## College Algebra 7th Edition

$x=5$
$\log x+\log(x-1)=\log(4x)$ $\log[x(x-1)]=\log(4x)$ $x(x-1)=4x$ $x^{2}-x=4x$ $x^{2}-x-4x=0$ $x^{2}-5x=0$ $x(x-5)=0$ $x=0$ 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).