Логарифмическое неравенство( ОДЗ!!)

$=================\\ \left \{ {{x+8\ \textgreater \ 0} \atop {6-x\ \textgreater \ 0}} \right. \\\\ \left \{ {{x\ \textgreater \ -8} \atop {x\ \textless \ 6}} \right. \\\\ -8 \ \textless \ x \ \textless \ 6\\ =================\\ log_3(x+8) \leq log_3(6-x)\\ x+8 \leq 6-x\\ 2x \leq -2\\ x \leq -1\\ \left \{ {{-8\ \textless \ x\ \textless \ 6} \atop {x \leq -1}} \right. \\ x \in (-8;1]$

Ответ: -8 < x <= -1