Answer
$x=5$
Work Step by Step
$\log x+\log(x-1)=\log(4x)$
$\log[x(x-1)]=\log(4x)$
$x(x-1)=4x$
$x^{2}-x=4x$
$x^{2}-x-4x=0$
$x^{2}-5x=0$
$x(x-5)=0$
$x=0$ or $x-5=0$
$x=0$ or $x=5$
But $x=0$ is not possible in the original equation, because we can't take a log of zero (undefined).