College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.6 - Rational Exponents - R.6 Exercises - Page 57: 67

Answer

$\text{The given expression simplifies to }\dfrac{1}{x^{10/3}}$

Work Step by Step

$\begin{array}{ l c } =\dfrac{\left( x^{2/3}\right)^{2}}{\left( x^{7/3}\right)^{2}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \left( a^{m}\right)^{n} =\left( a^{n}\right)^{m} \end{array}\\ & \\ =\left(\dfrac{x^{2/3}}{x^{7/3}}\right)^{2} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \dfrac{a^{n}}{b^{n}} =\left(\dfrac{a}{b}\right)^{n} \end{array}\\ & \\ =\left(\dfrac{1}{x^{7/3} \cdot x^{-2/3}}\right)^{2} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{n} =\dfrac{1}{a^{-n}} \end{array}\\ & \\ =\left(\dfrac{1}{x^{7/3-2/3}}\right)^{2} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{m} \cdot a^{-n} =a^{m-n} \end{array}\\ & \\ =\left(\dfrac{1}{x^{5/3}}\right)^{2} & \mathrm{Subtract\ exponents}\\ & \\ =\dfrac{1^{2}}{\left( x^{5/3}\right)^{2}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \left(\dfrac{a}{b}\right)^{n} =\dfrac{a^{n}}{b^{n}} \end{array}\\ & \\ =\dfrac{1}{x^{( 5/3) \cdot ( 2)}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \left( a^{m}\right)^{n} =a^{m\cdot m} \end{array}\\ & \\ =\dfrac{1}{x^{10/3}} & \mathrm{Simplify} \end{array}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.