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: 102

Answer

$\dfrac{3xy^{3}}{z^{3}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \left( \dfrac{9x^5y^4z^{-7}}{x^3y^{-2}z^{-1}} \right)^{\frac{1}{2}} ,$ 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{9x^5y^4z^{-7}}{x^3y^{-2}z^{-1}} \right)^{\frac{1}{2}} \\\\= \left( 9x^{5-3}y^{4-(-2)}z^{-7-(-1)} \right)^{\frac{1}{2}} \\\\= \left( 9x^{2}y^{6}z^{-6} \right)^{\frac{1}{2}} \\\\= \left( 3^2x^{2}y^{6}z^{-6} \right)^{\frac{1}{2}} .\end{array} 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} \left( 3^2x^{2}y^{6}z^{-6} \right)^{\frac{1}{2}} \\\\= 3^{2\cdot\frac{1}{2}}x^{2\cdot\frac{1}{2}}y^{6\cdot\frac{1}{2}}z^{-6\cdot\frac{1}{2}} \\\\= 3^{1}x^{1}y^{3}z^{-3} \\\\= 3xy^{3}z^{-3} .\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} 3xy^{3}z^{-3} \\\\= \dfrac{3xy^{3}}{z^{3}} .\end{array}
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