Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 8 - Section 8.2 - Multiplying and Dividing Radicals - Exercise Set: 104

Answer

$2\sqrt[5]{2}$

Work Step by Step

Using $\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}$ or the product rule of radicals, the given expression, $ \sqrt[5]{16}\cdot\sqrt[5]{4} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[5]{16(4)} \\\\= \sqrt[5]{64} \\\\= \sqrt[5]{32\cdot2} \\\\= \sqrt[5]{(2)^5\cdot2} \\\\= 2\sqrt[5]{2} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.