Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.4 - Properties of Logarithms - Exercise Set - Page 713: 91


a) $\dfrac{1}{2}$ b) $\log_{25}[\dfrac{x(x^2-1)^5}{5(x+1)}]$

Work Step by Step

a) Since, $\log_{25} {5}=\log _{25}(25)^{1/2}=\dfrac{1}{2}$ b) Now $\log_{25} x+5\log_{25}(x^2-1)-\log_{25}(x+1)-\dfrac{1}{2}=\log_{25} x+5\log_{25}(x^2-1)-\log_{25}(x+1)-\log_{25} {5}$ [From part (a), we have $\log_{25} {5}=\dfrac{1}{2}$] or, $=[\log_{25}x+\log_{25}(x^2-1)^5]-[\log_{25}(x+1)+\log_{25}5]$ or, $=\log_{25}[x(x^2-1)^5]-\log_{25}[5(x+1)]$ or, $=\log_{25}[\dfrac{x(x^2-1)^5}{5(x+1)}]$ Hence, a) $\dfrac{1}{2}$ b) $\log_{25}[\dfrac{x(x^2-1)^5}{5(x+1)}]$
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