Question

㏒²(-㏒x)+㏒㏒²x≤3 ( у всех логарифмов основание 2)

Answer 1

$log_2^2(-log_2x)+log_2\, log_2^2x \leq 3\; ;\\\\ ODZ:\; -log_2x\ \textgreater \ 0\; ,\; log_2x\ \textless \ 0\; ,\; \; \; \underline {0\ \textless \ x\ \textless \ 1}\\\\(-log_2x)^2=log_2^2x\; \; ;\; \; t=log_2x\; \; \to \\\\log_2^2(-t)+log_2((-t)^2)-3 \leq 0\\\\log_2^2(-t)+2\cdot log_2(-t)-3 \leq 0\\\\z=log_2(-t)\; ;\; \; z^2+2z-3 \leq 0\; ;\; -3 \leq z \leq 1\\\\ \left \{ {{log_2(-t) \leq 1} \atop {log_2(-t) \geq -3}} \right. \; \left \{ {{-t \leq 2} \atop {-t \geq \frac{1}{8}}} \right. \; \left \{ {{t \geq -2} \atop {t \leq -\frac{1}{8}} \right.$

$\left \{ {{log_2x \geq -2} \atop {log_2x \leq -\frac{1}{8}}} \right. \; \left \{ {{x \geq 2^{-2}} \atop {x \leq 2^{-\frac{1}{8}}}} \right. \; \left \{ {{x \geq \frac{1}{4}} \atop {x \leq \frac{1}{\sqrt[8]{2}}}} \right. \\\\Otvet:\quad x\in [\, \frac{1}{4}\; ;\; \frac{1}{\sqrt[8]{2}}\, ]$