Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises - Page 359: 36

Answer

$\log_{2}(\dfrac{s^{5}}{7t^2})=5\log_{2}s-\log_{2}7-2\log_{2}t$

Work Step by Step

$\log_{2}(\dfrac{s^{5}}{7t^2})$ We begin the expanding process, noting that a division can be expanded as a substraction: $\log_{2}(\dfrac{s^{5}}{7t^2})=\log_{2}(s^{5})-\log_{2}(7t^{2})$ We expand the product in $\log_{2}(7t^{2})$ as a sum: $...=\log_{2}(s^{5})-[\log_{2}(7)+\log_{2}(t^{2})]=...$ $...=\log_{2}(s^{5})-\log_{2}(7)-\log_{2}(t^{2})=...$ The exponents in $\log_{2}(s^{5})$ and $\log_{2}(t^{2})$ can be taken to the front of the logarithm to multiply: $...=5\log_{2}s-\log_{2}7-2\log_{2}t$
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