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

Published by Pearson

# Chapter 10 - Exponents and Radicals - Study Summary - Practice Exercises - Page 691: 13

#### Answer

$x^2\sqrt[6]{x}$

#### Work Step by Step

Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ then \begin{array}{l}\require{cancel} \dfrac{\sqrt{x^5}}{\sqrt[3]{x}} \\\\= \dfrac{x^{5/2}}{x^{1/3}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{x^{5/2}}{x^{1/3}} \\\\= x^{\frac{5}{2}-\frac{1}{3}} \\\\= x^{\frac{15}{6}-\frac{2}{6}} \\\\= x^{\frac{13}{6}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ then \begin{array}{l}\require{cancel} x^{\frac{13}{6}} \\\\= \sqrt[6]{x^{13}} .\end{array} Extracting the factor that is a perfect power of the index, then \begin{array}{l}\require{cancel} \sqrt[6]{x^{13}} \\\\= \sqrt[6]{x^{12}\cdot x} \\\\= \sqrt[6]{(x^{2})^6\cdot x} \\\\= x^2\sqrt[6]{x} .\end{array}

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