Functions Modeling Change: A Preparation for Calculus, 5th Edition

$\log(x-y)+\log(x+y)$
$x^2-y^2$ can be factored as $(x+y)(x-y)$ (FOIL the expression out to verify if needed). Therefore, the expression becomes $\log((x+y)(x-y))$. Since $\log(ab)=\log(a)+\log(b)$, the expression can be simplified to $\log(x-y)+\log(x+y)$.