Do branches of logarithms matter over real numbers? Suppose I start with $ln(x)$ where $x$ is a strictly real variable.  Other complex branches of $ln(x)$ are merely separated by an imaginary constant $2 i \pi k$. This seems to suggest that for all branches, the real line $x$ is mapped the same while it is only the imaginary component that changes. If this is true, then for purely real inputs of $x$, is it true that $ln(x) = ln(x)+2 i \pi k$? I feel like it shouldn't be quite that simple. and yet if I take the exponent, $$e^{ln(x)+2 \pi i k} = e^{ln(x)}e^{2 \pi i k}= e^{ln(x)}(1) = x$$ $x$ is always returned. If that's true, then does it follow that
$$e^{log(x)}+e^{log(x)+2 i \pi k} = x+x = 2x?$$ with no special domain restrictions beyond x being positive? 
Also, what if k isn't an integer? 
 A: If it seems too simple, remember it's not much different than asking for $\arccos(x)$.  When you've found one answer, you can add any integer multiple of $2\pi$ and get another answer.
This is because $\cos(x)$ is $2\pi$-periodic along the real line.  In other words, $\cos(x+2\pi k) = \cos(x)$ for integer $k$.  By comparison, $e^{z}$ is $2\pi$-periodic but "only in the complex direction."  If you take $e^{z+2\pi k}$ you won't get the same thing as $e^{z}$.  But, if you take $e^{z+2\pi i k}$ you will get the same thing.  

Edit: let me try to directly answer each of the questions directly, at risk of getting into semantics.


*

*Does $ln(x) = ln(x)+2 i \pi k$ for real $x$?  No.  The same way $\sin^{-1}(1/2) \neq \sin^{-1}(1/2)+2\pi$.  It happens that $\pi/6$ and $13\pi/6$ define "coterminal" angles, but they are different numbers.  Since the principal branch of $\sin$ is $[-\pi/2,\pi/2]$ I say that $\sin^{-1}(1/2) = \pi/6$ even though $\sin(13\pi/6) = 1/2$ as well.  The principal branch for $\ln$ is a 2d region since it is in the complex plane.  It is where $-\pi \leq \operatorname{Im}(z) \leq \pi$ and $z \neq 0$.

*Is it true that $$e^{log(x)}+e^{log(x)+2 i \pi k} = x+x = 2x?$$  Yes.  Just like $$\sin(\sin^{-1}(x)) + \sin(\sin^{-1}(x) + 2\pi) = x + x = 2x.$$  Neither $e^{z}$ nor $\sin(x)$ are one-to-one functions, so there are multiple inputs that produce the same results.  

*What if $k$ isn't an integer?  $e^{\ln(x) + 2\pi i k}$ gives you $x$ (positive real number, lying on the real axis) rotated by $k$ revolutions about the origin in the complex plane.  So if $k$ is an integer, you have a whole number of revolutions and you end up where you started.  If $k$ is not an integer, you will end up at some other $z\in\mathbb{C}$ where $|z| = x$.
