Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises: 45

Answer

$ \displaystyle \frac{3k^{3/2}}{8}$

Work Step by Step

$\displaystyle \frac{3k^{2}\cdot(4k^{-3})^{-1}}{4^{1/2}\cdot k^{7/2}} \qquad$ ....... use $(ab)^{m}=a^{m} b^{m}$ $=\displaystyle \frac{3k^{2}\cdot 4^{-1}(k^{-3})^{-1}}{4^{1/2}\cdot k^{7/2}}\qquad$ ....... use $(a^{m})^{n}=a^{mn}$, group like terms $=\displaystyle \frac{3\cdot 4^{-1}}{4^{1/2}}\cdot\frac{k^{2}\cdot k^{3}}{k^{7/2}} \qquad$ ....... use $a^{m}\cdot a^{n}=a^{m+n}$ $=3\displaystyle \cdot\frac{4^{-1}}{4^{1/2}}\cdot\frac{k^{5}}{k^{7/2}}\qquad$ ....... use $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ $=3(4^{-1-1/2})(k^{5-7/2)}$ $=3\cdot 4^{-3/2}\cdot k^{3/2}\qquad$ .......recognize: $4=2^{2}$ $=3(2^{2})^{-3/2}\cdot k^{3/2}\qquad$ ....... use $(a^{m})^{n}=a^{mn}$ $=3\cdot 2^{-3}\cdot k^{3/2}\qquad$ ....... use $a^{-n}=\displaystyle \frac{1}{a^{n}}=(\frac{1}{a})^{n}$ $= \displaystyle \frac{3k^{3/2}}{8}$
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