Помогите пожалуйста решить логарифмическое неравенство!!! Совсем забыла как решать....

Решите задачу:

$(log_2^2x-2log_2x)^2+36log_2x+45\ \textless \ 18log_2^2x\; ,\; \; ODZ:\; x\ \textgreater \ 0\\\\t=log_2x\; ,\; \; \; (t^2-2t)^2+36t-18t^2+45\ \textless \ 0\\\\t^4-4t^3+4t^2+36t-18t^2+45\ \textless \ 0\\\\t^4-4t^3-14t^2+36t+45\ \textless \ 0\\\\Pri\; t=5:\; \; 5^4-4\cdot 5^3-14\cdot 5^2+36\cdot 5+45=0\ \; \to \\\\t^4-4t^3-14t^2+36t+45=(t-5)(t^3+t^2-9t-9)=\\\\=(t-5)(t^2(t+1)-9(t+1))=(t-5)(t+1)(t-3)(t+3)\ \textless \ 0\\\\+++(-3)---(-1)+++(3)---(5)+++\\\\ \left [ {{-3\ \textless \ t\ \textless \ -1} \atop {3\ \textless \ t\ \textless \ 5}} \right.$

$\left [ {{-3\ \textless \ log_2x\ \textless \ -1} \atop {3\ \textless \ log_2x\ \textless \ 5}} \right. \; \left [ {{2^{-3}\ \textless \ x\ \textless \ 2^{-1}} \atop {2^3\ \textless \ x\ \textless \ 2^5}} \right. \; \left [ {{\frac{1}{8}\ \textless \ x\ \textless \ \frac{1}{2}} \atop {8\ \textless \ x\ \textless \ 32}} \right. \\\\x\in (\frac{1}{8};\frac{1}{2})\cup (8;32)$