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 448: 38

Answer

$\dfrac{32}{243}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents and the definition of rational exponents to simplify the given expression, $ \left( \dfrac{729}{64} \right)^{-5/6} .$ $\bf{\text{Solution Details:}}$ 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{729^{-5/6}}{64^{-5/6}} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{64^{5/6}}{729^{5/6}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(\sqrt[6]{64})^{5}}{(\sqrt[6]{729})^{5}} \\\\= \dfrac{(2)^{5}}{(3)^{5}} \\\\= \dfrac{32}{243} .\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.