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
$\log_5 (x+1)$
