Помогите решить логарифмическое неравенство, пожалуйста)

$(\frac{x}{4} ) ^{log_2x-1} \ \textless \ 4$

логарифмируем (к обоим частям неравенства приписываем log₂)

затем воспользуемся свойствами логарифмов:
1)logₐbⁿ=nlogₐb
2) logₐ(b/c)=logₐb - logₐc

$log_2 (\frac{x}{4} ) ^{log_2x-1} \ \textless \ log_24 \\ \\ (log_2x-1)*log_2(\frac{x}{4} ) \ \textless \ 2 \\ \\ (log_2x-1)(log_2x-log_24) \ \textless \ 2 \\ \\ (log_2x-1)(log_2x-2) \ \textless \ 2 \\ \\ log_2x=t \\ \\ (t-1)(t-2)\ \textless \ 2 \\ \\ t^2-2t-t+2\ \textless \ 2 \\ \\ t^2-3t\ \textless \ 0 \\ t(t-3)\ \textless \ 0 \\ \\ ++++(0)----(3)+++++\ \textgreater \ t$

$0\ \textless \ t\ \textless \ 3 \ \ \textless = \textgreater \ \left \{ {{t\ \textgreater \ 0} \atop {t\ \textless \ 3}} \right. \ \ \textless = \textgreater \ \ \left \{ {{log_2x\ \textgreater \ 0} \atop {log_2x\ \textless \ 3}} \right. \ \ \textless \ =\ \textgreater \ \\ \\ \ \textless = \textgreater \ \ \left \{ {{x\ \textgreater \ 2^0} \atop {x\ \textless \ 2^3}} \right.\ \ \textless \ =\ \textgreater \ \ \left \{ {{x\ \textgreater \ 1} \atop {x\ \textless \ 8}} \right. \\ \\ OTBET: \ x \in (1;8)$