Question

Помогите решить.
_______________

Answer 1

ОДЗ:
x>0

$3AMEHA: t=log_{0,5}(x) \\ \\ t^2-t-6\ \textless \ 0 \\ \\ (t-3)(t+2)\ \textless \ 0 \\ \\ +++++[-2]-----[3]+++++ \\ \\ t\in (-2;3) \\ \\ log_{0,5}(x)\in (-2;3) \\ \\ log_{0,5}(x)\ \textgreater \ -2 \\ log_{0,5}(x)\ \textless \ 3 \\ \\ x\ \textless \ 0,5^{-2} \\ x\ \textgreater \ 0,5^3 \\ \\ x\ \textless \ 4 \\ x\ \textgreater \ \frac{1}{8} \\ \\ \boxed {x\in( \frac{1}{8};4) }$

Answer 2

$\mathtt{\log_{\frac{1}{2}}^2x-\log_{\frac{1}{2}}x-6\ \textless \ 0;~\log_2^2x+\log_2x-6\ \textless \ 0;~}\\\\\mathtt{(\log_2x+3)(\log_2x-2)\ \textless \ 0;~(\log_2x-\log_2\frac{1}{8})(\log_2x-\log_24)\ \textless \ 0;~}\\\\\mathtt{(x-\frac{1}{8})(x-4)\ \textless \ 0,~\to~x\in(\frac{1}{8};4)}$

ОДЗ неравенства, кстати, никак не влияет на окончательный ответ: $\mathtt{x\in(\frac{1}{8};4)}$