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 1-14 - Cumulative Review - Final Exam - Page 930: 13

Answer

$25x^{4}y^{1/3}$

Work Step by Step

Using $\left(x^my^n\right)^p=x^{mp}y^{np}$ the given expression, $ \left(125x^6y^{1/2}\right)^{2/3} ,$ is equivalent to \begin{align*} & 125^{\frac{2}{3}}x^{6\cdot\frac{2}{3}}y^{\frac{1}{2}\cdot\frac{2}{3}} \\\\&= 125^{\frac{2}{3}}x^{4}y^{\frac{1}{3}} \\\\&= 125^{\frac{2}{3}}x^{4}y^{1/3} .\end{align*} Using $x^{\frac{m}{n}}=\sqrt[n]{x^m}=\left(\sqrt[n]{x}\right)^m$ the expression above is equivalent to \begin{align*} & \left(\sqrt[3]{125}\right)^{2}x^{4}y^{1/3} \\\\&= \left(5\right)^{2}x^{4}y^{1/3} \\\\&= 25x^{4}y^{1/3} .\end{align*} Hence, the expression $\left(125x^6y^{1/2}\right)^{2/3}$ simplifies to $25x^{4}y^{1/3}$.
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