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: 44

Answer

$a)$ $\Big(\dfrac{s^{2}t^{-4}}{5s^{-1}t}\Big)^{-2}=\dfrac{25t^{10}}{s^{6}}$ $b)$ $\Big(\dfrac{xy^{-2}z^{-3}}{x^{2}y^{3}z^{-4}}\Big)^{-3}=\dfrac{x^{3}y^{15}}{z^{3}}$

Work Step by Step

$a)$ $\Big(\dfrac{s^{2}t^{-4}}{5s^{-1}t}\Big)^{-2}$ $\Big(\dfrac{s^{2}t^{-4}}{5s^{-1}t}\Big)^{-2}=\Big(\dfrac{5s^{-1}t}{s^{2}t^{-4}}\Big)^{2}=\dfrac{25s^{-2}t^{2}}{s^{4}t^{-8}}=25s^{-2-4}t^{2-(- 8)}=...$ $...=25s^{-6}t^{10}=\dfrac{25t^{10}}{s^{6}}$ $b)$ $\Big(\dfrac{xy^{-2}z^{-3}}{x^{2}y^{3}z^{-4}}\Big)^{-3}$ $\Big(\dfrac{xy^{-2}z^{-3}}{x^{2}y^{3}z^{-4}}\Big)^{-3}=\Big(\dfrac{x^{2}y^{3}z^{-4}}{xy^{-2}z^{-3}}\Big)^{3}=\dfrac{x^{6}y^{9}z^{-12}}{x^{3}y^{-6}z^{-9}}=...$ $...=x^{6-3}y^{9-(-6)}z^{-12-(-9)}=x^{3}y^{15}z^{-3}=\dfrac{x^{3}y^{15}}{z^{3}}$
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