# What is wrong with this logarithm?

#### rahulpsharma

Joined Sep 5, 2010
60
I am unable to spot the error in this calculation...!! What's wrong here...??

Thanks and regards,
Rahul

#### ericgibbs

Joined Jan 29, 2010
19,076
hi,
Check out this video.
E

#### Wolframore

Joined Jan 21, 2019
2,610
The mistake is here

#### WBahn

Joined Mar 31, 2012
30,286
So what is the mistake?

Is it that (a^x)^2 isn't a^(2x)? Or is it that (-1)^2 isn't 1?

If you are being careful, it should be noted that

$$e^{ix} \; = \; e^{i\left(x+2n\pi\right)}$$

I would say that the problem is here:

$$ln \left( e^{2 \pi i} \right) \; = \; 2 \pi i$$

This should be

$$ln \left( e^{2 \pi i} \right) \; = ln \left( e^{2n \pi i} \right) \; = \; 2n \pi i \;\; \text{(n is an integer)}$$

So now in the next set you have

$$2n \pi i \; = \; 0$$

which is true if we choose n=0

#### Wolframore

Joined Jan 21, 2019
2,610
solve for (e^(i pi))^2

the issue is you're using complex "i"

#### Wolframore

Joined Jan 21, 2019
2,610
Would it help to use Euler's identity?