The product law of logarithms

Let $u=\log_a x$ and $v=\log_a y$. By the definition of logarithm (it is the inverse of the exponential), this means exactly $a^u=x$ and $a^v=y$.

Multiply these two equations: $xy = a^u \cdot a^v$. By the exponent law $a^{u+v}=a^u\cdot a^v$, the right-hand side equals $a^{u+v}$, so $xy=a^{u+v}$.

By definition of logarithm again, $xy=a^{u+v}$ means exactly $\log_a(xy)=u+v$ — but this uses that an exponent producing a given output is unique, i.e. $a^u=a^v \Rightarrow u=v$ whenever $a>0,a\ne 1$ (the exponential function never gives the same output for two different exponents, since it is strictly increasing for $a>1$ or strictly decreasing for $0<a<1$).

Substituting back $u=\log_a x$ and $v=\log_a y$ gives $\log_a(xy) = \log_a x + \log_a y$, which is $\log_a(xy) = \log_a x + \log_a y$.