Решите уравнение 3/2log((1/4),(x+2)^2)-3=log((1/4),(4-x)^3)+log((1/4),(x+6)^3)

$\frac{3}{2} log_{1/4}(x+2)^2-3= log_{1/4}(4-x)^3+ log_{1/4}(x+6)^3$
О.Д.З.: $\begin{cases} x+2 \neq 0 \\ 4-x\ \textgreater \ 0 \\ x+6\ \textgreater \ 0 \end{cases}=\ \textgreater \ \begin{cases} x\ \textgreater \ -6 \\ x \neq -2 \\ x\ \textless \ 4 \end{cases}\ =\ \textgreater \ \boxed{x \in (-6;-2)\cup(-2;-4)}$
$3log_{1/4}(x+2)-3= 3og_{1/4}(4-x)+ 3log_{1/4}(x+6) \\ log_{1/4}(x+2)-1= log_{1/4}(4-x)+ log_{1/4}(x+6) \\ 4(x+2)=(4-x)(x+6)\\ 4x+8=-x^2-2x+24\\ x^2+6x-16=0\\ x_1=-8 \notin O.D.3.\\ x_2=2 \in O.D.3.$
Ответ: 2.