Why is $\ln{(-1 \cdot(a-1))} \ne \ln{(-1)} +\ln{(a-1)}$ 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?
 A: 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 $e^{i \pi /2} = i$
A: $ \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.
A: 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:
$${\rm Log}(1-a)={\rm Log}(-1)+{\rm Log}(a-1)$$
