Question

Solve the following inequality:
${\log }_2(2^x-1)\cdot {\log }_{1/2}(2^{x+1}-2)>-2$

Answer 1

$-{\log }_2(2^x-1){\log }_2(2^{x+1}-2)>-2-{\log }_2(2^x-1)({\log }_2(2^x-1)+1)>-2{\left({\log }_2(2^x-1)\right)}^2+{\log }_2(2^x-1)<2{\left({\log }_2(2^x-1)\right)}^2+{\log }_2(2^x-1)-2<0{\left({\log }_2(2^x-1)\right)}^2+2{\log }_2(2^x-1)-{\log }_2(2^x-1)-2<0{\log }_2(2^x-1)({\log }_2(2^x-1)+2)-1({\log }_2(2^x-1)+2)<0({\log }_2(2^x-1)-1)({\log }_2(2^x-1)+2)<0{\log }_2(2^x-1)\in (-2,1)(2^x-1)\in (2^{-2},2)2^x\in (1+2^{-2},3)x\in ({\log }_2(1+2^{-2}),{\log }_{2}3)$

$2^x-1>02^x>1x>0$ and $2^{x+1}-2>02^{x+1}>2x+1>1x>0$

Taking intersection, we get $x\in ({\log }_2(1+2^{-2}),{\log }_{2}3)$