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

HINT:
Multiply both side by $2.$ You need to show that:
$$2 \ln (x-2) -2 \cdot \frac{1}{2} \ln (x-1) = 2\cdot\frac{1}{2} \ln \frac{(x-2)^2}{x-1}$$
Use the rule
$$ \ln x - \ln y= \ln~(x/ y)$$
Next get rid of $\ln ..$ ( by exponentiation of both sides). What went before integration is not asked, but only about last step redoing..
