## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises - Page 448: 60

#### Answer

$w^{7/12}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents and the laws of exponents to convert the given expression, $\dfrac{\sqrt[4]{w^3}}{\sqrt[6]{w}} ,$ to exponential form. $\bf{\text{Solution Details:}}$ Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{w^{\frac{3}{4}}}{w^{\frac{1}{6}}} .\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} w^{\frac{3}{4}-\frac{1}{6}} .\end{array} To simplify the expression, $\dfrac{3}{4}-\dfrac{1}{6} ,$ find the $LCD$ of the denominators $\{ 4,6 \}.$ The $LCD$ is $12$ since it is the lowest number that can be divided by both denominators. Multiplying both the numerator and the denominator of each term by the constant that will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} w^{\frac{3}{4}\cdot\frac{3}{3}-\frac{1}{6}\cdot\frac{2}{2}} \\\\= w^{\frac{9}{12}-\frac{2}{12}} \\\\= w^{\frac{7}{12}} \\\\= w^{7/12} .\end{array}

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