To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.
This leads us to the
Quotient to a Negative Power Property .
Quotient to a negative exponent property
If
$a\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b$ are real numbers,
$a\ne 0,b\ne 0,$ and
$n$ is an integer, then
${\left(\frac{a}{b}\right)}^{\text{\u2212}n}={\left(\frac{b}{a}\right)}^{n}$ .
Simplify:
ⓐ
${\left(\frac{5}{7}\right)}^{\mathrm{2}}$
ⓑ
${\left(\frac{2x}{y}\right)}^{\mathrm{3}}.$
Solution

ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{5}{7}\right)}^{\mathrm{2}}\hfill \\ \text{Use the Quotient to a Negative Exponent Property,}\phantom{\rule{0.2em}{0ex}}{\left(\frac{a}{b}\right)}^{\text{\u2212}n}={\left(\frac{b}{a}\right)}^{n}.\hfill & & & \\ \text{Take the reciprocal of the fraction and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{7}{5}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{49}{25}\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{2x}{y}\right)}^{\mathrm{3}}\hfill \\ \text{Use the Quotient to a Negative Exponent Property,}\phantom{\rule{0.2em}{0ex}}{\left(\frac{a}{b}\right)}^{\text{\u2212}n}={\left(\frac{b}{a}\right)}^{n}.\hfill & & & \\ \text{Take the reciprocal of the fraction and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{y}{2x}\right)}^{3}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{{y}^{3}}{8{x}^{3}}\hfill \end{array}$
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When simplifying an expression with exponents, we must be careful to correctly identify the base.
Simplify:
ⓐ
${\left(\mathrm{3}\right)}^{\mathrm{2}}$
ⓑ
$\text{\u2212}{3}^{\mathrm{2}}$
ⓒ
${\left(\frac{1}{3}\right)}^{\mathrm{2}}$
ⓓ
$\text{\u2212}{\left(\frac{1}{3}\right)}^{\mathrm{2}}.$
Solution

ⓐ Here the exponent applies to the base
$\mathrm{3}$ .
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\mathrm{3}\right)}^{\mathrm{2}}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(\mathrm{3}\right)}^{\mathrm{2}}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{9}\hfill \end{array}$

ⓑ The expression
$\text{\u2212}{3}^{\mathrm{2}}$ means “find the opposite of
${3}^{\mathrm{2}}$ ”. Here the exponent applies to the base 3.
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\text{\u2212}{3}^{\mathrm{2}}\hfill \\ \text{Rewrite as a product with}\phantom{\rule{0.2em}{0ex}}\mathrm{1}.\hfill & & & \phantom{\rule{4em}{0ex}}\mathrm{1}\xb7{3}^{\mathrm{2}}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}\mathrm{1}\xb7\frac{1}{{3}^{2}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{9}\hfill \end{array}$

ⓒ Here the exponent applies to the base
${\left(\frac{1}{3}\right)}^{}$ .
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\frac{1}{3}\right)}^{\mathrm{2}}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(\frac{3}{1}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}9\hfill \end{array}$

ⓓ The expression
$\text{\u2212}{\left(\frac{1}{3}\right)}^{\mathrm{2}}$ means “find the opposite of
${\left(\frac{1}{3}\right)}^{\mathrm{2}}$ ”. Here the exponent applies to the base
$\left(\frac{1}{3}\right)$ .
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\text{\u2212}{\left(\frac{1}{3}\right)}^{\mathrm{2}}\hfill \\ \text{Rewrite as a product with}\phantom{\rule{0.2em}{0ex}}\mathrm{1}.\hfill & & & \phantom{\rule{4em}{0ex}}\mathrm{1}\xb7{\left(\frac{1}{3}\right)}^{\mathrm{2}}\hfill \\ \text{Take the reciprocal of the base and change the sign of the exponent.}\hfill & & & \phantom{\rule{4em}{0ex}}\mathrm{1}\xb7{\left(\frac{3}{1}\right)}^{2}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\mathrm{9}\hfill \end{array}$
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Simplify:
ⓐ
${\left(\mathrm{5}\right)}^{\mathrm{2}}$
ⓑ
$\text{\u2212}{5}^{\mathrm{2}}$
ⓒ
${\left(\frac{1}{5}\right)}^{\mathrm{2}}$
ⓓ
$\text{\u2212}{\left(\frac{1}{5}\right)}^{\mathrm{2}}.$
ⓐ
$\frac{1}{25}$
ⓑ
$\frac{1}{25}$
ⓒ 25
ⓓ
$\mathrm{25}$
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Simplify:
ⓐ
${\left(\mathrm{7}\right)}^{\mathrm{2}}$
ⓑ
$\text{\u2212}{7}^{\mathrm{2}}$ ,
ⓒ
${\left(\frac{1}{7}\right)}^{\mathrm{2}}$
ⓓ
$\text{\u2212}{\left(\frac{1}{7}\right)}^{\mathrm{2}}.$
ⓐ
$\frac{1}{49}$
ⓑ
$\frac{1}{49}$
ⓒ 49
ⓓ
$\mathrm{49}$
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We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.
Simplify:
ⓐ
$4\xb7{2}^{\mathrm{1}}$
ⓑ
${\left(4\xb72\right)}^{\mathrm{1}}.$
Solution

ⓐ
$\begin{array}{cccc}\text{Do exponents before multiplication.}\hfill & & & \phantom{\rule{4em}{0ex}}4\xb7{2}^{\mathrm{1}}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}{a}^{\text{\u2212}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}4\xb7\frac{1}{{2}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}2\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(4\xb72\right)}^{\mathrm{1}}\hfill \\ \text{Simplify inside the parentheses first.}\hfill & & & \phantom{\rule{4em}{0ex}}{\left(8\right)}^{\mathrm{1}}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}{a}^{\text{\u2212}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{8}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{8}\hfill \end{array}$
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When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are nonzero.
Simplify:
ⓐ
${x}^{\mathrm{6}}$
ⓑ
${\left({u}^{4}\right)}^{\mathrm{3}}.$
Solution

ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{x}^{\mathrm{6}}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{\u2212}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{x}^{6}}\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left({u}^{4}\right)}^{\mathrm{3}}\hfill \\ \text{Use the definition of a negative exponent,}\phantom{\rule{0.2em}{0ex}}{a}^{\text{\u2212}n}=\frac{1}{{a}^{n}}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left({u}^{4}\right)}^{3}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{u}^{12}}\hfill \end{array}$
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When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the Order of Operations, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.
Simplify:
ⓐ
$5{y}^{\mathrm{1}}$
ⓑ
${\left(5y\right)}^{\mathrm{1}}$
ⓒ
${\left(\mathrm{5}y\right)}^{\mathrm{1}}.$
Solution

ⓐ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}5{y}^{\mathrm{1}}\hfill \\ \begin{array}{c}\text{Notice the exponent applies to just the base}\phantom{\rule{0.2em}{0ex}}y.\hfill \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}5\xb7\frac{1}{{y}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{5}{y}\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(5y\right)}^{\mathrm{1}}\hfill \\ \begin{array}{c}\text{Here the parentheses make the exponent apply to the base}\phantom{\rule{0.2em}{0ex}}5y.\hfill \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}5y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(5y\right)}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{5y}\hfill \end{array}$

ⓒ
$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{\left(\mathrm{5}y\right)}^{\mathrm{1}}\hfill \\ \begin{array}{cc}\text{The base here is}\phantom{\rule{0.2em}{0ex}}\mathrm{5}y.\hfill & \\ \text{Take the reciprocal of}\phantom{\rule{0.2em}{0ex}}\mathrm{5}y\phantom{\rule{0.2em}{0ex}}\text{and change the sign of the exponent.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{{\left(\mathrm{5}y\right)}^{1}}\hfill \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{\mathrm{5}y}\hfill \\ \text{Use}\phantom{\rule{0.2em}{0ex}}\frac{a}{\text{\u2212}b}=\frac{a}{b}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{1}{5y}\hfill \end{array}$
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