Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents - Page 446: 16

Answer

$4^{\frac{11}{15}}$

Work Step by Step

We can use the product rule to simplify, which holds that $a^{m}\times a^{n}=a^{m+n}$ (where a is a real number, and m and n are positive integers). $4^{\frac{1}{3}}\times4^{\frac{2}{5}}=4^{\frac{1}{3}+\frac{2}{5}}=4^{\frac{5}{15}+\frac{6}{15}}=4^{\frac{11}{15}}$
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