$log_ax+log_ay= E_1+E_2--> log_a(xy)$

the logs equals to the exponent have in the original function and the arguments equals to the outcome of the original function,

$log_ax +log_ay ⇒a^{E_1}+a^{E_2}$

$log_a xy ⇒ a^{E_1+E_2}$

, therefore the exponent can add if it was a same base for the combined outcome.

However

$log_a(x+y)$ is $a^{E_3}=x+y$