*The function \(\log x\) satisfies the functional equation* \[\begin{equation*} f(xy) = f(x) + f(y). \tag{1} \end{equation*}\] For, making the substitution \(t = yu\), we see that \[\begin{aligned} \log xy &= \int_{1}^{xy} \frac{dt}{t} = \int_{1/y}^{x} \frac{du}{u} = \int_{1}^{x} \frac{du}{u} – \int_{1}^{1/y} \frac{du}{u}\\ &= \log x – \log(1/y) = \log x + \log y,\end{aligned}\] which proves the theorem.

$\leftarrow$ 196–197. The logarithmic function | Main Page | 199–201. The behaviour of $\log x$ as $x$ tends to infinity or to zero $\rightarrow$ |