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: 79

Answer

$\dfrac{c^{11/3}}{b^{11/4}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \left( \dfrac{b^{-3/2}}{c^{-5/3}} \right)^2 \left( b^{-1/4}c^{-1/3} \right)^{-1} .$ $\bf{\text{Solution Details:}}$ 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{b^{-\frac{3}{2}\cdot2}}{c^{-\frac{5}{3}\cdot2}} \right) b^{-\frac{1}{4}\cdot(-1)}c^{-\frac{1}{3}\cdot(-1)} \\\\= \left( \dfrac{b^{-3}}{c^{-\frac{10}{3}}} \right) b^{\frac{1}{4}}c^{\frac{1}{3}} \\\\= \dfrac{b^{-3}b^{\frac{1}{4}}c^{\frac{1}{3}}}{c^{-\frac{10}{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{b^{-3+\frac{1}{4}}c^{\frac{1}{3}}}{c^{-\frac{10}{3}}} \\\\= \dfrac{b^{-\frac{12}{4}+\frac{1}{4}}c^{\frac{1}{3}}}{c^{-\frac{10}{3}}} \\\\= \dfrac{b^{-\frac{11}{4}}c^{\frac{1}{3}}}{c^{-\frac{10}{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} b^{-\frac{11}{4}}c^{\frac{1}{3}-\left( -\frac{10}{3} \right)} \\\\= b^{-\frac{11}{4}}c^{\frac{1}{3}+\frac{10}{3}} \\\\= b^{-\frac{11}{4}}c^{\frac{11}{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} \dfrac{c^{\frac{11}{3}}}{b^{\frac{11}{4}}} \\\\= \dfrac{c^{11/3}}{b^{11/4}} .\end{array}
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