College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises - Page 90: 60



Work Step by Step

$\sqrt[4]8\times\sqrt[4]10=\sqrt[4](8\times10)=\sqrt[4]80=\sqrt[4](16\times5)=\sqrt[4]16\times\sqrt[4]5=2\sqrt[4]5$ We know that $\sqrt[4]16=2$, because $(2)^{4}=2\times2\times2\times2=16$.
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