Let's say we want to make a function L(x) with the property that L(xy) = L(x) + L(y) for all x and y in the domain. Let's investigate its properties and then form a definition to encompass those properties.

First of all, if 0 is in the domain of L(x), then we'd have L(x*0) = L(0) = L(0) + L(x) and L(x) = 0 for all x. Thus, to make a more interesting function, let's forbid 0 from being in the domain.

Now, we see that L(x*1) = L(x) = L(x) + L(1), so L(1) = 0.
Furthermore, L(-1*-1) = L(1) = L(-1) + L(-1), so L(-1) = 0 as well.
In fact, for any x in the domain of L,
L(x*x) = L(x^2) = L(-x*-x) = 2L(x) = 2L(-x), so L(x) = L(-x) and the function L is an even function.

Now, let's consider a L(x) which is differentiable and examine its properties.

We know that for any fixed y, we can differentiate with respect to x to get ' = ' = L'(x) = yL'(xy), where the second part comes from the Chain Rule. By now substituting x = 1, we get L'(1) = yL'(y), or L'(y) = L'(1) / y. Since L(x) is an even function, it has an odd derivative, so L'(y) = L'(1) / y holds for all y which are not 0.

Let K = L'(1). We know from the Fundamental Theora of Calculus that the integral from 1 to y of Kdx / x is L(y) - L(1) = L(y), if y > 0.
So for positive y, we have L(y) = K dx / x.

If y < 0, then L(y) = L(|y|) = K dx / x, since L is an even function. This is the most general rule which describes ALL logarithms. If we let K=1, and we restrict our domain to positive numbers, then we get the natural logarithm function. However, note that if K = 1 / ln b for some positive b, then we get logarithms to the base b. It's interesting to see that, as a consequence of this reason, the function L(x) = K dt / t is the MOST GENERAL FUNCTION that fulfills L(xy) = L(x) + L(y). Hence, the only functions which do this are logarithms!

e is defined as the unique number > 0 such that L(e) = 1. This is unique, because L(x) is strictly increasing for x > 0. Once you develop Taylor Polynomial approximations, then you can prove that
0> (1 + ax)^(1/ax) = e^a, and this justifies the formula that most people know for e.