Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.2 Rational Numbers as Exponents - 10.2 Exercise Set: 63

Answer

$\dfrac{a^3}{3^2\sqrt{3}\sqrt[3]{b^7}}$

Work Step by Step

Using the laws of exponents, the given expression, $ 3^{-5/2}a^3b^{-7/3} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{a^3}{3^{5/2}b^{7/3}} .\end{array} Using $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$, the expression, $ \dfrac{a^3}{3^{5/2}b^{7/3}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{a^3}{\sqrt{3^5}\sqrt[3]{b^7}} \\\\= \dfrac{a^3}{\sqrt{3^4\cdot3}\sqrt[3]{b^7}} \\\\= \dfrac{a^3}{\sqrt{(3^2)^2\cdot3}\sqrt[3]{b^7}} \\\\= \dfrac{a^3}{3^2\sqrt{3}\sqrt[3]{b^7}} .\end{array}
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