Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises: 68

Answer

$x^{1/15}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ x^{2/5}\cdot x^{-1/3} .$ $\bf{\text{Solution Details:}}$ 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}{5}+\left(-\frac{1}{3}\right)} \\\\= x^{\frac{2}{5}-\frac{1}{3}} .\end{array} To simplify the expression, $ \dfrac{2}{5}-\dfrac{1}{3} ,$ find the $LCD$ of the denominators $\left\{ 5,3 \right\}.$ The $LCD$ is $ 15 $ since it is the lowest number that can be exactly divided by the denominators. Multiplying both the numerator and denominator of each terms by an expression equal to $1$ which will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} x^{\frac{2}{5}\cdot\frac{3}{3}-\frac{1}{3}\cdot\frac{5}{5}} \\\\= x^{\frac{6}{15}-\frac{5}{15}} \\\\= x^{\frac{1}{15}} \\\\= x^{1/15} .\end{array}
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