Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises - Page 134: 17

Answer

$a)$ $b^{14}$ $b)$ $12xy^{8}$

Work Step by Step

$a)$ $(a^{2})^{-3}(a^{3}b)^{2}(b^{3})^{4}$ Evaluate the powers: $(a^{2})^{-3}(a^{3}b)^{2}(b^{3})^{4}=(a^{-6})(a^{6}b^{2})(b^{12})=...$ Evaluate the products: $...=a^{-6+6}b^{2+12}=a^{0}b^{14}=b^{14}$ $b)$ $(3xy^{2})^{3}(\frac{2}{3}x^{-1}y)^{2}$ Evaluate the powers: $(3xy^{2})^{3}(\frac{2}{3}x^{-1}y)^{2}=(27x^{3}y^{6})(\frac{4}{9}x^{-2}y^{2})=...$ Evaluate the product: $...=(27)\Big(\dfrac{4}{9}\Big)x^{3-2}y^{6+2}=12xy^{8}$
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