Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set: 49

Answer

$\dfrac{16}{x^2}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{(4x^3)^2}{x^8} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{4^{1(2)}x^{3(2)}}{x^8} \\\\= \dfrac{4^{2}x^{6}}{x^8} \\\\= 16x^{6-8} \\\\= 16x^{-2} \\\\= \dfrac{16}{x^2} .\end{array}
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