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 - Mid-Chapter Check Point - Page 541: 20

Answer

$\sqrt[6] {32}$.

Work Step by Step

The given function is $=\sqrt{2}\cdot \sqrt[3] 2$ Rewrite as an exponential expression. $=(2)^{\frac{1}{2}}\cdot (2)^{\frac{1}{3}}$ Use power rule. $=(2)^{\frac{1}{2}+\frac{1}{3}}$ Simplify. $=(2)^{\frac{3}{6}+\frac{2}{6}}$ $=(2)^{\frac{3+2}{6}}$ $=(2)^{\frac{5}{6}}$ Rewrite in radical notation. $=\sqrt[6] {2^5}$ Simplify. $=\sqrt[6] {32}$.
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.