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 23: 62

Answer

$a)$ $16b^{3/4}$ $b)$ $45a^{2}$

Work Step by Step

$a)$ $(4b)^{1/2}(8b^{1/4})$ Raise each factor inside the first parentheses to the power of $\dfrac{1}{2}$: $(4b)^{1/2}(8b^{1/4})=(4^{1/2})(b^{1/2})(8b^{1/4})=...$ Knowing that $4^{1/2}=\sqrt{4}=2$, evaluate the products and simplify: $...=(2)(b^{1/2})(8b^{1/4})=16b^{1/2+1/4}=16b^{3/4}$ $b)$ $(3a^{3/4})^{2}(5a^{1/2})$ Square each factor inside the first parentheses: $(3a^{3/4})^{2}(5a^{1/2})=(3^{2})(a^{3/2})(5a^{1/2})=(9)(a^{3/2})(5a^{1/2})=...$ Evaluate the products and simplify: $...=45a^{3/2+1/2}=45a^{4/2}=45a^{2}$
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