How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$ 
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$

Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} dx$ but I need to combine the expression to the right hand side to continue with the steps in the question I am attempting

$$\ln(x-2)=\ln(((x-2)^2)^{1/2})=\frac{1}{2}\ln(x-2)^2$$
Can you continue?
