Log1/3(x^2-2x-8+3>0 Помогите решить!!!!!!!!

Решите задачу:

$log_{\frac{1}{3}} (x^2-2x-8)+3\ \textgreater \ 0 \\\\ x^2-2x-8\ \textgreater \ 0; (x-4)(x+2)\ \textgreater \ 0; \left \{ {{x\ \textgreater \ 4} \atop {x\ \textless \ -2}} \right. \\ x\in (-\infty;-2)\cup(4;+\infty) \\\\ log_{\frac{1}{3}} (x^2-2x-8)\ \textgreater \ -3 \\ log_{\frac{1}{3}} (x^2-2x-8)\ \textgreater \ log_{\frac{1}{3}} 27 \\ x^2-2x-8\ \textless \ 27 \\ x^2-2x-35 \ \textless \ 0 \\ (x-7)(x+5)\ \textless \ 0 \\ -5\ \textless \ x\ \textless \ 7 \\\\ x\in (-5;-2)\cup(4;7)$