help with log of negative numbers

Was just curious and surfing math stuff and I stumbled across deriving ln(-1) = ipi with e^(i pi) +1 = 0

I understand how this makes sense, but I have an inconsistency I came across I need an explaination for.

ln(-1) = i pi
2ln(-1) = ln[(-1)^2] = ln 1 = 0 != 2i pi
This doesnt make much sense to me, there must be an error in reasoning. Any help? Also are there some real life applications of logs of negative values, or is it just pure theory?
you can't take logs of non-positive real numbers until you learn in complex analysis course how to deal with it. and even if you know how to deal with it, many rules are still "broken" or at least different from positive real world. just keep in mind all the rules you learn now only applies to positive reals and if you ever want to generalize you need to prove it before you can use it
Let's simplify your problem a bit:
You showed that 2ln(-1) = ln(1)

That statement an be written as e^(2i*pi) = e^(0).
Are you sure that we can conclude that 2i*pi = 0?

Let us give an easier example to understand.
1⁵=1⁶ does not mean that 5=6 because functions like 1^(x),0^(x), -1^(x) can have the same values for different values of x (eg -1²=-1⁴=1)

Let us take this specific example,
e^(i*pi*2) = e^(i*pi*0)  => (e^(i*pi))^(2)=(e^(i*pi))^(0)
Substituting e^(i*pi)=(-1); We have -1^(2) = -1^(0)
We already know from a previous example that -1^(x) can have same value for different x so the same applies for e^(i*pi*x).
Hence, we can avoid ln(-1) as it can have multiple different answers. Better to avoid it as we do for log base 1, 0.

This may not be the best way to approach this problem nor the most accurate way either. I haven't ever dealt with complex analysis but this is the way I like to see it and I think this is intuitive for a layman or a novice like me.