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 - Page 57: 63

Answer

$\text{The simplified form of the given expression is }k^{2/3}$.

Work Step by Step

$\begin{array}{ l l } =\dfrac{k^{1/3} \cdot k}{k^{2/3}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \dfrac{1}{a^{-n}} =a^{n} \end{array}\\ & \\ =\dfrac{k^{\frac{1}{3}+1}}{k^{2/3}} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ a^{b} \cdot a^{c} =a^{b+c} \end{array}\\ & \\ =\dfrac{k^{\frac{1}{3}+\frac{3}{3}}}{k^{2/3}} & \mathrm{Express} \ 1\ \mathrm{as\ } \frac{3}{3}.\\ & \\ =\dfrac{k^{4/3}}{k^{2/3}} & \mathrm{Add\ Exponents}\\ & \\ =k^{4/3-2/3} & \begin{array}{l} \mathrm{Apply\ the\ rule}\\ \dfrac{a^{m}}{a^{n}} =a^{m-n} \end{array}\\ & \\ =k^{2/3} & \mathrm{Subtract\ exponents} \end{array}$
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