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
 A: $$\ln (x-2) - \frac{1}{2} \ln (x-1)$$
$$=\frac{2\ln (x-2) -\ln (x-1)}{2}$$
$$=\frac{\ln (x-2)^2 -\ln (x-1)}{2}$$
$$=\frac{\ln\frac{ (x-2)^2}{ x-1}}{2}$$
A: If $a,b>0$ and $p\in\mathbb R$, then you can apply the well-known logarithm rules:

*

*$\ln a-\ln b=\ln \dfrac ab$


*$\ln a^p=p \ln a$.

However, what I want to say in this answer is a bit different.
Note that, if $1<x<2$, then the identity fails, because $\ln \frac {(x-2)^2}{x-1}$ is defined, since $\frac {(x-2)^2}{x-1}>0$; but $\ln (x-2)-\frac 12 \ln (x-1)$ is obviously undefined.
To summarize, although the domain of $\ln (x-2) - \frac 12 \ln (x-1)$  is $(2,+\infty)$, the domain of $\ln \frac{(x-2)^2}{x-1}$ is the larger set: $(1,2)\cup (2,+\infty)$.
This difference is not related to the invalidity of the logarithm rules, but shows that when applying the real-valued logarithm rules, these rules are valid only in the set they are defined.
Therefore, remember that
$$\ln (x-2) - \frac 12 \ln (x-1) = \frac 12 \ln \frac{(x-2)^2}{x-1}$$
holds, iff $x>2$.
Maybe, it's a small detail, but I wanted to note it anyway.
A: $$\ln(x-2)=\ln(((x-2)^2)^{1/2})=\frac{1}{2}\ln(x-2)^2$$
Can you continue?
A: 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..
