Question

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

Answer 1

${\log }_2{x}^2+{\log }_{\sqrt{2}}(x-1)<{\log }_{\sqrt{2}}{\log }_{\sqrt{2}}2$

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

$2{\log }_2\left(x(x-1)\right)<2{\log }_{2}2x>0x>1$

$2{\log }_2\left(x(x-1)\right)<2$

${\log }_2\left(x(x-1)\right)$

$x(x-1)<2$

$x^2-x-2<0$

$(x-2)(x+1)<0$

$xϵ(-1,2)$

Taking intersection we get $xϵ(1,2)$