# Euler's Identity - What's the falacy?

We know that $$e^{i\pi} = -1$$. So, $$\ln(-1) = i\pi$$. Well, $$\ln(1) = 0$$. So, I guess we can write $\ln(-1) + \ln(-1) = \ln(-1\cdot -1) = \ln(1) = 0$. But $\ln(-1) + \ln(-1) = i\pi + i\pi = 2i\pi$

What's the falacy?

Note by Gabriel Laurentino
4 years ago

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the identity you used is applicable for numbers belonging to the set (0,infinity) only

- 4 years ago

May be because the properties of ln is applicable only for numbers within its domain..

- 4 years ago

I thought the same. Thanks!

- 4 years ago