Помогите решить!!!Срочно!!!!!!!!

ОДЗ:
$\left \{ {{x^2-4x\ \textgreater \ 0} \atop {x-3\ \textgreater \ 0, x-3\neq1}} \right. \\ \left \{ {{x(x-4)\ \textgreater \ 0} \atop {x\ \textgreater \ 3, x\neq4}} \right. \\ \left \{ {{x(x-4)\ \textgreater \ 0} \atop {x\ \textgreater \ 3, x\neq4}} \right. \Rightarrow x\in (4;+\infty)\\$
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$log_{x-3}(x^2-4x) \leq 2\\ log_{x-3}(x^2-4x) \leq 2log_{x-3}(x-3)\\ log_{x-3}(x^2-4x) \leq log_{x-3}(x-3)^2\\ x^2-4x \leq (x-3)^2\\ x^2-4x \leq x^2-6x+9\\ x^2-x^2-4x+6x \leq 9\\ 2x \leq 9\\ x \leq 4.5\\ -----------\\ \left \{ {{x\ \textgreater \ 4} \atop {x \leq 4.5}} \right. \Rightarrow x\in (4;4.5]$
Ответ: $x\in (4;4.5]$