Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises - Page 449: 82



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \left( \dfrac{2^{-2}w^{-3/4}x^{-5/8}}{ w^{3/4}x^{-1/2}} \right)^{-3} .$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2^{-2\cdot(-3)}w^{-\frac{3}{4}\cdot(-3)}x^{-\frac{5}{8}\cdot(-3)}}{ w^{\frac{3}{4}\cdot(-3)}x^{-\frac{1}{2}\cdot(-3)}} \\\\= \dfrac{2^{6}w^{\frac{9}{4}}x^{\frac{15}{8}}}{ w^{-\frac{9}{4}}x^{\frac{3}{2}}} \\\\= \dfrac{64w^{\frac{9}{4}}x^{\frac{15}{8}}}{ w^{-\frac{9}{4}}x^{\frac{3}{2}}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 64w^{\frac{9}{4}-\left(-\frac{9}{4}\right)}x^{\frac{15}{8}-\frac{3}{2}} \\\\= 64w^{\frac{9}{4}+\frac{9}{4}}x^{\frac{15}{8}-\frac{3}{2}} \\\\= 64w^{\frac{18}{4}}x^{\frac{15}{8}-\frac{3}{2}} \\\\= 64w^{\frac{9}{2}}x^{\frac{15}{8}-\frac{3}{2}} .\end{array} Changing the exponent to similar fractions, the expression above is equivalent to \begin{array}{l}\require{cancel} 64w^{\frac{9}{2}}x^{\frac{15}{8}-\frac{12}{8}} \\\\= 64w^{\frac{9}{2}}x^{\frac{3}{8}} \\\\= 64w^{9/2}x^{3/8} .\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.