# 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

• Why don’t you write, $\dfrac{1}{2} \ln \dfrac{(x-2)^2}{x-1}=\dfrac{1}{2}\ln\left(x-2\right)^2- \dfrac{1}{2}\ln\left(x-1\right)$, and proceed.
– YNK
Nov 16, 2022 at 16:17
• Can you write the steps of integration$?$ Nov 16, 2022 at 16:20

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

• Typo never leaves me Nov 16, 2022 at 16:14
• That's not completely correct. The domain of the first expression is $(2,\infty)$ and the domain of the last expression is $(1,2) \cup (2,\infty)$. Nov 16, 2022 at 18:56
• The second and third expressions aren't necessarily equal because they have different domains. But since this is part of an indefinite integral, I think this answer is alright. Nov 16, 2022 at 22:47

If $$a,b>0$$ and $$p\in\mathbb R$$, then you can apply the well-known logarithm rules:

1. $$\ln a-\ln b=\ln \dfrac ab$$

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

However, what I want to say in this answer is a bit different.

Note that, if $$1, then the identity fails, because $$\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.

• You seem to be the only one who has realized the problem with the domain of the two expressions in question. +1 Nov 16, 2022 at 18:53

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..