How logarithm is properly defined on a field? Given a field $(F,+,\times)$, an exponential function is  defined as a function $E:F\to F$ s.t. $E(x+y)=E(x)E(y)$ and $E(0)=1$ where $0$ is the additive identity and $1$ is the multiplicative identity. 
I am curious how logarithm is properly defined for $F$ and how the connection with the exponential function is made? Is it defined as a $L:F \to F$ function s.t. $L(xy)=L(x)+L(y)$ and $L(1) = 0$? Any explanation and reference is welcome. Thank you!
 A: I suppose that the function defined by the functional equation is continuous, so to avoid ''wild'' solutions. In this case, to see when we can define an inverse (logarithm) function, define:
$$
A_0=\{x\in F : E(x)=1\}
$$ 
Now we can prove that if $A_0=\{0\}$ that $E(x)$ is invertible.
Prove by contraposition: 
If we have $x_1,x_2 \in F$ such that $E(x_1)=E(x_2)$ than $E(x_1-x_2)=E(x_1)E(-x_2)=E(x_1)E(x_2)^{-1}=E(x_1)E(x_1)^{-1}=1$, so $x_1-x_2 \in A_0$: contradiction.
In this case we can define an inverse of the $E$ function:
$$
L=E^{-1}:E(F)\to F \quad L(a)=x \quad \mbox{such that}\quad  E(x)=a
$$
and we can prove that $L(ab)=L(E(x)E(y))=L(E(x+y))=x+y=L(a)+L(b)$. This is the case if $F=\mathbb{R}$.
But, if there is $x_0\ne0 \in A_0$ than we can prove that $E(x)$ is a periodic function because: $E(x+nx_0)=E(x)E(nx_0)=E(x)E(x_0)^n=E(x)\cdot 1^n= E(x)$. So the function $E$ is not invertible and if we want define a ''logarithm'' we have to chose one period that fix a ''principal value'' for the inverse function. This is the case if $F=\mathbb{C}$.
