Why $\log xy=\log x+\log y$? It is of course well known and basic formula. I am just curious. Is there a proof for it?
How to prove that $\log xy=\log x+\log y$?

We have $$\ln (xy) = \int_{1}^{xy}\frac{dt}{t}=\int_{1}^{x}\frac{dt}{t}+\int_{x}^{xy}\frac{dt}{t}=\ln(x)+\int_{x}^{xy}\frac{dt}{t}.$$ For the last integral, we substitute $u=\frac{t}{x}$ to get $du=\frac{dt}{x}$, thus reducing the last integral to $\int_{1}^{y}\frac{du}{u}=\ln (y)$. Hence, $\ln(xy)=\ln(x)+\ln(y)$.
Dividing both sides by $\log (a)$, we have $\log_{a}(xy)=\log_{a}(x)+\log_{a}(y)$ for any base $a\neq 1.$
