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: 81

Answer

$\dfrac{q^{5/3}}{9p^{7/2}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \left( \dfrac{p^{-1/4}q^{-3/2}}{ 3^{-1}p^{-2}q^{-2/3}} \right)^{-2} .$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{p^{-\frac{1}{4}\cdot(-2)}q^{-\frac{3}{2}\cdot(-2)}}{3^{-1\cdot(-2)}p^{{-2}\cdot(-2)}q^{-\frac{2}{3}\cdot(-2)}} \\\\= \dfrac{p^{\frac{1}{2}}q^{3}}{3^{2}p^{4}q^{\frac{4}{3}}} \\\\= \dfrac{p^{\frac{1}{2}}q^{3}}{9p^{4}q^{\frac{4}{3}}} .\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} \dfrac{p^{\frac{1}{2}-4}q^{3-\frac{4}{3}}}{9} .\end{array} Changing the exponents to similar fractions, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{p^{\frac{1}{2}-\frac{8}{2}}q^{\frac{9}{3}-\frac{4}{3}}}{9} \\\\= \dfrac{p^{-\frac{7}{2}}q^{\frac{5}{3}}}{9} .\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} \dfrac{q^{\frac{5}{3}}}{9p^{\frac{7}{2}}} \\\\= \dfrac{q^{5/3}}{9p^{7/2}} .\end{array}
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