Пожалуйста, помогите решить (если можно, то не применяя метод замены множителей) (log₂²x...

Решите задачу:

$(log_2^2x-3log_2x)^2+66log_2x+72\ \textless \ 22log_2^2x\\\\t=log_2x\; ;\; \; (t^2-3t)^2+66t+72-22t^2\ \textless \ 0\\\\t^4-6t^3+9t^2+66t+72-22t^2\ \textless \ 0\\\\t^4-6t^3-13t^2+66t+72=(t+3)(t^3-9t^2+14t+24)=\\\\=(t+3)(t-4)(t^2-5t-6)=(t+3)(t-4)(t-6)(t+1)\\\\(t+3)(t-4)(t-6)(t+1)\ \textless \ 0\\\\+++(-3)---(-1)+++(4)---(6)+++\\\\t\in (-3,-1)\cup (4,6)\; \to \; \; \; \left [ {{-3\ \textless \ log_2x\ \textless \ -1} \atop {4\ \textless \ log_2x\ \textless \ 6}} \right.$

$a)\quad\left \{ {{log_2x\ \textless \ -1} \atop {log_2x\ \textless \ -3}} \right. \; \left \{ {{x\ \textless \ \frac{1}{2}} \atop {x\ \textgreater \ \frac{1}{8}}} \right. ;\quad \frac{1}{8}\ \textless \ x\ \textless \ \frac{1}{2}$

$b)\quad \left \{ {{log_2x\ \textgreater \ 4} \atop {log_2x\ \textless \ 6}} \right. \; \left \{ {{x\ \textgreater \ 16} \atop {x\ \textless \ 64}} \right. \; ,\quad 16\ \textless \ x\ \textless \ 64\\\\c)\quad x\in (\frac{1}{8};\frac{1}{2})\cup (16;64)$