Answer
$x=1, 6$.
Work Step by Step
We know that $\log_a x^n=n\cdot \log_a x$, hence $2\log_2(x-3)=\log_2(x-3)^2$. Thus the equation is: $\log_2(x+3)=\log_2(x-3)^2$
($\log_a{M}=\log_a{N} \longrightarrow M=N$.) Thus $x+3=(x-3)^2=x^2-6x+9\\x^2-7x+6=0\\(x-1)(x-6)=0$
Thus $x=1$ or $x=6$
