Answer
$x=20$
Work Step by Step
$\log x+\log(x-15)=2$
$\log x(x-15)=2$
$\log(x^2-15x)=2$
$10^{\log(x^2-15x)}=10^2$
$x^2-15x=100$
$x^2-15x-100=0$
$x^2-20x+5x-100=0$
$x(x-20)+5(x-20)=0$
$(x+5)(x-20)=0$
$x+5=0$
$x=-5$
$x-20=0$
$x=20$
But, the domain of $\log x$ is $(0,∞)$. Also, the domain of $\log (x-15)$ is $(15,∞)$. So, $x=-5$ is not a valid solution.
