invested at an interest rate, $r$, for $t$ years, the new balance, $A$, will depend on how often the interest is compounded, i.e., how often the interest is calculated and then added to the new balance. If the interest is compounded $n$ times a year we use the formula $A=P\left(1+\dfrac{r}{n}\right)^{n t}$. If the interest is compounded continuously, we use the formula $A=Pe^{rt}$. These are the formulas for compound interest.

Definitions and Formulae $\PageIndex{3}$

Compound Interest is interest that accumulates on the interest earned.

The principal is the amount invested.

Interest is compounded when the interest is calculated and then added to the new balance.

For a principal, $P$, invested at an interest rate, $r$, for $t$ years, the new balance is:

$\begin{array}{ll}{P\left(1+\dfrac{r}{n}\right)^{n t}} & {\text { when compounded } n \text { times a year. }} \\ {P e^{r t}} & {\text { when continuously. }}\end{array}$

To see why the first formula works, work out some examples by writing down how the compounding works in detail. As you work with the Interest formulas, it is often helpful to identify the values of the variables first and then substitute them into the formula.

Example $\PageIndex{4}$

A total of $$10,000$ was invested in a college fund for a new grandchild. If the interest rate is $5$%, how much will be in the account in $18$ years by each method of compounding?

a. compound quarterly

b. compound monthly

c. compound continuously

Solution:


Identify the values of each variable in the formulas. Remember to express the percent as a decimal.	$\begin{aligned} P &=\$ 10,000 \\ r &=0.05 \\ t &=18 \text { years } \end{aligned}$
For quarterly compounding, $n=4$. There are $4$ quarters in a year.	$P\left(1+\dfrac{r}{n}\right)^{n t}$
Substitute the values in the formula.	$10,000\left(1+\dfrac{0.05}{4}\right)^{4 \cdot 18}$
Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator.	The amount in the account will be $\$ 24,459.20$.


For monthly compounding, $n=12$.There are $12$ months in a year.	$P\left(1+\dfrac{r}{n}\right)^{n t}$
Substitute the values in the formula.	$10,000\left(1+\dfrac{0.05}{12}\right)^{12 \cdot 18}$
Compute the amount.	The amount in the account will be $\$ 24,550.08$.


For compounding continuously,	$P e^{r t}$
Substitute the values in the formula.	$10,000 e^{0.05 \cdot 18}$
Compute the amount.	The amount in the account will be \(\$ 24,596.03\

Try It $\PageIndex{5}$

Angela invested $$15,000$ in a savings account. If the interest rate is $4$%, how much will be in the account in $10$ years by each method of compounding?

a. compound quarterly

b. compound monthly

c. compound continuously

Answer

a. $$22,332.96$

b. $$22,362.49$

c. $$22,377.37$

Try It $\PageIndex{6}$

Allan invested $$10,000$ in a mutual fund. If the interest rate is $5$%, how much will be in the account in $15$ years by each method of compounding?

a. compound quarterly

b. compound monthly

c. compound continuously

Answer

a. $$21,071.81$

b. $$21,137.04$

c. $$21,170.00$

Other topics that are modeled by exponential equations involve growth and decay. Both also use the expression $Pe^{rt}$ we used for the growth of money. For growth and decay, generally we use $A_{0}$ (Amount at time 0), as the original amount instead of calling it $P$, the principal. We see that exponential growth has a positive rate $r$ of growth and exponential decay has a negative rate $r$ of growth.

Definition $\PageIndex{7}$

Exponential Growth and Decay

For an original amount, $A_{0}$, that grows or decays at a rate, $r$, for a certain time, $t$, the final amount is:

$A_{0} e^{r t}$

Exponential growth is typically seen in the growth of populations of humans or animals or bacteria. Our next example looks at the growth of a virus.

Example $\PageIndex{8}$

Chris is a researcher at the Center for Disease Control and Prevention and he is trying to understand the behavior of a new and dangerous virus. He starts his experiment with $100$ of the virus that grows at a rate of $25$% per hour. He will check on the virus in $24$ hours. How many viruses will he find?

Solution:


Identify the values of each variable in the formulas. Be sure to put the percent in decimal form. Be sure the units match--the rate is per hour and the time is in hours.	$\begin{aligned}A_{0} &=100 \\ r &=0.25 / \text { hour } \\ t &=24 \text { hours } \end{aligned}$
Substitute the values in the expression: $A_{0} e^{r t}$.	$100 e^{0.25 \cdot 24}$
Compute the amount.	$40,342.88$
Round to the nearest whole virus.	$40,343$
Conclude.	The researcher will find $40,343$ viruses.

Try It $\PageIndex{9}$

Another researcher at the Center for Disease Control and Prevention, Lisa, is studying the growth of a bacteria. She starts his experiment with $50$ of the bacteria that grows at a rate of $15\%$ per hour. He will check on the bacteria every $8$ hours. How many bacteria will he find in $8$ hours?

Answer: She will find $166$ bacteria.

Try It $\PageIndex{10}$

Maria, a biologist is observing the growth pattern of a virus. She starts with $100$ of the virus that grows at a rate of $10\%$ per hour. She will check on the virus in $24$ hours. How many viruses will she find?

Answer: She will find $1,102$ viruses.

Writing Exercises $\PageIndex{11}$

Why are we dividing by $n$ in the interest formula?
Why is $1$ being added to $r/n$ in the formula?

Exit Question

How much do you have to invest today at $2\%$ compounded monthly to obtain $\$50,000$ in return in $3$ years? What if it is compounded continuously? Which one is better?

Key Concepts

Compound Interest: For a principal, $P$, invested at an interest rate, $r$, for $t$ years, the new balance is
$\begin{array}{ll}{P\left(1+\dfrac{r}{n}\right)^{n t}} & {\text { when compounded } n \text { times a year. }} \\ {P e^{r t}} & {\text { when compounded continuously. }}\end{array}$
Exponential Growth and Decay: For an original amount, $A_{0}$ that grows or decays at a rate, $r$, for a certain time $t$, the final amount is $A_{0}e^{rt}$.

Glossary

exponential expression: An exponential expression, where $a>0$ and $a≠1$, is an expression involving an expression of the form $a^{x}$.

natural base: The number $e$ is an irrational number that appears in many applications. $e≈2.718281827...$


Identify the values of each variable in the formulas. Remember to express the percent as a decimal.	\(\begin{aligned} P &=\$ 10,000 \\ r &=0.05 \\ t &=18 \text { years } \end{aligned}\)
For quarterly compounding, \(n=4\). There are \(4\) quarters in a year.	\(P\left(1+\dfrac{r}{n}\right)^{n t}\)
Substitute the values in the formula.	\(10,000\left(1+\dfrac{0.05}{4}\right)^{4 \cdot 18}\)
Compute the amount. Be careful to consider the order of operations as you enter the expression into your calculator.	The amount in the account will be \(\$ 24,459.20\).


For monthly compounding, \(n=12\).There are \(12\) months in a year.	\(P\left(1+\dfrac{r}{n}\right)^{n t}\)
Substitute the values in the formula.	\(10,000\left(1+\dfrac{0.05}{12}\right)^{12 \cdot 18}\)
Compute the amount.	The amount in the account will be \(\$ 24,550.08\).


Identify the values of each variable in the formulas. Be sure to put the percent in decimal form. Be sure the units match--the rate is per hour and the time is in hours.	\(\begin{aligned}A_{0} &=100 \\ r &=0.25 / \text { hour } \\ t &=24 \text { hours } \end{aligned}\)
Substitute the values in the expression: \(A_{0} e^{r t}\).	\(100 e^{0.25 \cdot 24}\)
Compute the amount.	\(40,342.88\)
Round to the nearest whole virus.	\(40,343\)
Conclude.	The researcher will find \(40,343\) viruses.

Search

Definition \(\PageIndex{1}\)

Note \(\PageIndex{2}\)

Note

Definitions and Formulae \(\PageIndex{3}\)

Example \(\PageIndex{4}\)

Try It \(\PageIndex{5}\)

Try It \(\PageIndex{6}\)

Definition \(\PageIndex{7}\)

Example \(\PageIndex{8}\)

Try It \(\PageIndex{9}\)

Try It \(\PageIndex{10}\)

Writing Exercises \(\PageIndex{11}\)

Exit Question


For compounding continuously,	\(P e^{r t}\)
Substitute the values in the formula.	\(10,000 e^{0.05 \cdot 18}\)
Compute the amount.	The amount in the account will be \(\$ 24,596.03\