Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 50


$\frac{1}{3}\cdot \log_5{(x^2+1)}-\log_5{x+1}-log_5{x-1}$

Work Step by Step

Recall: (1) $\sqrt[m]{a}=a^{\frac{1}{m}}$ (2) $\log_a {x^n}=n\cdot \log_a {x}$. (3) $\log_a{xy}=\log_a{x} +\log_a{y}$ (4) $\log_a{\frac{x}{y}}=\log_a{x} -\log_a{y}$ Use Rule (4) to obtain $\log_5{\frac{\sqrt[3] {x^2+1}}{x^2-1}}=\log_5{\sqrt[3] {x^2+1}}-\log_5{(x^2-1)}$. Use Rule (2) to obtain $\log_5{\sqrt[3] {x^2+1}}-\log_5{(x^2-1)}=\log_5{(x^2+1)^{\frac{1}{3}}}-\log_5{(x+1)(x-1)}.$ Use Rule (3) to obtain: $\log_5{(x^2+1)^{\frac{1}{3}}}-\log_5{(x+1)(x-1)}=\frac{1}{3}\cdot \log_5{(x^2+1)}-\log_5{x+1}-log_5{x-1}$
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