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If you have the log of a modulus, (like after integration), how do the log laws work?

So if you have $a\ln\left|2x-3\right|$ does it become: $\ln\left|(2x-3)^a\right|$ or $\ln(\left|2x-3\right|)^a$, or does it not matter?

And what about $\ln\left|x+1|\right| - \ln(10)$? does it become $\ln\left|\frac{x+1}{10}\right|$ or $\ln\frac{\left|x+1\right|}{10}$?

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you cant define $\ln(-10)$ –  Maisam Hedyelloo Jan 27 '13 at 17:17
    
of course, thanks :) –  Jonathan. Jan 27 '13 at 17:18

2 Answers 2

Hint: For any $a$, it is obvious that $|2x-3|^a$ and $|(2x-3)^a|$ both are positive. So excluding $x\neq\frac{3}{2}$, both $\ln$'s can be evaluated. Moreover, both $|2x-3|^a$ and $|(2x-3)^a|$ are the same, provided we consider them in the same domains. For the other as @Maisam commented, we are not allowed to write and think about $\ln(-10)$ since logarithms are considered only of positive real numbers. If we have $\ln(10)$ instead, what have you tried at the last of the question looks coorect.

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Nice hint, nice points! +1 –  amWhy Feb 1 '13 at 1:26
    
@amWhy: Thanks after 9 exhausting hours. –  Babak S. Feb 1 '13 at 6:02

Hints:

  1. I assume you are asking for $a$ and $x$ $\in \mathbb{R}$ and restrictions on $a$

  2. Go back to first principles for $| f(x) |$

For example, take $f(x)$ positive and take it negative domain, what does the absolute value do?

Now, for each case, how would the exponent work?

You could also try plotting your attempts, but it is much more preferable to do the analysis above and derive which case is correct.

Regards

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@BabakSorouh: Thank you! I editted my answer. ;-) Regards –  Amzoti Jan 27 '13 at 17:53
    
Great suggestions and helpful hints ;-) –  amWhy May 6 '13 at 0:31

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