## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 145

#### Answer

$\dfrac{4k^{17}}{125}$

#### Work Step by Step

Using the laws of exponents, the given expression, $\dfrac{(2k)^2k^3}{k^{-1}k^{-5}}(5k^{-2})^{-3} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{4k^2k^3(5^{-3}k^{-2(-3)})}{k^{-1}k^{-5}} \\\\= \dfrac{4k^{2+3}(5^{-3}k^{6})}{k^{-1+(-5)}} \\\\= \dfrac{4k^{5}(k^{6})}{5^{3}k^{-6}} \\\\= \dfrac{4k^{5+6-(-6)}}{5^{3}} \\\\= \dfrac{4k^{5+6+6}}{125} \\\\= \dfrac{4k^{17}}{125} .\end{array}

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