6.7 Integer exponents and scientific notation (Page 2/10)

Page 2 / 10

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 $a and b$ are real numbers, $a \neq 0, b \neq 0,$ and $n$ is an integer, then ${(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n}$ .

Simplify: ⓐ ${(\frac{5}{7})}^{−2}$ ⓑ ${(- \frac{2 x}{y})}^{−3} .$

Solution

ⓐ
$\begin{matrix} {(\frac{5}{7})}^{−2} \\ Use the Quotient to a Negative Exponent Property, {(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n} . \\ Take the reciprocal of the fraction and change the sign of the exponent. & {(\frac{7}{5})}^{2} \\ Simplify. & \frac{49}{25} \end{matrix}$
ⓑ
$\begin{matrix} {(- \frac{2 x}{y})}^{−3} \\ Use the Quotient to a Negative Exponent Property, {(\frac{a}{b})}^{− n} = {(\frac{b}{a})}^{n} . \\ Take the reciprocal of the fraction and change the sign of the exponent. & {(- \frac{y}{2 x})}^{3} \\ Simplify. & - \frac{y^{3}}{8 x^{3}} \end{matrix}$

Got questions? Get instant answers now!

Simplify: ⓐ ${(\frac{2}{3})}^{−4}$ ⓑ ${(- \frac{6 m}{n})}^{−2} .$

ⓐ $\frac{81}{16}$ ⓑ $\frac{n^{2}}{36 m^{2}}$

Got questions? Get instant answers now!

Simplify: ⓐ ${(\frac{3}{5})}^{−3}$ ⓑ ${(- \frac{a}{2 b})}^{−4} .$

ⓐ $\frac{125}{27}$ ⓑ $\frac{16 b^{4}}{a^{4}}$

Got questions? Get instant answers now!

When simplifying an expression with exponents, we must be careful to correctly identify the base.

Simplify: ⓐ ${(−3)}^{−2}$ ⓑ $− 3^{−2}$ ⓒ ${(- \frac{1}{3})}^{−2}$ ⓓ $− {(\frac{1}{3})}^{−2} .$

Solution

ⓐ Here the exponent applies to the base $−3$ .
$\begin{matrix} {(−3)}^{−2} \\ Take the reciprocal of the base and change the sign of the exponent. & \frac{1}{{(−3)}^{−2}} \\ Simplify. & \frac{1}{9} \end{matrix}$
ⓑ The expression $− 3^{−2}$ means “find the opposite of $3^{−2}$ ”. Here the exponent applies to the base 3.
$\begin{matrix} − 3^{−2} \\ Rewrite as a product with −1 . & −1 \cdot 3^{−2} \\ Take the reciprocal of the base and change the sign of the exponent. & −1 \cdot \frac{1}{3^{2}} \\ Simplify. & - \frac{1}{9} \end{matrix}$
ⓒ Here the exponent applies to the base ${(- \frac{1}{3})}^{}$ .
$\begin{matrix} {(- \frac{1}{3})}^{−2} \\ Take the reciprocal of the base and change the sign of the exponent. & {(- \frac{3}{1})}^{2} \\ Simplify. & 9 \end{matrix}$
ⓓ The expression $− {(\frac{1}{3})}^{−2}$ means “find the opposite of ${(\frac{1}{3})}^{−2}$ ”. Here the exponent applies to the base $(\frac{1}{3})$ .
$\begin{matrix} − {(\frac{1}{3})}^{−2} \\ Rewrite as a product with −1 . & −1 \cdot {(\frac{1}{3})}^{−2} \\ Take the reciprocal of the base and change the sign of the exponent. & −1 \cdot {(\frac{3}{1})}^{2} \\ Simplify. & −9 \end{matrix}$

Got questions? Get instant answers now!

Simplify: ⓐ ${(−5)}^{−2}$ ⓑ $− 5^{−2}$ ⓒ ${(- \frac{1}{5})}^{−2}$ ⓓ $− {(\frac{1}{5})}^{−2} .$

ⓐ $\frac{1}{25}$ ⓑ $- \frac{1}{25}$ ⓒ 25 ⓓ $−25$

Got questions? Get instant answers now!

Simplify: ⓐ ${(−7)}^{−2}$ ⓑ $− 7^{−2}$ , ⓒ ${(- \frac{1}{7})}^{−2}$ ⓓ $− {(\frac{1}{7})}^{−2} .$

ⓐ $\frac{1}{49}$ ⓑ $- \frac{1}{49}$ ⓒ 49 ⓓ $−49$

Got questions? Get instant answers now!

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: ⓐ $4 \cdot 2^{−1}$ ⓑ ${(4 \cdot 2)}^{−1} .$

Solution

ⓐ
$\begin{matrix} Do exponents before multiplication. & 4 \cdot 2^{−1} \\ Use a^{− n} = \frac{1}{a^{n}} . & 4 \cdot \frac{1}{2^{1}} \\ Simplify. & 2 \end{matrix}$
ⓑ
$\begin{matrix} {(4 \cdot 2)}^{−1} \\ Simplify inside the parentheses first. & {(8)}^{−1} \\ Use a^{− n} = \frac{1}{a^{n}} . & \frac{1}{8^{1}} \\ Simplify. & \frac{1}{8} \end{matrix}$

Got questions? Get instant answers now!

Simplify: ⓐ $6 \cdot 3^{−1}$ ⓑ ${(6 \cdot 3)}^{−1} .$

ⓐ $2$ ⓑ $\frac{1}{18}$

Got questions? Get instant answers now!

Simplify: ⓐ $8 \cdot 2^{−2}$ ⓑ ${(8 \cdot 2)}^{−2} .$

ⓐ 2 ⓑ $\frac{1}{16}$

Got questions? Get instant answers now!

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 non-zero.

Simplify: ⓐ $x^{−6}$ ⓑ ${(u^{4})}^{−3} .$

Solution

ⓐ
$\begin{matrix} x^{−6} \\ Use the definition of a negative exponent, a^{− n} = \frac{1}{a^{n}} . & \frac{1}{x^{6}} \end{matrix}$
ⓑ
$\begin{matrix} {(u^{4})}^{−3} \\ Use the definition of a negative exponent, a^{− n} = \frac{1}{a^{n}} . & \frac{1}{{(u^{4})}^{3}} \\ Simplify. & \frac{1}{u^{12}} \end{matrix}$

Got questions? Get instant answers now!

Simplify: ⓐ $y^{−7}$ ⓑ ${(z^{3})}^{−5} .$

ⓐ $\frac{1}{y^{7}}$ ⓑ $\frac{1}{z^{15}}$

Got questions? Get instant answers now!

Simplify: ⓐ $p^{−9}$ ⓑ ${(q^{4})}^{−6} .$

ⓐ $\frac{1}{p^{9}}$ ⓑ $\frac{1}{q^{24}}$

Got questions? Get instant answers now!

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.

Solution

ⓐ
$\begin{matrix} 5 y^{−1} \\ \begin{matrix} Notice the exponent applies to just the base y . \\ Take the reciprocal of y and change the sign of the exponent. \end{matrix} & 5 \cdot \frac{1}{y^{1}} \\ Simplify. & \frac{5}{y} \end{matrix}$
ⓑ
$\begin{matrix} {(5 y)}^{−1} \\ \begin{matrix} Here the parentheses make the exponent apply to the base 5 y . \\ Take the reciprocal of 5 y and change the sign of the exponent. \end{matrix} & \frac{1}{{(5 y)}^{1}} \\ Simplify. & \frac{1}{5 y} \end{matrix}$
ⓒ
$\begin{matrix} {(−5 y)}^{−1} \\ \begin{matrix} The base here is −5 y . \\ Take the reciprocal of −5 y and change the sign of the exponent. \end{matrix} & \frac{1}{{(−5 y)}^{1}} \\ Simplify. & \frac{1}{−5 y} \\ Use \frac{a}{− b} = - \frac{a}{b} . & - \frac{1}{5 y} \end{matrix}$

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 2

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Elementary algebra' conversation and receive update notifications?

Ask

	MAPEH TEST By Edgar Delgado Start Test
	44 Biology 44 Ecology and the Biosphere MCQ By OpenStax Start Quiz
	Advertising Promotion BUS210 By Melinda Salzer Start Quiz
	Introductory statistics By OpenStax Read Online Course
	How well do you know 5SOS? By Wey Hey Start Quiz
	4 Psychology MCQ 2009 Final Exam By John Gabrieli Start Exam
	19 Biology 19 The Evolution of Populations MCQ By OpenStax Start Quiz
	American Government MCQ By Saylor Foundation Start Quiz
	6 Microeconomics 06 Elasticity By OpenStax Start Flashcards
	2 Timeshare 2 By Jams Kalo Start Quiz