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I'm having some problems explaining myself the following ambiguity.

According to logarithm rules:

  • $\ln6=\ln(2\cdot3)=\color\red{\ln2+\ln3}$
  • $\ln6=\ln((-2)\cdot(-3))=\ln(-2)+\ln(-3)=\color\red{\ln2+\ln3+2i\pi}$

Therefore, $\ln2+\ln3=\ln2+\ln3+2i\pi$.

Therefore, $2i\pi=0$.

Therefore, $i=0$.

I realize I must be doing some kind of rookie mistake like "$x=2x\implies1=2$".

So I would appreciate if anyone could point it out.

Thanks

barak manos
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    $\ln ab=\ln a+\ln b$ does not hold when $a,b<0$. – S.C.B. May 08 '16 at 07:49
  • I hope you know that logarithm of negative number does not exist.ou just Check the graph of log to any base. – Murtuza Vadharia May 08 '16 at 07:51
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    @MurtuzaVadharia It exists. Check here. It's just complex. It's defined. – S.C.B. May 08 '16 at 07:54
  • @barak If you want to talk of complex logarithm please do tag your question accordingly, otherwise the usual, standard assumtion is you mean the real one. – DonAntonio May 08 '16 at 07:57
  • @MXYMXY It is defined only within the realm of complex functions. – DonAntonio May 08 '16 at 07:58
  • @Joanpemo Well, it still exists, does it not (in the realm of complex functions)? – S.C.B. May 08 '16 at 07:58
  • @MXYMXY Thank you. Well, yes if the asker meant that, otherwise we're reduced to work within the real numbers and then the logarithm, in any base, of a negative number isn't defined. – DonAntonio May 08 '16 at 07:59
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    @Joanpemo The tag says "questions related to real and complex logarithms". It is fine. – MathematicsStudent1122 May 08 '16 at 08:00
  • @MathematicsStudent1122 Thank you. Yes, and that's possible to know only if one gets into the tag's definition, which doesn't appear in the open question. I think that it could probably be better to make a special tag for the complex logarithm as some people assume by default that it is the real one. – DonAntonio May 08 '16 at 08:03
  • @Joanpemo: Thanks for your comments. Apart from the tag being associated with complex numbers, there is also an explicit "$i$" in the question, so I think it's clear enough that the question applies to the realm of complex numbers. I didn't put the explicit tag, because the question itself is about logarithms, not about complex numbers (I could just as well tag it under arithmetic operations or something similar, because there are a couple of "$+$"s in there). But thank you for reading through, I appreciate it nonetheless. – barak manos May 08 '16 at 08:39
  • @barakmanos Thank you. The problem of knowing what the tag includes and what not...Anyway, the answer talks of positive elements, which make not sense in general within the complex numbers (the intention, though, seems to be for real complex numbers and etc.). Also, $;\log(ab)=\log a+\log b;$ still remains true in the complex numbers if we agree on always keeping the argument withing the domain that remains after a branch cut is provided. – DonAntonio May 08 '16 at 08:42
  • @Joanpemo: Thanks. So, are you saying that there's a problem with the answer below? If it holds for complex numbers $x$ and $y$, then the expression "$x,y>0$" is kind of undefined. – barak manos May 08 '16 at 08:51
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    @barakmanos I don't think there's a problem with the answer below: it all depends on what we're talking about. If we restrict ourselves to real arguments for the complex logarithm then we're done, and we don't even need to remark the positive real numbers as the usual branch cut of the complex logarithm is the non-positive real axis... Now, if we want to make the argument as general as possible then we could add what I wrote above...I think it all depends on what context we're working. – DonAntonio May 08 '16 at 08:58
  • @barakmanos Similar consideration: $\sqrt{-m}=i\sqrt{m}$ is true only when $m>0$. Otherwise $+\sqrt{4}=\sqrt{(-1)(-4)}=i\sqrt{-4}=i\times 2i=-2$ – Ng Chung Tak May 08 '16 at 14:17
  • @NgChungTak: But when calculating the $n$th root, we may have $n$ different (possibly complex) solutions. In your example, when calculating $\sqrt[2]{4}$, we have $2$ different solutions ($\pm2$). So I don't see any problem in this example. With logarithm rules, on the other hand, I am not aware of anything equivalent which could be used in similar reasoning. – barak manos May 08 '16 at 14:25
  • @barakmanos Yes, Wolframalpha do gives a complex value of $\sqrt[3]{-1}$ which regard as the principal root. Usually, we use Log as complex logarithm while ln is for real domain only so it's single-valued. Last but not least, the log identity comes from laws of indices. – Ng Chung Tak May 08 '16 at 14:31
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    @barakmanos Actually, you're not rookie at all. You play around the mathematical paradoxes very well. An interesting book: http://www.prometheusbooks.com/images/MagnificentMistakesinMath.jpg – Ng Chung Tak May 08 '16 at 14:38

1 Answers1

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Your error comes from a misunderstanding in the properties of logarithims.

$$\displaystyle\log_a(xy)=\log_ax+\log_ay$$

The above identity holds only for $x,y>0$ and $a\not =1$, $a>0$.

S.C.B.
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