Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises - Page 22: 48

Answer

$a)$ $\sqrt[4]{x^{4}y^{2}z^{2}}=x\sqrt{yz}$ $b)$ $\sqrt[3]{\sqrt{64x^{6}}}=2x$

Work Step by Step

$a)$ $\sqrt[4]{x^{4}y^{2}z^{2}}$ Take the 4th root of each factor. $\sqrt[4]{x^{4}y^{2}z^{2}}=(\sqrt[4]{x^{4}})(\sqrt[4]{y^{2}z^{2}})=x\sqrt{yz}$ $b)$ $\sqrt[3]{\sqrt{64x^{6}}}$ Evaluate $\sqrt{64x^{6}}$ $\sqrt[3]{\sqrt{64x^{6}}}=\sqrt[3]{8x^{3}}=...$ Take the cubic root of each factor: $...=(\sqrt[3]{8})(\sqrt[3]{x^{3}})=2x$
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