## Algebra and Trigonometry 10th Edition

$x=20$
$\log x+\log(x-15)=2$ $\log x(x-15)=2$ $\log(x^2-15x)=2$ $10^{\log(x^2-15x)}=10^2$ $x^2-15x=100$ $x^2-15x-100=0$ $x^2-20x+5x-100=0$ $x(x-20)+5(x-20)=0$ $(x+5)(x-20)=0$ $x+5=0$ $x=-5$ $x-20=0$ $x=20$ 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.