Помогите решить неравенство [latex]log^23(x)-|log3(x)|\ \textless \ 2[/latex] Решить систему уравнений [latex] \left \{ {{2^{x+3}*3^{y+2}=8} \atop {x-y=2}} \right. [/latex]

[latex]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.[/latex] [latex] 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. [/latex] Ответ: [latex]x[/latex] ∈ [latex]( \frac{1}{9} ;3)[/latex] U [latex]( \frac{1}{3} ;9)[/latex]