## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises: 54

#### Answer

$\log_5 (x+1)$

#### Work Step by Step

$Combine$ $the$ $expression$: $\log_5 (x^2-1)$ - $\log_5 (x-1)$ Apply the Second Law of Logarithms $\log_5 (x^2-1)$ - $\log_5 (x-1)$ = $\log_5 \frac{(x^2-1)}{(x-1)}$ We know that x$^2$-1 is a difference of two squares since x$^2$ and 1 are perfect squares. Factor (x$^2$-1) $\log_5 \frac{(x-1)(x+1)}{(x-1)}$ Simplify $\log_5 (x+1)$

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