Помогите решить неравенство Решить систему уравнений

$log_3^2x-|log_3x|\ \textless \ 2 \\ log_3x=a \\ a^2-|a|-2\ \textless \ 0 \\ \\ 1)a \geq 0 \\a^2-a-2\ \textless \ 0 \\ (a+1)(a-2)\ \textless \ 0 \\ -1\ \textless \ a\ \textless \ 2 \\ -1\ \textless \ log_3x\ \textless \ 2 \\ \left \{ {{log_3x\ \textgreater \ -1} \atop {log_3x\ \textless \ 2}} \right. \left \{ {{log_3x\ \textgreater \ log_3{ \frac{1}{3} }} \atop {log_3x\ \textless \ log_39}} \right. \left \{ {{x\ \textgreater \ \frac{1}{3}} \atop {x\ \textless \ 9}} \right.$
$2)a\ \textless \ 0 \\ a^2+a-2\ \textless \ 0 \\ (a+2)(a-1)\ \textless \ 0 \\ -2\ \textless \ a\ \textless \ 1 \\ -2\ \textless \ log_3x\ \textless \ 1 \\ \left \{ {{log_3x\ \textgreater \ -2} \atop {log_3x\ \textless \ 1}} \right. \left \{ {{log_3x\ \textgreater \ log_3{ \frac{1}{9}} } \atop {log_3x\ \textless \ log_33}} \right. \left \{ {{x\ \textgreater \ \frac{1}{9}} \atop {x\ \textless \ 3}} \right.$

Ответ: $x$ ∈ $( \frac{1}{9} ;3)$ U $( \frac{1}{3} ;9)$