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

[latex] log_{2}(x-3)^2+ log_{ \sqrt{2} } (3-x)\ \textless \ 12[/latex] ОДЗ:   [latex] \left \{ {{(x-3)^2\ \textgreater \ 0} \atop {3-x\ \textgreater \ 0}} \right. [/latex] [latex] \left \{ {{x \neq 3} \atop {x\ \textless \ 3}} \right. [/latex] [latex]x[/latex] ∈ [latex](-[/latex] ∞ [latex];3)[/latex] [latex] log_{2}(3-x)^2+ log_{ 2_{ \frac{1}{2} } } } (3-x)\ \textless \ 12[/latex] [latex] 2log_{2}(3-x)+ 2log_{ 2_ } } (3-x)\ \textless \ 12[/latex] [latex]4log_{ 2_ } } (3-x)\ \textless \ 12[/latex] [latex]log_{ 2_ } } (3-x)\ \textless \ 3[/latex] [latex]log_{ 2_ } } (3-x)\ \textless \ log_{ 2_ } } 8[/latex] [latex]3-x\ \textless \ 8[/latex] [latex]-x\ \textless \ 8-3[/latex] [latex]-x\ \textless \ 5[/latex] [latex]x\ \textgreater \ -5[/latex] --------(-5)------------------               //////////////////// ----------------------(3)------ //////////////////////// Ответ:  [latex](-5;3)[/latex]