The Functional Equation of the Logarithm

The functional equation of the logarithm is well-known:

\[ \log(xy) = \log(x) + \log(y) \]

But why is this so?

As a matter of fact, it follows directly from the definition of the logarithm. The logarithm is defined as a definite integral:

\[ \log(x) = \int_1^x \! \frac{dt}{t} \]

Hence:

\[ \log(xy) = \int_1^{xy} \! \frac{dt}{t} \]

Now we simply split the range of integration:

\[ \int_1^{xy} \! \frac{dt}{t} = \int_1^{x} \! \frac{dt}{t} + \int_x^{xy} \! \frac{dt}{t} \]

Notice that it doesn’t matter whether $y$ is greater or smaller than $x$, or whether it is greater or smaller than 1. We just split mechanically. (Both $x$ and $y$ must be greater than 0 for the logarithm to be well-defined; this is always assumed.)

The first integral on the right we recognize to be equal to $\log x$. In the second integral, we perform the following variable substitution:

\[ u = x t \qquad \Longrightarrow \qquad t = \frac{u}{x} \quad \text{and} \quad dt = \frac{dt}{du} du \quad \text{or} \quad dt = \frac{du}{x} \]

The change of variables also necessitates a change of the limits on the integral, hence we have:

\[ \int_x^{xy} \! \frac{dt}{t} = \int_1^y \! \frac{1}{u/x} \frac{du}{x} = \int_1^y \! \frac{du}{u} = \log y \]

This completes the proof.