Собъяснением : log^2 0,5 x + log0,5 x - 2=< 0

ОДЗ: x>0

a²+a-2≤0
(a+2)(a-1)≤0 
           +                       -                           +
________[-2]_____________[1]_________
a∈[-2;1]

\left \{ {log_{0,5}x \geq -2} \atop {log_{0,5}x \leq 1}} \right. =\ \textgreater \ \left \{ {{log_{0,5}x \geq log_{0,5}4} \atop {log_{0,5}x \leq log_{0,5}0,5}} \right.\; \; \; \; \; \; 0\ \textless \ 0,5\ \textless \ 1\\\\ \left \{ {{x \leq 4} \atop {x \geq 0,5}} \right. " alt="a=log_{0,5}x" />

a²+a-2≤0
(a+2)(a-1)≤0 
           +                       -                           +
________[-2]_____________[1]_________
a∈[-2;1]

\left \{ {log_{0,5}x \geq -2} \atop {log_{0,5}x \leq 1}} \right. =\ \textgreater \ \left \{ {{log_{0,5}x \geq log_{0,5}4} \atop {log_{0,5}x \leq log_{0,5}0,5}} \right.\; \; \; \; \; \; 0\ \textless \ 0,5\ \textless \ 1\\\\ \left \{ {{x \leq 4} \atop {x \geq 0,5}} \right. " />

ответ: [0,5;4]