Question

logx(2) *log2x(2)*log2(4x)>1

Answer 1

$displaystyle log_x2*log_{2x}2*log_24x>1\\ODZ:left { {{x>0} atop {xneq 1; 2x}neq1} right. ; left { {{x>0} atop {xneq 1; xneq 1/2 }} right. \\frac{1}{log_2x}* frac{1}{log_22x}*(log_24+log_2x)>1\\ frac{1}{log_2x*(log_22+log_2x)}*(2+log_x2)>1\\log_2x=t\\ frac{2+t}{t(1+t)}>1\\ frac{2+t-t(1+t)}{t(1+t)}>0\\ frac{2+t-t-t^2}{t(1+t)}>0\\ frac{-(t^2-2)}{t(1+t)}>0$

__ -√2____  -1 ____ 0  _____√2____

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1)-√2<t<-1

$displaystyle -sqrt{2}<t<-1\\- sqrt{2}<log_2x<-1\\ frac{1}{2^{sqrt{2}}}<x< frac{1}{2}$

2) 0<t<√2

$displaystyle 0<t<sqrt{2}\\0<log_2x< sqrt{2}\\1<x<2^{sqrt{2}}$