Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.5 Properties of Logarithms - 4.5 Assess Your Understanding - Page 331: 61


$\log_4 \left(\dfrac{x-1}{(x+1)^4} \right)$

Work Step by Step

Recalll: $\log_a M^r = r \log_a M$ Using the rule above gives: $5 \log_4 (x+1) = \log_4 (x+1)^5$ Thus, $\log_4 (x^2-1)-5 \log_4 (x+1) = \log_4 (x^2-1)-\log_4 (x+1)^5$ Recall also that: $\log_a\left(\dfrac{M}{N}\right) = \log_a M-\log_a N$ Hence, $\log_4 (x^2-1)-\log_4 (x+1)^5 = \log_4 \left(\dfrac{x^2-1}{(x+1)^5} \right)$ Factor the numerator using the formula $a^-b^2=(a-b)(a+b)$ to obtain: \begin{align*} \require{cancel} \log_4 \left(\dfrac{x^2-1}{(x+1)^5} \right) &= \log_4 \left(\dfrac{(x-1)(x+1)}{(x+1)^5} \right) \\\\ &= \log_4 \left(\dfrac{(x-1)\cancel{(x+1)}}{(x+1)^\cancel{5}{^4}} \right) \\\\ &= \log_4 \left(\dfrac{x-1}{(x+1)^4} \right) \end{align*} Therefore, $\log_4 (x^2-1)-5 \log_4 (x+1) = \boxed{\log_4 \left(\dfrac{x-1}{(x+1)^4} \right)}$
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