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: $(-5;3)$