Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises: 44

Answer

$\color{blue}{2k^{5}}$

Work Step by Step

RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ (4) $(ab)^m = a^mb^m$ (5) $(a^m)^n=a^{mn}$ (6) $a^m \cdot a^n = a^{m+n}$ (7) $a^0=1, a\ne0$ Use rule (5) above to obtain: $=\dfrac{12k^{-2}(k^{-3\cdot (-4)})}{6k^5} \\=\dfrac{12k^{-2}(k^{12})}{6k^5}$ Use rule (6) above to obtain: $=\dfrac{12k^{-2+12}}{6k^5} \\=\dfrac{12k^{10}}{6k^5}$ Divide the coefficients by cancelling common factors to obtain: $\require{cancel} \\=\dfrac{\cancel{12}2k^{10}}{\cancel{6}k^5} \\=\dfrac{k^{10}}{k^5}$ Use rule (2) above to obtain: $=2k^{10-5} \\=\color{blue}{2k^{5}}$
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