Algebraic Manipulation of Euler's Identity Leads to a Strange Result

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I was playing around with Euler's identity the other day. I came across something that seems contradictory to everything else I know, but I can't really explain it.

I started with

[tex]e^{i\pi} = -1[/tex].

I rewrote this as

[tex]ln[-1] = i\pi[/tex]

Multiplying by a constant, we have

[tex]kln[-1] = ki\pi[/tex]

and using log properties I arrived at

[tex]ln[-1^{k}] = ki\pi[/tex]

Now if I set k equal to any even number, I have

[tex]ln[1] = 0 = ki\pi[/tex]

This seems to imply that [tex]i\pi[/tex] is 0, however it is not. Furthermore, any even value of k gives the same answer. Why is this?
 

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ln(-1) = (2/2)*ln(-1) = (1/2)*2*ln(-1) = (1/2)*ln(-1^2) = (1/2)*ln(1) = (1/2)*0 = 0

Therefore, ln(-1) = 0.

In order for properties of logs to work, you must be taking the log of a valid number. You can not take the log of a negative number, therefore, when you apply the properties of logs to something that is undefined, you get ludicrousness.
 
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Thanks!
 

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