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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

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    $\begingroup$ 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. $\endgroup$
    – YNK
    Nov 16, 2022 at 16:17
  • $\begingroup$ Can you write the steps of integration$?$ $\endgroup$
    – Vanessa
    Nov 16, 2022 at 16:20

3 Answers 3

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

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  • $\begingroup$ Typo never leaves me $\endgroup$
    – Vanessa
    Nov 16, 2022 at 16:14
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    $\begingroup$ That's not completely correct. The domain of the first expression is $(2,\infty)$ and the domain of the last expression is $(1,2) \cup (2,\infty)$. $\endgroup$ Nov 16, 2022 at 18:56
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    $\begingroup$ 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. $\endgroup$ Nov 16, 2022 at 22:47
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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 $1<x<2$, 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.

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

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