5.3.2: Properties of Exponents with Zero and Negative Exponents

Learning Outcomes

Explain the meaning of a zero exponent
Explain the meaning of a negative exponent
Apply the power of product rule for exponents
Apply the power of quotient rule for exponents
Simplify exponential expressions
Use properties of exponents to write equivalent exponential functions in standard form

Zero Exponents

The quotient rule for exponents can be used to determine the meaning of $x^0$ :

The quotient rule for exponents tells us that by subtracting the exponents:

$\dfrac{x^n}{x^n}=x^0$

But since the numerator and denominator are identical, we can cancel the terms by division:

$\dfrac{x^n}{x^n}=1$ provided $x\neq0$ since we can’t divide by 0.

Therefore,

$x^0=1$ , $x\neq0$

exponent of zero

For all real numbers $a\neq0$ ,

$a^0=1$

For example,

$\begin{aligned}2022^0&=1\\(pq)^0&=1,\;p,\;q\neq 0 \\(2050xy)^{0}&=1,\;x,\;y\neq 0\end{aligned}$

The sole exception is the expression

0^{0}

${0}^{0}$ . This appears later in more advanced courses, but for now, we will consider the value to be undefined.

TIP FOR SIMPLIFYING EXPONENTIAL EXPRESSIONS

When simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression

\frac{5 a^{m} z^{2}}{a^{m} z}

$\dfrac{5 a^m z^2}{a^mz}$ using exponent rules, it may be helpful to break the fraction up into a product of fractions, then simplify:

$\begin{aligned}\dfrac{5 a^m z^2}{a^mz}&=5\cdot\dfrac{a^m}{a^m}\cdot\dfrac{z^2}{z}&&\text{Separate into fractions}\\&=5 \cdot a^{m-m}\cdot z^{2-1}&&\text{Subtract the exponents}\\&=5\cdot a^0 \cdot z^1 &&\text{Simplify }a^0=1\text{ and }z^1=z\\&=5z\end{aligned}$

Example 1

Simplify each expression.

$\dfrac{{c}^{3}}{{c}^{3}}$
$\dfrac{-3{x}^{5}}{{x}^{5}}$
$\dfrac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}$
$\dfrac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}$

Solution

We can apply the zero exponent rule and other rules to simplify each expression:

$\begin{align}\frac{c^{3}}{c^{3}}&=c^{3-3}&&\text{Apply the quotient rule: subtract exponents} \\ & =c^{0}&&\text{Apply the zero exponent rule} \\ & =1\end{align}$

$\begin{align} \frac{-3{x}^{5}}{{x}^{5}}& = -3\cdot \frac{{x}^{5}}{{x}^{5}} \\ & = -3\cdot {x}^{5 - 5}&&\text{Apply the quotient rule: subtract exponents} \\ & = -3\cdot {x}^{0}&&\text{Apply the zero exponent rule}\\ & = -3\cdot 1 \\ & = -3 \end{align}$

$\begin{align} \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}} && \text{Use the product rule in the denominator. The base is }(j^2k). \\ & = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}} && \text{Use the quotient rule}. \\ & = {\left({j}^{2}k\right)}^{4 - 4} \\ & = {\left({j}^{2}k\right)}^{0} && \text{Use the zero exponent rule}. \\ & = 1 \end{align}$

$\begin{align} \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& = 5{\left(r{s}^{2}\right)}^{2 - 2}&&\text{Use the quotient rule}. \\ & = 5{\left(r{s}^{2}\right)}^{0}&&\text{Use the zero exponent rule}. \\ & = 5\cdot 1 \\ & = 5 \end{align}$

Try It 1

Simplify each expression using the zero exponent rule of exponents.

$\dfrac{{t}^{7}}{{t}^{7}}$
$\dfrac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}$
$\dfrac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}$
$\dfrac{-5{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}$

Show Answer

$1$
$\dfrac{1}{2}$
$1$
$-5$

Negative Exponents

The quotient rule for exponents can also be used to determine what it means to have a negative exponent $x^{-n}$ . If [latex]m $\begin{align}\frac{x^{2}}{x^{4}} &=\frac{x\cdot x}{x\cdot x\cdot x\cdot x} \\[1mm] &=\frac{\cancel{x}\cdot\cancel{x}}{\cancel{x}\cdot\cancel{x}\cdot x\cdot x} \\[1mm] & =\frac{1}{x^2}\end{align}$

Consequently, $x^{-2}=\dfrac{1}{x^2}$ .

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator. This can be generalized to $a^{-n}=\dfrac{1}{a^n}$ .

If the negative exponent is on the denominator, $\dfrac{1}{x^{-n}}$ , we can use division of fractions to simplify it:

$\begin{aligned}\dfrac{1}{x^{-n}}&=1\div x^{-n}\\&=1\div \dfrac{1}{x^n}\\&=1\times \dfrac{x^n}{1}\\&=x^n\end{aligned}$

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from denominator to numerator.

NEGATIVE EXPONENTS

For any real numbers $a\neq0$ and $n$ ,

$a^{-n}=\dfrac{1}{a^n}$

and

$\dfrac{1}{{a}^{-n}}=a^n$

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar.

Example 2

Simplify the expressions. Write answers with positive exponents.

$\dfrac{x^3}{x^{10}}$
$\dfrac{z^2\cdot z}{z^4}$
$\dfrac{{\left(-5t^3\right)}^4}{\left(-5t^3\right)^8}$

Solution

$1.\\ \begin{aligned}\dfrac{x^3}{x^{10}}&=x^{3 - 10}&&\text{Quotient rule}\\&=x^{-7}&&\text{Negative exponent rule}\\&=\dfrac{1}{x^7}\end{aligned}$	$2.\\ \begin{aligned}\dfrac{z^2\cdot z}{z^4}&=\dfrac{z^{2+1}}{z^4}&&\text{Product rule}\\&=\dfrac{z^3}{z^4}&&\text{Quotient rule}\\&={z}^{3 - 4}\\&={z}^{-1}&&\text{Negative exponent rule}\\&=\dfrac{1}{z}\end{aligned}$
$3.\\ \begin{aligned}\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}&={\left(-5{t}^{3}\right)}^{4 - 8}&&\text{Product rule: the base is }-5t^3\\&={\left(-5{t}^{3}\right)}^{-4}&&\text{Negative exponent rule}\\&=\dfrac{1}{{\left(-5{t}^{3}\right)}^{4}}\end{aligned}$

Try It 2

Simplify the expressions. Write answers with positive exponents.

$\dfrac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}$
$\dfrac{{f}^{47}}{{f}^{49}\cdot f}$
$\dfrac{2{k}^{4}}{5{k}^{7}}$
$\dfrac{\left(-3x^4\right)^5}{\left(-3x^4\right)^{12}}$
$\dfrac{5y^{-8}}{y^{-6}}$

Show Answer

$\dfrac{1}{{\left(-3t\right)}^{6}}$
$\dfrac{1}{{f}^{3}}$
$\dfrac{2}{5{k}^{3}}$
$\dfrac{1}{\left(-3x^4\right)^7}$
$\dfrac{5}{y^2}$

Example 3

Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.

${b}^{2}\cdot {b}^{-8}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}$
$\dfrac{-7z}{{\left(-7z\right)}^{5}}$

Solution

$\begin{aligned}{b}^{2}\cdot {b}^{-8}&={b}^{2 + (-8)}&&\text{Product rule}\\&={b}^{-6}\\&=\frac{1}{{b}^{6}}&&\text{Negative exponent rule}\end{aligned}$

$\begin{aligned}{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}&={\left(-x\right)}^{5+(-5)}&&\text{Product rule}\\&={\left(-x\right)}^{0}\\&=1&&\text{Zero exponent rule}\end{aligned}$

$\begin{aligned}\dfrac{-7z}{{\left(-7z\right)}^{5}}&=\dfrac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}\\&={\left(-7z\right)}^{1 - 5}&&\text{Quotient rule}\\&={\left(-7z\right)}^{-4}\\&=\dfrac{1}{{\left(-7z\right)}^{4}}&&\text{Negative exponent rule}\end{aligned}$

Try It 3

Simplify. Write answers with positive exponents.

${t}^{-11}\cdot {t}^{6}$
$\dfrac{{25}^{12}}{{25}^{13}}$

Show Answer

${t}^{-5}=\dfrac{1}{{t}^{5}}$
$\dfrac{1}{25}$

The Power of a Product Rule

To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For example, consider ${\left(pq\right)}^{3}$ . We begin by using the associative and commutative properties of multiplication to regroup the factors:

\begin{matrix} {(p q)}^{3} & = 3 factors (p q) \cdot (p q) \cdot (p q) = p \cdot q \cdot p \cdot q \cdot p \cdot q = 3 factors p \cdot p \cdot p \cdot 3 factors q \cdot q \cdot q = p^{3} \cdot q^{3} \end{matrix}

$\begin{align} {\left(pq\right)}^{3}& = \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}} \\ & = p\cdot q\cdot p\cdot q\cdot p\cdot q \\& = \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}} \\ & = {p}^{3}\cdot {q}^{3} \end{align}$

In other words, ${\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}$ .

The Power of a Product Rule

For any real numbers $a,\;b$ and $n$ , the power of a product rule of exponents states that

{(a b)}^{n} = a^{n} b^{n}

$\large{\left(ab\right)}^{n}={a}^{n}{b}^{n}$

Example 4

Simplify each of the following products as much as possible. Write answers with positive exponents.

${\left(a{b}^{2}\right)}^{3}$
${\left(2t\right)}^{15}$
${\left(-2{w}^{3}\right)}^{3}$
$\dfrac{1}{{\left(-7z\right)}^{4}}$
${\left({e}^{-2}{f}^{2}\right)}^{7}$

Solution

We can use the product and quotient rules and the new definitions to simplify each expression. If a number is raised to a power, we can evaluate it.

$\begin{aligned}{\left(a{b}^{2}\right)}^{3}&={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}&&\text{Power of a product rule}\\&={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}\\&={a}^{3}{b}^{6}\end{aligned}$

$\begin{aligned}(2{t})^{15}&={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}&&\text{Power of a product rule}\\&={2}^{15}{t}^{15}&&\text{Evaluate }2^{15}\text{ using a calculator}\\&=32,768{t}^{15}\end{aligned}$

$\begin{aligned}{\left(-2{w}^{3}\right)}^{3}&={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}&&\text{Power of a product rule}\\&=-8\cdot {w}^{3\cdot 3}&&{(-2)}^3=-8\\&=-8{w}^{9}\end{aligned}$

$\begin{aligned}\dfrac{1}{{\left(-7z\right)}^{4}}&=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}&&\text{Power of a product rule}\\&=\dfrac{1}{2,401{z}^{4}}&&{(-7)}^4\text{ is evaluated}\end{aligned}$

$\begin{aligned}{\left({e}^{-2}{f}^{2}\right)}^{7}&={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}&&\text{Power of a product rule}\\&={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}\\&={e}^{-14}{f}^{14}&&\text{Negative exponent rule}\\&=\dfrac{{f}^{14}}{{e}^{14}}\end{aligned}$

Try It 4

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

${\left({g}^{2}{h}^{3}\right)}^{5}$
${\left(5t\right)}^{3}$
${\left(-3{y}^{5}\right)}^{3}$
$\dfrac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}$
${\left({r}^{3}{s}^{-2}\right)}^{4}$

Show Answer

${g}^{10}{h}^{15}$
$125{t}^{3}$
$-27{y}^{15}$
$\dfrac{1}{{a}^{18}{b}^{21}}$
$\dfrac{{r}^{12}}{{s}^{8}}$

The Power of a Quotient Rule

To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors.The power of a quotient rule is an extension of the power of a product rule since a quotient can be written as a product:

$\begin{aligned}\left(\dfrac{a}{b}\right)^n&=\left(a\times\dfrac{1}{b}\right)^n\\&=a^n\times \left(b^{-1}\right)^n\\&=a^n\times b^{-n}\\&=a^n\times \dfrac{1}{b^n}\\&=\dfrac{a^n}{b^n}\end{aligned}$ .

The Power of a Quotient Rule

For any real numbers $a,\;b$ and $n$ , provided $b\neq0$ , the power of a quotient rule of exponents states that

$\large{\left(\dfrac{a}{b}\right)}^{n}=\dfrac{{a}^{n}}{{b}^{n}}$

Example 5

Simplify each of the following quotients as much as possible. Write answers with positive exponents.

${\left(\dfrac{4}{{z}^{11}}\right)}^{3}$
${\left(\dfrac{p}{{q}^{3}}\right)}^{6}$
${\left(\dfrac{-1}{{t}^{2}}\right)}^{27}$
${\left({j}^{3}{k}^{-2}\right)}^{4}$
${\left({m}^{-2}{n}^{-2}\right)}^{3}$

Solution

$\begin{aligned}{\left(\dfrac{4}{{z}^{11}}\right)}^{3}&=\dfrac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}&&\text{Quotient to a power rule}\\&=\dfrac{64}{{z}^{11\cdot 3}}&&\text{Evaluate }4^3=64\text{. Power to a power rule.} \\&=\dfrac{64}{{z}^{33}}\end{aligned}$

$\begin{aligned}{\left(\dfrac{p}{{q}^{3}}\right)}^{6}&=\dfrac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}&&\text{Power of a quotient rule}\\&=\dfrac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}&&\text{Power to a power rule}\\&=\dfrac{{p}^{6}}{{q}^{18}}\end{aligned}$

$\begin{aligned}{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}&=\dfrac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}&&\text{Power of a quotient rule}\\&=\dfrac{-1}{{t}^{2\cdot 27}}&&\text{Power of a power rule}\\&=\dfrac{-1}{{t}^{54}}&&\text{Put the negative sign in front of the fraction}\\&=-\dfrac{1}{{t}^{54}}\end{aligned}$

$\begin{aligned}{\left({j}^{3}{k}^{-2}\right)}^{4}&={\left(\dfrac{{j}^{3}}{{k}^{2}}\right)}^{4}&&\text{Negative exponent rule}\\&=\dfrac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}&&\text{Power of a quotient rule}\\&=\dfrac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}&&\text{Power to a power rule}\\&=\dfrac{{j}^{12}}{{k}^{8}}\end{aligned}$

$\begin{aligned}{\left({m}^{-2}{n}^{-2}\right)}^{3}&={\left(\dfrac{1}{{m}^{2}{n}^{2}}\right)}^{3}&&\text{Negative exponent rule}\\&=\dfrac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}&&\text{Power of a quotient rule}\\&=\dfrac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}&&\text{Evaluate }1^3=1\\&=\dfrac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}&&\text{Power to a power rule}\\&=\dfrac{1}{{m}^{6}{n}^{6}}\end{aligned}$

Try It 5

Simplify each of the following quotients as much as possible. Write answers with positive exponents.

${\left(\dfrac{{b}^{5}}{c}\right)}^{3}$
${\left(\dfrac{5}{{u}^{8}}\right)}^{4}$
${\left(\dfrac{-1}{{w}^{3}}\right)}^{35}$
${\left({p}^{-4}{q}^{3}\right)}^{8}$
${\left({c}^{-5}{d}^{-3}\right)}^{4}$

Show Answer

$\dfrac{{b}^{15}}{{c}^{3}}$
$\dfrac{625}{{u}^{32}}$
$\dfrac{-1}{{w}^{105}}$
$\dfrac{{q}^{24}}{{p}^{32}}$
$\dfrac{1}{{c}^{20}{d}^{12}}$

Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combining terms or exponents. Evaluating an expression means to get a numerical answer. The rules for exponents can be combined to simplify expressions.

Example 6

Simplify each expression and write the answer with positive exponents only.

${\left(6{m}^{2}{n}^{-1}\right)}^{3}$
${17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}$
${\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}$
$\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)$
${\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}$
$\dfrac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}$

Solution

$\begin{align} {\left(6{m}^{2}{n}^{-1}\right)}^{3}& = {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}&& \text{Power of a product rule} \\ & = {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}&& \text{Power rule} \\ & = 216{m}^{6}{n}^{-3}&&\text{Evaluate: }2^6=216\text{ and simplify}. \\ & = \frac{216{m}^{6}}{{n}^{3}}&& \text{Negative exponent rule} \end{align}$

$\begin{align} {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}&=17^{5 +(-4)+(-3)}&& \text{Product rule} \\ & = {17}^{-2}&&\text{Negative exponent rule}\\&=\dfrac{1}{{17}^{2}}&&\text{Evaluate}\\&=\dfrac{1}{289}\end{align}$

$\begin{align} {\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}&=\dfrac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}&&\text{Power of a quotient rule}\\&=\dfrac{{u}^{-2}{v}^{2}}{{v}^{-2}}&&\text{Power of a product rule}\\&={u}^{-2}{v}^{2-(-2)}&&\text{Quotient rule}\\&={u}^{-2}{v}^{4}&&\text{Evaluate: }2-(-2)=2+2=4\\&=\dfrac{{v}^{4}}{{u}^{2}}&&\text{Negative exponent rule}\end{align}$

$\begin{align}\left(-2a^3b^{-1}\right)\left(5a^{-2}b^2 \right)&=\left(-2\cdot 5\right)\left({a}^{3}\cdot {a}^{-2}\right)\left( {b}^{-1}\cdot {b}^{2}\right)&& \text{Commutative/associative properties} \\ & = -10{a}^{3+(-2)}{b}^{-1+2}&& \text{Product rule} \\ & = -10a^1b^1\\&=-10ab \end{align}$

$\begin{align} {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& = {\left({x}^{2}\sqrt{2}\right)}^{4+(-4)} && \text{Product rule: base is }(x^2\sqrt{2}) \\ & = {\left({x}^{2}\sqrt{2}\right)}^{0}&& \text{Zero exponent rule}\\ & = 1 \end{align}$

$\begin{align} \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& = \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}&& \text{Power of a product rule} \\ & = \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}&& \text{Power rule} \\ & = \frac{243{w}^{10}}{36{w}^{-4}} && \text{Evaluate: }3^5=243\text{ and }6^2=36 \\ & = \frac{27{w}^{10-\left(-4\right)}}{4}&& \text{Quotient rule and simplify fraction} \\ & = \frac{27{w}^{14}}{4}\end{align}$

Try It 6

Simplify each expression and write the answer with positive exponents only.

${\left(2u{v}^{-2}\right)}^{-3}$
${x}^{8}\cdot {x}^{-12}\cdot x$
${\left(\dfrac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}$
$\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)$
${\left(\dfrac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\dfrac{4}{9}t{w}^{-2}\right)}^{3}$
$\dfrac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}$

Show Answer

$\dfrac{{v}^{6}}{8{u}^{3}}$
$\dfrac{1}{{x}^{3}}$
$\dfrac{{e}^{4}}{{f}^{4}}$
$\dfrac{27r}{s}$
$1$
$\dfrac{16{h}^{10}}{49}$

TRY IT 7

Try It 8

Try It 9

Try It 10

Simplifying Exponential Functions

Each of the properties of exponents can be used to simplify and write equivalent exponential functions. For example, the function $f(x)=2^{x+3}$ can be written as the equivalent exponential function $f(x)=8\left(2^x\right)$ :

$\begin{aligned}f(x)&=2^{x+3}\\&=2^x\cdot 2^3\\&=2^x\cdot 8\\&=8\left(2^x\right)\end{aligned}$

Being able to simplify an exponential function into the standard form $f(x)=ar^{x-h}+k$ makes the function easier to graph and allows us to determine the transformations that were made from the parent function $f(x)=r^x$ . In the case of $f(x)=2^{x+3}=8\left(2^x\right)$ , the 8 tells us that the graph of $f(x)=2^x$ has been stretched by a factor of 8.

Example 7

Use exponential rules to write equivalent exponential functions.

$f(x)=3^{x+2}$
$g(x)=5^{2x+1}$
$h(x)=4^{3x-2}$

Solution

$\begin{aligned}f(x)&=3^{x+2}\\&=3^x\cdot 3^2&&\text{Product rule (in reverse)}\\&=3^x\cdot 9&&\text{Evaluate }3^2=9\\&=9\left(3^x\right)&&\text{Write in standard form}\end{aligned}$

$\begin{aligned}g(x)&=5^{2x+1}\\&=5^{2x}\cdot 5^1&&\text{Product rule (in reverse)}\\&=\left(5^2\right)^x\cdot 5&&\text{Evaluate }5^1=5\\&=5\left(25^x\right)&&\text{Write in standard form}\end{aligned}$

$\begin{aligned}h(x)&=4^{3x-2}\\&=4^{3x}\cdot 4^{-2}&&\text{Product rule (in reverse)}\\&=\left(4^3\right)^x\cdot \dfrac{1}{4^2}&&\text{Power to a power rule (in reverse) and negative exponent rule}\\&=64^x\cdot \dfrac{1}{16}&&\text{Evaluate: }4^3=64\text{ and }4^2=16\\&=\dfrac{1}{16}\left(64^x\right)&&\text{Write in standard form}\end{aligned}$

Try It 11

Use exponential rules to write equivalent exponential functions.

$f(x)=2^{x+4}$
$g(x)=3^{x-2}$
$h(x)=5^{2x-1}$

Show Answer

$f(x)=16\left(2^x\right)$
$g(x)=\dfrac{1}{9}\left(3^x\right)$
$h(x)=\dfrac{1}{5}\left(25^{x}\right)$

Chapter 5: Exponential Functions