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: 23

Answer

$xy\sqrt[10]{x^7y^3}$

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.
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.