how to find $\ln(-e)$ I don't really understand the step by step procedure on how to calculate this? That $-$ confuses me, because I don't understand how there can be an $\ln$ of a negative number. Can someone please explain? Thanks
 A: It looks like you are referring to the complex logarithmic function which is defined as:
$\ln(z) = \ln(r) + i(\theta+2\pi k)$, where $z = re^{i\theta}, k\in \mathbb Z$.
Applying that here, since $-e = e\times e^{i\pi}$, then $\ln(-e) = \ln(e)+ i(\pi+ 2\pi k) = 1 + i(\pi+ 2\pi k)$.
A: Indeed, from Euler's equation and more specifically Euler's identity you can define a logarithm of a negative number.
If you had $\ln (-1)$, all you would have to do is add it to $\ln (n)$ to get $\ln (-n)$. The axioms defined for logs remain the same.
Euler's equation: $e^{i\pi}=-1 \implies \ln({-1})=i\pi$
I know there's a periodicity of $2\pi$ that I haven't accounted for yet but I think that's a fair enough point made.
I assume you're not that familiar with complex numbers so like any good fellow, I wish you good tidings on the yellow brick road you will trod in complex numbers.
What they are: https://www.youtube.com/playlist?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF
Why $e$ shows up: https://youtu.be/mvmuCPvRoWQ
