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I am currently struggeling with a little logarithm problem. A basic rule of the logarithm is $$\ln(a\cdot b) = \ln(a) + \ln(b)$$ Now I have $$\ln{(-1 \cdot (a-1))}$$ which I formed to $$\ln{(-1)} + \ln{(a-1)}$$ But this seems not to be correct. I am confused. What have I overlooked?

Jimmy R.
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Fruchtzwerg
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    The rule works for positive $a$ and $b$. There are ways to define logs for negative numbers, but you have to be careful. – T.J. Gaffney Dec 02 '16 at 11:41
  • "The rule works for positive a and b. There are ways to define logs for negative numbers, but" then the rule does not apply anymore to every a and b. – Did Dec 02 '16 at 11:46
  • Why is the WA page you are linking to supposed to show that $\log(-(a-1))\ne\log(-1)+\log(a-1)$? Actually, once suitable definitions and restrictions are agreed upon, it shows exactly that. – Did Dec 02 '16 at 11:51
  • @Did The rule would not work for a branch cut, but it should still work for a multivalued function or the log defined on a Riemann surface. – T.J. Gaffney Dec 02 '16 at 11:53
  • @Gaffney I know that, thanks. :-) The point is to determine the framework the OP is operating in, if one must explain. – Did Dec 02 '16 at 11:54
  • @Gaffney And actually your comment is even offtopic, please read the WA page. – Did Dec 02 '16 at 11:55
  • Let me rephrase my previous comment (and hope for an answer to it from the OP, this time): Why is the WA page you are linking to, supposed to show that log(−(a−1))≠log(−1)+log(a−1)? Actually, once suitable definitions and restrictions are agreed upon, it shows exactly that log⁡(−(a−1))=log⁡(−1)+log⁡(a−1). – Did Dec 02 '16 at 11:58
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    I interpreted the $\neq$ because typically the output is "true" like: https://www.wolframalpha.com/input/?i=ln(a+*+b)+%3D+ln(a)+%2B+ln(b) – Fruchtzwerg Dec 02 '16 at 12:01
  • But WA says that $\log(-(a-1))=i\pi+\log(a-1)$, which is true for example if $a-1$ is a positive real number and if one chooses the so called principal determination of the complex logarithm so, once again, why do you think WA's answer is showing that log(−(a−1))≠log(−1)+log(a−1)? The fact that WA declares log(ab)=log(a)+log(b) to be true when a and b are positive real does not imply this identity cannot hold for other as and bs. – Did Dec 02 '16 at 16:09

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Clearly Wolfram re-wrote to $\ln(1-a)$. However, note that one way of writing the relation is \begin{align} \ln (-(a-1)) &= \ln i^2 + \ln(a-1) \\ &= 2 \ln i + \ln (a-1) \\ &= i \pi + \ln (a-1) \end{align} the last step is achieved by noting that $e^{i \pi /2} = i$

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    Much more probably, WA considers directly that $\log(-1)=i\pi$, which is correct if one sticks to the (complex) logarithm based on the so-called principal determination of the argument, with values in $(-\pi,\pi]$. Thus, no need to invoke $\ln(i^2)=2\ln i$ here. – Did Dec 02 '16 at 16:19
  • Ah, clean slate. Nice answer. +1 – DonAntonio Dec 02 '16 at 16:20
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$ \ln $ is only defined for positive numbers. The rule

$\ln(a * b) = \ln(a) + \ln(b)$

therefore holds for $a,b>0$. We have $-1<0$

Thats all.

Fred
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  • Sorry but the fact that the rule only applies for a and b positive and real is (not true and) not due to the fact that log is only defined for positive numbers (which is not true either). – Did Dec 02 '16 at 11:48
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For a real valued logarithm the statement isnt true because negative logarithms are not defined. In the case of complex logarithm suppose that $a\in\Bbb C$, then

$$\ln (1-a)=\ln |1-a|+i\arg (1-a)$$

but

$$\ln (-1)+\ln (a-1)=\ln |1|+i\arg(-1)+\ln |a-1|+i\arg(a-1)=\\=0+i\pi+\ln |1-a|+i\arg(a-1)$$

where $|\arg(1-a)-\arg(a-1)|=\pi$. Because $\arg$ is the principal argument of $(1-a)$ and is defined in $(-\pi,\pi]$ observe that the result could be different in both cases, i.e.

$$\ln(1-a)=\ln |1-a|+\color{red}{i\arg (1-a)}\neq \ln |1-a|+\color{red}{i(\pi+\arg (a-1))}=\ln(-1)+\ln (a-1)$$

Choosing, by example, $\arg(1-a)=0$ then we have that $\arg(a-1)=\pi$ but then we have that $0\neq 2\pi$. However, using the set valued complex logarithm defined as

$${\rm Log}:\Bbb C\to\mathcal P(\Bbb C),\quad z\mapsto \ln|z|+i(\arg(z)+2\pi\Bbb Z)$$

we have that the equality holds:

$${\rm Log}(1-a)={\rm Log}(-1)+{\rm Log}(a-1)$$

Masacroso
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