What is wrong with this reasoning?
$\ln(4)=\ln((-2)^2)=2\ln(-2)$
We can obvisouly work out $\ln(4)$, but we can't with $\ln(-2)$. The reason I am asking is because I have a situation in another problem where I arrived to something similiar to $2\ln(-2)$ and I don't know if I should just write $\ln(4)$.
On
I am pretty sure you did something wrong with your calculation and here is the proof assume $\exists$ number $x<0$ so that $\ln(x)=y$ with $y\in\mathbb{R}$ then
$$\ln(x)=y$$
$$x=e^y $$ this leads us that $e^y$ is negative which is obviously a contradiction to its definition as it is always postive .
This is actually a reason why mathematicians used to think that logarithms of negative numbers are equal to logarithms of positive numbers (it used to be somewhere on the Wikipedia, I'll try to find it). However, logarithms of negative numbers can indeed be defined, but they are not so simple.
$$4=e^{2\ln(-2)}=e^{2\ln(2)}$$
This equality is true if we define $\ln(-2)$, and such is explained in this post. However, the inverse is not true:
$$\ln(-2)\ne\ln(2)$$
This is because there just so happens to be the case that
$$e^{2\pi i}=1$$
And
$$2\ln(-2)=2\ln(2)+2k\pi i$$
for some whole number $k$.
Thus, we end up with
$$e^{2\ln(-2)}=e^{2\ln(2)+2k\pi i}=e^{2\ln(2)}e^{2k\pi i}=e^{2\ln(2)}$$