Answer
$\frac{4}{3}ln(x-4)-\frac{2}{3}ln(x+1)-\frac{2}{3}ln(x-1)$.
Work Step by Step
$ln[\frac{(x-4)^2}{x^2-1}]^{2/3}=\frac{2}{3}ln\frac{(x-4)^2}{(x+1)(x-1)}=\frac{2}{3}[2ln(x-4)-ln(x+1)-ln(x-1)]=\frac{4}{3}ln(x-4)-\frac{2}{3}ln(x+1)-\frac{2}{3}ln(x-1)$.
