CR.22: Laws of Exponents (Negative Exponents)

Learning Outcomes

Simplify expressions with negative exponents.
Simplify exponential expressions.

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that $m>n$ in the quotient rule even further. For example, can we simplify $\dfrac{{h}^{3}}{{h}^{5}}$ ? When [latex]mnegative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, $\dfrac{{h}^{3}}{{h}^{5}}$ .

\begin{aligned} \frac{h^{3}}{h^{5}} & = \frac{h \cdot h \cdot h}{h \cdot h \cdot h \cdot h \cdot h} \\ = \frac{h \cdot h \cdot h}{h \cdot h \cdot h \cdot h \cdot h} \\ = \frac{1}{h \cdot h} \\ = \frac{1}{h^{2}} \end{aligned}

$\begin{align} \frac{{h}^{3}}{{h}^{5}}& = \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h} \\ & = \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h} \\ & = \frac{1}{h\cdot h} \\ & = \frac{1}{{h}^{2}} \end{align}$

If we were to simplify the original expression using the quotient rule, we would have

\begin{aligned} \frac{h^{3}}{h^{5}} & = h^{3 - 5} \\ = h^{- 2} \end{aligned}

$\begin{align} \frac{{h}^{3}}{{h}^{5}}& = {h}^{3 - 5} \\ & = {h}^{-2} \end{align}$

Putting the answers together, we have ${h}^{-2}=\dfrac{1}{{h}^{2}}$ . This is true for any nonzero real number, or any variable representing a nonzero real number.

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 or vice versa.

a^{- n} = \frac{1}{a^{n}} and a^{n} = \frac{1}{a^{- n}}

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

We have shown that the exponential expression ${a}^{n}$ is defined when $n$ is a natural number, 0, or the negative of a natural number. That means that ${a}^{n}$ is defined for any integer $n$ . Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n$ .

A General Note: The Negative Rule of Exponents

For any nonzero real number $a$ and natural number $n$ , the negative rule of exponents states that

a^{- n} = \frac{1}{a^{n}} and a^{n} = \frac{1}{a^{- n}}

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

Example: Using the Negative Exponent Rule

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

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

Show Solution

$\dfrac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\dfrac{1}{{\theta }^{7}}$
$\dfrac{{z}^{2}\cdot z}{{z}^{4}}=\dfrac{{z}^{2+1}}{{z}^{4}}=\dfrac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\dfrac{1}{z}$
$\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\dfrac{1}{{\left(-5{t}^{3}\right)}^{4}}$

Try It

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

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

Show Solution

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

Watch this video to see more examples of simplifying expressions with negative exponents.

Example: Using the Product and Quotient Rules

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

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

Show Solution

${b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}$
${\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1$
$\dfrac{-7z}{{\left(-7z\right)}^{5}}=\dfrac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\dfrac{1}{{\left(-7z\right)}^{4}}$

Try It

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

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

Show Solution

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

Finding the Power of a Product

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 instance, consider ${\left(pq\right)}^{3}$ . We begin by using the associative and commutative properties of multiplication to regroup the factors.

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

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

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

A General Note: The Power of a Product Rule of Exponents

For any real numbers $a$ and $b$ and any integer $n$ , 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: Using the Power of a Product Rule

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

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

Show Solution

Use the product and quotient rules and the new definitions to simplify each expression.

${\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}$
$2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}$
${\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}$
$\dfrac{1}{{\left(-7z\right)}^{4}}=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\dfrac{1}{2,401{z}^{4}}$
${\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\dfrac{{f}^{14}}{{e}^{14}}$

Try It

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 Solution

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

In the following video we show more examples of how to find hte power of a product.

Finding the Power of a Quotient

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. For example, let’s look at the following example.

{(e^{- 2} f^{2})}^{7} = \frac{f^{14}}{e^{14}}

${\left({e}^{-2}{f}^{2}\right)}^{7}=\dfrac{{f}^{14}}{{e}^{14}}$

Let’s rewrite the original problem differently and look at the result.

\begin{aligned} {(e^{- 2} f^{2})}^{7} & = {(\frac{f^{2}}{e^{2}})}^{7} \\ = \frac{f^{14}}{e^{14}} \end{aligned}

$\begin{align} {\left({e}^{-2}{f}^{2}\right)}^{7}& = {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7} \\[1mm] & = \frac{{f}^{14}}{{e}^{14}} \\ \text{ } \end{align}$

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

\begin{aligned} {(e^{- 2} f^{2})}^{7} & = {(\frac{f^{2}}{e^{2}})}^{7} \\ = \frac{{(f^{2})}^{7}}{{(e^{2})}^{7}} \\ = \frac{f^{2 \cdot 7}}{e^{2 \cdot 7}} \\ = \frac{f^{14}}{e^{14}} \end{aligned}

$\begin{align} {\left({e}^{-2}{f}^{2}\right)}^{7}& = {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7} \\[1mm] & = \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}} \\[1mm] & = \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}} \\[1mm] & = \frac{{f}^{14}}{{e}^{14}} \\ \text{ } \end{align}$

A General Note: The Power of a Quotient Rule of Exponents

For any real numbers $a$ and $b$ and any integer $n$ , the power of a quotient rule of exponents states that

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

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

Example: Using the Power of a Quotient Rule

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

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

Show Solution

${\left(\dfrac{4}{{z}^{11}}\right)}^{3}=\dfrac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\dfrac{64}{{z}^{11\cdot 3}}=\dfrac{64}{{z}^{33}}$
${\left(\dfrac{p}{{q}^{3}}\right)}^{6}=\dfrac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\dfrac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\dfrac{{p}^{6}}{{q}^{18}}$
${\left(\dfrac{-1}{{t}^{2}}\right)}^{27}=\dfrac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}=\dfrac{-1}{{t}^{2\cdot 27}}=\dfrac{-1}{{t}^{54}}=-\dfrac{1}{{t}^{54}}$
${\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\dfrac{{j}^{3}}{{k}^{2}}\right)}^{4}=\dfrac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\dfrac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\dfrac{{j}^{12}}{{k}^{8}}$
${\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\dfrac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\dfrac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\dfrac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\dfrac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\dfrac{1}{{m}^{6}{n}^{6}}$

Try It

Simplify each of the following quotients as much as possible using the power of a quotient rule. 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 Solution

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

Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.

Example: Simplifying Exponential Expressions

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

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

Show Solution

$\begin{align} {\left(6{m}^{2}{n}^{-1}\right)}^{3}& = {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}&& \text{The power of a product rule} \\ & = {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}&& \text{The power rule} \\ & = 216{m}^{6}{n}^{-3}&& \text{Simplify}. \\ & = \frac{216{m}^{6}}{{n}^{3}}&& \text{The negative exponent rule} \end{align}$
$\begin{align} {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}& =& {17}^{5 - 4-3}&& \text{The product rule} \\ & = {17}^{-2}&& \text{Simplify}. \\ & = \frac{1}{{17}^{2}}\text{ or }\frac{1}{289}&& \text{The negative exponent rule} \end{align}$
$\begin{align} {\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}& = \frac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}&& \text{The power of a quotient rule} \\ & = \frac{{u}^{-2}{v}^{2}}{{v}^{-2}}&& \text{The power of a product rule} \\ & = {u}^{-2}{v}^{2-\left(-2\right)}&& \text{The quotient rule} \\ & = {u}^{-2}{v}^{4}&& \text{Simplify}. \\ & = \frac{{v}^{4}}{{u}^{2}}&& \text{The negative exponent rule} \end{align}$
$\begin{align} \left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)& =& -2\cdot 5\cdot {a}^{3}\cdot {a}^{-2}\cdot {b}^{-1}\cdot {b}^{2}&& \text{Commutative and associative laws of multiplication} \\ & = -10\cdot {a}^{3 - 2}\cdot {b}^{-1+2}&& \text{The product rule} \\ & = -10ab&& \text{Simplify}. \end{align}$
$\begin{align} {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& = {\left({x}^{2}\sqrt{2}\right)}^{4 - 4} && \text{The product rule} \\ & = {\left({x}^{2}\sqrt{2}\right)}^{0}&& \text{Simplify}. \\ & = 1&& \text{The zero exponent rule} \end{align}$
$\begin{align} \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& = \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}&& \text{The power of a product rule} \\ & = \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}&& \text{The power rule} \\ & = \frac{243{w}^{10}}{36{w}^{-4}} && \text{Simplify}. \\ & = \frac{27{w}^{10-\left(-4\right)}}{4}&& \text{The quotient rule and reduce fraction} \\ & = \frac{27{w}^{14}}{4}&& \text{Simplify}. \end{align}$

Try It

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

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

Show Solution

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

In the following video we show more examples of how to find the power of a quotient.

Chapter CR: Corequisite Review