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: 69

Answer

$\color{blue}{k^{2/3}}$

Work Step by Step

RECALL: (1) $a^m \cdot a^n = a^{m+n}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $a^{1/n} = \sqrt[n]{a}$ (4) When $n$ is odd, $\sqrt[n]{a^n}=a$ (5) $a^{m/n} = \left(\sqrt[n]{a}\right)^m$ Use rule (1) above to obtain: $=\dfrac{k^{1/3}}{k^{2/3+(-1)}} \\=\dfrac{k^{1/3}}{k^{2/3+(-3/3)}} \\=\dfrac{k^{1/3}}{k^{-1/3}}$ Use rule (2) above to obtain: $=k^{1/3 - (-1/3)} \\=k^{1/3+1/3} \\=\color{blue}{k^{2/3}}$
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