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: 101

Answer

$\dfrac{4mn^{4}}{p}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \left( \dfrac{16m^3n^6p}{mn^{-2}p^3} \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{16m^3n^6p}{mn^{-2}p^3} \right)^{\frac{1}{2}} \\\\= \left( 16m^{3-1}n^{6-(-2)}p^{1-3} \right)^{\frac{1}{2}} \\\\= \left( 16m^{2}n^{8}p^{-2} \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( 16m^{2}n^{8}p^{-2} \right)^{\frac{1}{2}} \\\\= \left( 4^2m^{2}n^{8}p^{-2} \right)^{\frac{1}{2}} \\\\= 4^{2\cdot\frac{1}{2}}m^{2\cdot\frac{1}{2}}n^{8\cdot\frac{1}{2}}p^{-2\cdot\frac{1}{2}} \\\\= 4^{1}m^{1}n^{4}p^{-1} \\\\= 4mn^{4}p^{-1} .\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} 4mn^{4}p^{-1} \\\\= \dfrac{4mn^{4}}{p^1} \\\\= \dfrac{4mn^{4}}{p} .\end{array}
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