Answer
$x=5$
Work Step by Step
$\log x+\log(x-3)=1$
$\log[x(x-3)]=1$
$\log_{10}[x^2-3x]=1$
$x^{2}-3x=10^1$
$x^{2}-3x-10=0$
$(x+2)(x-5)=0$
$(x+2)=0$ or $(x-5)=0$
$x=-2$ or $x=5$
However, $x=-2$ does not work in the original equation because we can't take the log of a negative number (undefined). So we throw this solution out.