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 - Mid-Chapter Review - Mixed Review - Page 663: 23



Work Step by Step

Using $x^{m/n}=\sqrt[n]{x^m}=\left(\sqrt[n]{x} \right)^m,$ then \begin{array}{l}\require{cancel} \sqrt{x^3y}\sqrt[5]{xy^4} \\\\= (x^3y)^{1/2}(xy^4)^{1/5} \\\\= (x^3y)^{5/10}(xy^4)^{2/10} \\\\= \sqrt[10]{(x^3y)^5}\cdot\sqrt[10]{(xy^4)^2} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[10]{(x^3y)^5}\cdot\sqrt[10]{(xy^4)^2} \\\\= \sqrt[10]{x^{15}y^5}\cdot\sqrt[10]{x^2y^8} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt[10]{x^{15}y^5(x^2y^8)} \\\\= \sqrt[10]{x^{15+2}y^{5+8}} \\\\= \sqrt[10]{x^{17}y^{13}} .\end{array} Extracting the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[10]{x^{17}y^{13}} \\\\= \sqrt[10]{x^{10}y^{10}\cdot x^7y^3} \\\\= \sqrt[10]{(xy)^{10}\cdot x^7y^3} \\\\= xy\sqrt[10]{x^7y^3} .\end{array} Note that all variables are assumed to be positive.
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