Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 129

Answer

$\dfrac{4}{a^2}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{4a^5(a^{-1})^3}{(a^{-2})^{-2}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4a^5(a^{-1(3)})}{(a^{-2(-2)})} \\\\= \dfrac{4a^5(a^{-3})}{a^{4}} \\\\= \dfrac{4a^{5+(-3)}}{a^{4}} \\\\= \dfrac{4a^{2}}{a^{4}} \\\\= 4a^{2-4} \\\\= 4a^{-2} \\\\= \dfrac{4}{a^2} .\end{array}
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