College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.6 - Rational Exponents - R.6 Exercises: 36

Answer

$2k^{5}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{12k^{-2}(k^{-3})^{-4}}{6k^5} .$ $\bf{\text{Solution Details:}}$ Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{12k^{-2}k^{-3(-4)}}{6k^5} \\\\= \dfrac{\cancel6(2)k^{-2}k^{12}}{\cancel6k^5} \\\\= \dfrac{2k^{-2}k^{12}}{k^5} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2k^{-2+12}}{k^5} \\\\= \dfrac{2k^{10}}{k^5} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 2k^{10-5} \\\\= 2k^{5} .\end{array}
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