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!

  • $\begingroup$ By using formal power series. $\endgroup$
    – user26857
    Sep 28, 2016 at 17:04
  • $\begingroup$ @user26857 if $F$ has a topology ... $\endgroup$
    – user251257
    Sep 28, 2016 at 18:39
  • $\begingroup$ @user251257 I have no idea why need a topology. Characteristic zero should be enough. $\endgroup$
    – user26857
    Sep 28, 2016 at 18:44
  • 1
    $\begingroup$ @user26857 what do you do with that formal series? $\endgroup$
    – user251257
    Sep 28, 2016 at 18:46
  • $\begingroup$ Logarithm has a Taylor expansion which can be thought of as a definition over fields of characteristic zero. $\endgroup$
    – user26857
    Sep 29, 2016 at 6:30

1 Answer 1


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 $F=\mathbb{C}$.

  • $\begingroup$ Where do you need continuity? $\endgroup$
    – user251257
    Sep 28, 2016 at 23:55
  • $\begingroup$ If we don't require continuity ( or at least misurability) we can have many different functions that satisfies the functional equation $\endgroup$ Sep 29, 2016 at 7:52
  • $\begingroup$ yeah. But you haven't shown that there is at most one solution. So you don't really need that for what you have written in your answer, do you? $\endgroup$
    – user251257
    Sep 29, 2016 at 7:59

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