Question

㏒²_{2}x+8$㏒_{2}$x+15>0

Answer 1

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

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