Answer
$x=6$
Work Step by Step
Use the rule $m\cdot\log_a{b} =\log_a{b^m}$ to obtain:
$$\log_{2}(x+3)=\log_{2}(x-3)^{2}$$
Use the rule $\log_a{x}=\log_a{y} \implies x=y$ to obtain:
$x+3=(x-3)^{2}$
$x+3=x^{2}-6 x+9$
$0=x^{2}-6 x+9-x-3$
$0=x^{2}-7 x+6=0$
Factor the trinomial by grouping. Look for factors of $6$ whose sum is $-7$.
Since $6=(-6)(-1)$ and $(-6)+(-1)=-7$, then the factors were are loking for are $-6$ and $-1$.
Thus, factoring by grouping gives:
$x^{2}-6 x-x+6=0$
$x(x-6)-1(x-6)=0$
$(x-6)(x-1)=0$
Using the Zero-Product Property gives:
$x-6=0 \quad\quad\text { or } \quad\quad x-1=0$
$x=6 \quad\quad \quad\text{ or }\quad \quad\quad x=1$
Note, that when $x=1$, then
$\log_2{1-3} \longrightarrow \log_2{-1}$, which is undefined.
Therefore, the only solution is $x=6$.
