I know that e^(2pi i)=1 so by taking the natural logarithm on both sides ln(e^(2pi i))=ln(1)=0 however, why isn't this 2pi i as expected? Is it beacuse logarithms can only provide real values?

The exponential function is not one-to-one in complex numbers, which is an issue because $\exp ( z + 2 \pi i ) = \exp z$. As a result, it is no longer possible to define a suitable inverse for it. For example, if I only tell you a number's exponential, you won't know how many multiples of $2 \pi i$ I've added.
This might be fixed in a few different ways. One is to think of the logarithm as a relation rather than a multi-valued function, so that $\log 1 "=" 2 \pi i k$ for every integer $k$. This can be useful at times, such as when solving equations.
Another option is to define a logarithm rather than the logarithm itself, which entails selecting any imaginary component at random to provide a useful, well-defined function. You may, for example, declare you're interested only in value with imaginary part between $- \pi$ and $\pi$ (That the lower bound is exclusive and the upper bound is inclusive.) or $0$ and $2 \pi$, or anything else you like. Sadly, this forces your logarithm function to be discontinuous, but if you slice a line out of the complex plane from $0$ down the negative reals, you can at least make the logarithm continuous everywhere other than on that line – you'll have something like imaginary part $- \pi$ on one side of the line and imaginary part $\pi$ on the other side. You are selecting a "branch" of the logarithm in what is referred to as a "branch cut."

Note that
$e^{2 \pi i }= e^{2 \pi k i }$
hence, you do
$\ln ( e^{2 \pi i }) = 2 \pi k i , k \in \mathbb{Z}$
use the primary value for $k = 0$