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: 57

Answer

$\dfrac{9}{8x^{5}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (2x^3y^{-4})^{-3}(3x^2y^{-6})^2 ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents, which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2x^3y^{-4})^{-3}(3x^2y^{-6})^2 \\\\ \left( 2^{-3}x^{3(-3)}y^{-4(-3)} \right) \left( 3^2x^{2(2)}y^{-6(2)} \right) \\\\ \left( 2^{-3}x^{-9}y^{12} \right) \left( 3^2x^{4}y^{-12} \right) .\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} \left( 2^{-3}x^{-9}y^{12} \right) \left( 3^2x^{4}y^{-12} \right) \\\\= (2^{-3})(3^2)x^{-9+4}y^{12+(-12)} \\\\= (2^{-3})(9)x^{-5}y^{0} .\end{array} Since any expression (except $0$) raised to the $0$ power is $1,$ then the expression above simplifies to \begin{array}{l}\require{cancel} (2^{-3})(9)x^{-5}y^{0} \\\\ (2^{-3})(9)x^{-5}(1) \\\\ (2^{-3})(9)x^{-5} .\end{array} Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2^{-3})(9)x^{-5} \\\\= \dfrac{9}{2^{3}x^{5}} \\\\= \dfrac{9}{8x^{5}} .\end{array}
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