Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set - Page 522: 123

Answer

$ \frac{x^{3}}{y^{2}} $.

Work Step by Step

The given expression is $\left ( \frac{x^{-\frac{5}{4}}y^{\frac{1}{3}}}{x^{-\frac{3}{4}}} \right )^{-6}$ Use $\frac{1}{a^n}=a^{-n}$ $=\left ( x^{-\frac{5}{4}}y^{\frac{1}{3}}\cdot x^{\frac{3}{4}} \right )^{-6}$ Use $a^m\cdot a^n= a^{m+n}$. $=\left ( x^{-\frac{5}{4}+\frac{3}{4}}y^{\frac{1}{3}} \right )^{-6}$ $=\left ( x^{\frac{-5+3}{4}}y^{\frac{1}{3}} \right )^{-6}$ $=\left ( x^{\frac{-2}{4}}y^{\frac{1}{3}} \right )^{-6}$ $=\left ( x^{\frac{-1}{2}}y^{\frac{1}{3}} \right )^{-6}$ Use $(ab)^m=a^mb^m$. $ =x^{\frac{6}{2}}y^{\frac{-6}{3}} $ $ =x^{3}y^{-2} $ Use $a^{-n}=\frac{1}{a^n}$ $ =\frac{x^{3}}{y^{2}} $.
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.