Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review: 47



Work Step by Step

$(\frac{b^{\frac{3}{4}}}{a^{-\frac{1}{2}}})^{8}=(a^{\frac{1}{2}}b^{\frac{3}{4}})^{8}$ To simplify, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $(a^{\frac{1}{2}}b^{\frac{3}{4}})^{8}=a^{\frac{1}{2}\times8}b^{\frac{3}{4}\times8}=a^{\frac{8}{2}}b^{\frac{24}{4}}=a^{4}b^{6}$
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