Question

㏒²_{2}x+8$㏒_{2}$x+15>0

Answer 1

$\sf\displaystyle \log ^2_2x+8\log _2x+15>0\\OD3:x>0\\\log_2x=y\\y^2+8x+15>0\\\\y^2+8x+15=0\\D=64-60=2^2\\y_{1,2}=\frac{-8\pm 2}{2}=\left |{ {{-3} \atop {-5}} \right. \\+++(-5)---(-3)+++>y\\\\ \left [{ {{y>-3} \atop {y<-5}} \right. \left [{ {{\log_2x>-3} \atop {\log_2x<-5}} \right. \left [{ {{x>\frac{1}{8}} \atop {0<x<\frac{1}{32}}} \right. \Leftrightarrow x\in(0;\frac{1}{32})\cup (\frac{1}{8};+\infty)$

Ответ: $(0;\frac{1}{32})\cup (\frac{1}{8};+\infty)$