proof of $\log(xy) =\log (x) + \log (y)$

Let $r=\log(x), s=\log(y)$ and $t=\log(xy)$.

Then $x=10^r,y=10^s$ and $xy=10^t=10^r10^s=10^{r+s}$.

Therefore, $t=r+s$, that is, $\log(xy)=\log(x)+\log(y)$.

Using a different definition of the logarithm, $\ln(x) := \int_1^x \frac{dt}{t}$, we can prove this without a definition of exponentiation.

$$ \ln(ab) = \int_1^{ab}\frac{dt}{t} = \int_1^a\frac{dt}{t} + \int_a^{ab} \frac{dt}{t} = \ln(a) + \int_a^{ab} \frac{dt}{t} $$ Next we can do a u-substitution with $u= \frac ta$, meaning the limits of the remaining integral can be re-expressed as $1$ and $b$. Also, note that $a du = dt$ as well as $au = t$, meaning we find that $ \int_a^{ab} \frac{dt}{t} = \int_1^b \frac{a\,du}{au} = \int_1^b \frac{du}{u} = \ln(b)$.

Now we find that $\ln(ab) = \ln(a) + \ln(b)$, using just basic properties of integrals.

Considering that $\log(x)$ is the inverse function of $e^x$ you have

$$e^{\log(xy)} = xy = e^{\log(x)} \cdot e^{\log(y)} = e^{\log(x)+\log(y)}$$

But since the exponential function is injective you get $\log(xy) = \log(x)+\log(y)$