The logarithm to base $b$ of a number $x,$ written $y=log_b(x),$ is the number $y$ such that $b^y=x.$ It is the inverse function to exponentiation.

When restricted to real number, logarithms are only defined on positive numbers. Logarithms are never defined at zero, since $x^y$ is never 0 (except in the trivial case where $x=0$ and $y$ is non-zero).

There are some standard rules for logarithms, similar to, and derived from, the rules for exponents and indices. Thus we have:

$log(1)=0$

$log_b(b)=1$ where $b>0$

$log(ab)=log(a)+log(b)$ where $a>0$ and $b>0$

$log(a^n)=n.log(a)$ where $a>0$

When ranging over complex numbers the logarithm is a multi-valued function (and hence not a function at all).

Consider the complex number $z.$ Because of the polar representation of a complex number we can write $z~=~r.e^{i\theta}.$ However, we can add $2\pi$ to $\theta$ and get the same result, so we also have $z~=~r.e^{i(\theta+2\pi)}.$ Thus $r.e^{i\theta}~=~r.e^{i(\theta+2\pi)}.$

Taking logarithms base $e$ of both sides we have:

$log(r.e^{i\theta})~=~log(r.e^{i(\theta+2\pi)}).$

Using the rules of logarithms we can proceed as follows:

$log(r)+log(e^{i\theta})~=~log(r)+log(e^{i(\theta+2\pi)}).$

$log(e^{i\theta})~=~log(e^{i(\theta+2\pi)}).$

$i\theta~=~i(\theta+2\pi).$

Dividing by $i$ and subtracting $\theta$ we then have $0=2\pi,$ which is nonsense.

The error is in assuming that even if $x=y$ it must then follow that $log(x)=log(y).$ This is not a valid deduction if $log$ is not a single valued function - it's akin to assuming that just because $x^2=y^2$ then $x=y,$ or that if $\sin(x)=\sin(y)$ then $x=y.$

However, if we consider logarithms on an appropriate Riemann Surface then the analogous function is a true, single-valued function.

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