Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Mid-Chapter Check Point - Page 541: 18

Answer

$x\sqrt[7] {x^2y^{2}}$.

Work Step by Step

The given expression is $=\frac{\sqrt[7] {x^4y^9}}{\sqrt [7]{x^{-5}y^7}}$ Divide the radicands and retain the common index. $=\sqrt[7] {\frac{x^4y^9}{x^{-5}y^7}}$ Divide factors in the radicand. Subtract exponents on common bases. $=\sqrt[7] {x^{4+5}y^{9-7}}$ Simplify. $=\sqrt[7] {x^{9}y^{2}}$ Factor the radicands as a power of $7$. $=\sqrt[7] {x^{7}x^2y^{2}}$ Factor into two radicals. $=\sqrt[7] {x^{7}}\sqrt[7] {x^2y^{2}}$ Simplify. $=x\sqrt[7] {x^2y^{2}}$.
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