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: 80

Answer

$\dfrac{a^{5/2}}{m^{23/12}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \left( \dfrac{m^{-2/3}}{a^{-3/4}} \right)^4 \left( m^{-3/8}a^{1/4} \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} \left( \dfrac{m^{-\frac{2}{3}\cdot4}}{a^{-\frac{3}{4}\cdot4}} \right) m^{-\frac{3}{8}\cdot(-2)}a^{\frac{1}{4}\cdot(-2)} \\\\= \left( \dfrac{m^{-\frac{8}{3}}}{a^{-3}} \right) m^{\frac{3}{4}}a^{-\frac{1}{2}} \\\\= \dfrac{m^{-\frac{8}{3}}m^{\frac{3}{4}}a^{-\frac{1}{2}}}{a^{-3}} .\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} \dfrac{m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}}}{a^{-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} m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}-(-3)} \\\\= m^{-\frac{8}{3}+\frac{3}{4}}a^{-\frac{1}{2}+3} .\end{array} Changing the exponents to similar fractions, the expression above is equivalent to \begin{array}{l}\require{cancel} m^{-\frac{32}{12}+\frac{9}{12}}a^{-\frac{1}{2}+\frac{6}{2}} \\\\= m^{-\frac{23}{12}}a^{\frac{5}{2}} .\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{a^{\frac{5}{2}}}{m^{\frac{23}{12}}} \\\\= \dfrac{a^{5/2}}{m^{23/12}} .\end{array}
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