## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises - Page 449: 67

#### Answer

$x^{5/12}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $x^{2/3}\cdot x^{-1/4} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^{\frac{2}{3}+\left(-\frac{1}{4}\right)} \\\\= x^{\frac{2}{3}-\frac{1}{4}} .\end{array} To simplify the expression, $\dfrac{2}{3}-\dfrac{1}{4} ,$ find the $LCD$ of the denominators $\left\{ 3,4 \right\}.$ The $LCD$ is $12$ since it is the lowest number that can be exactly divided by the denominators. Multiplying both the numerator and denominator of each terms by the constant that will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} x^{\frac{2}{3}\cdot\frac{4}{4}-\frac{1}{4}\cdot\frac{3}{3}} \\\\= x^{\frac{8}{12}-\frac{3}{12}} \\\\= x^{\frac{5}{12}} \\\\= x^{5/12} .\end{array}

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