Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 693: 21



Work Step by Step

Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the given expression is equivalent to \begin{array}{l}\require{cancel} (x^{-2/3})^{3/5} \\\\= x^{-\frac{2}{3}\cdot\frac{3}{5}} \\\\= x^{-\frac{2}{\cancel3}\cdot\frac{\cancel3}{5}} \\\\= x^{-\frac{2}{5}} .\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} x^{-\frac{2}{5}} \\\\= \dfrac{1}{x^{\frac{2}{5}}} \\\\= \dfrac{1}{\sqrt[5]{x^2}} .\end{array} Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{1}{\sqrt[5]{x^2}} \\\\= \dfrac{1}{\sqrt[5]{x^2}}\cdot\frac{\sqrt[5]{x^3}}{\sqrt[5]{x^3}} \\\\= \dfrac{\sqrt[5]{x^3}}{\sqrt[5]{x^5}} \\\\= \dfrac{\sqrt[5]{x^3}}{x} .\end{array}
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