Why is $\ln{(-1 \cdot(a-1))} \ne \ln{(-1)} +\ln{(a-1)}$

Question

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?

Answer 1

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 $e^{i \pi /2} = i$

Answer 2

$ \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.

Answer 3

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:

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