Log2(x-2)+log2(x-3)>1

$log_{2}(x-2) + log_{2}(x-3)\ \textgreater \ 1$

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

$log_{2}(x-2)+log_{2}(x-3)\ \textgreater \ 1 \\ \\ log_{2}((x-2)(x-3))\ \textgreater \ log_{2}2 \\ \\ (x-2)(x-3)\ \textgreater \ 2 \\ \\ x^{2} -3x-2x+6-2\ \textgreater \ 0 \\ x^{2} -5x+4\ \textgreater \ 0 \\ x_{1} = 4 \\ x_{2} =1 \\ (-\infty;1)u(4:\infty)$

Ответ: $(4;\infty)$