Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set - Page 522: 69

Answer

$32x$

Work Step by Step

Use the rule $(ab)^m= a^{m}b^{m}$ to obtain: $=2^5(x^{\frac{1}{5}})^5 \\=2(2)(2)(2)(2)(x^{\frac{1}{5}})^5 \\=32(x^{\frac{1}{5}})^5$ Use the rule $(a^{m})^n = a^{mn}$ to obtain: $=32x^{\frac{1}{5} \cdot 5} \\=32x$
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