Answer
$x=4$
Work Step by Step
$\log_{2}x+\log_{2}(x-3)=2$
$\log_{2}[x(x-3)]=2$
$\log_{2}[x^2-3x]=2$
$x^{2}-3x=2^{2}=4$
$x^{2}-3x-4=0$
$(x-4)(x+1)$
$(x-4)=0$ or $(x+1)=0$
$x=4$ or $x=-1$
However, $x=-1$ 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.