Пожалуйста!!!!! log2 (x-3)^2+log корень из 2(3-x)<12

$log_{2}(x-3)^2+ log_{ \sqrt{2} } (3-x)\ \textless \ 12$

ОДЗ:
$\left \{ {{(x-3)^2\ \textgreater \ 0} \atop {3-x\ \textgreater \ 0}} \right.$

$\left \{ {{x \neq 3} \atop {x\ \textless \ 3}} \right.$

$x$ ∈ $(-$ ∞ $;3)$

$log_{2}(3-x)^2+ log_{ 2_{ \frac{1}{2} } } } (3-x)\ \textless \ 12$

$2log_{2}(3-x)+ 2log_{ 2_ } } (3-x)\ \textless \ 12$

$4log_{ 2_ } } (3-x)\ \textless \ 12$

$log_{ 2_ } } (3-x)\ \textless \ 3$

$log_{ 2_ } } (3-x)\ \textless \ log_{ 2_ } } 8$

$3-x\ \textless \ 8$

$-x\ \textless \ 8-3$

$-x\ \textless \ 5$

$x\ \textgreater \ -5$

--------(-5)------------------
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----------------------(3)------
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Ответ: $(-5;3)$