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: 78

Answer

$z^{1/3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{ z^{1/3}z^{-2/3}z^{1/6}}{\left( z^{-1/6} \right)^{3}} .$ $\bf{\text{Solution Details:}}$ Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{ z^{1/3}z^{-2/3}z^{1/6}}{z^{-\frac{1}{6}\cdot3}} \\\\= \dfrac{ z^{1/3}z^{-2/3}z^{1/6}}{z^{-\frac{3}{6}}} .\end{array} 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} \dfrac{ z^{\frac{1}{3}+\left( -\frac{2}{3} \right)+\frac{1}{6}}}{z^{-\frac{3}{6}}} \\\\= \dfrac{ z^{\frac{1}{3}-\frac{2}{3}+\frac{1}{6}}}{z^{-\frac{3}{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} z^{\frac{1}{3}-\frac{2}{3}+\frac{1}{6}-\left( -\frac{3}{6} \right)} \\\\ z^{\frac{1}{3}-\frac{2}{3}+\frac{1}{6}+\frac{3}{6}} \\\\ z^{\frac{2}{6}-\frac{4}{6}+\frac{1}{6}+\frac{3}{6}} \\\\ z^{\frac{2}{6}} \\\\ z^{\frac{1}{3}} \\\\ z^{1/3} .\end{array}
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