Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.1 Rules for Exponents - 3.1 Exercises - Page 234: 100



Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \left( \dfrac{-27k^8v^7}{8k^2v} \right)^{\frac{1}{3}} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ 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} \left( \dfrac{-27k^8v^7}{8k^2v} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{-27k^{8-2}v^{7-1}}{8} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{-27k^{6}v^{6}}{8} \right)^{\frac{1}{3}} \\\\ \left( \dfrac{(-3)^3k^{6}v^{6}}{2^3} \right)^{\frac{1}{3}} .\end{array} Using the Power of a Quotient Rule of the laws of exponents which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{(-3)^3k^{6}v^{6}}{2^3} \right)^{\frac{1}{3}} \\\\= \dfrac{(-3)^{3\cdot\frac{1}{3}}k^{6\cdot\frac{1}{3}}v^{6\cdot\frac{1}{3}}}{2^{3\cdot\frac{1}{3}}} \\\\= \dfrac{(-3)^{1}k^{2}v^{2}}{2^{1}} \\\\= \dfrac{-3k^{2}v^{2}}{2} .\end{array}
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