Answer
$\log_4 \left(\dfrac{x-1}{(x+1)^4} \right)$
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,
$\log_4 (x^2-1)-5 \log_4 (x+1) = \boxed{\log_4 \left(\dfrac{x-1}{(x+1)^4} \right)}$
