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: 130

Answer

$2k^{5}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{12k^{-2}(k^{-3})^{-4}}{6k^5} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{\cancel{6}(2)k^{-2}(k^{-3(-4)})}{\cancel{6}k^5} \\\\= \dfrac{2k^{-2}(k^{12})}{k^5} \\\\= 2k^{-2+12-5} \\\\= 2k^{5} .\end{array}
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