Question

If the solution set for the inequality $2{\log }_{2}x+{\log }_{\sqrt{2}}(x-1)<{\log }_{\sqrt{2}}{\log }_{\sqrt{2}}2$ is $(a,b)$, then value of $a+b$ is

Answer 1

$2{\log }_{2}x+{\log }_{\sqrt{2}}(x-1)<{\log }_{\sqrt{2}}{\log }_{\sqrt{2}}2$
For $\log$ to be defined, $x>1...\left(1\right)$
Now, $2{\log }_{2}x+2{\log }_2(x-1)<2{\log }_{2}2{\log }_{2}2$
$\Rightarrow 2{\log }_{2}x(x-1)<2$
$\Rightarrow {\log }_{2}x(x-1)<1$
$\Rightarrow x(x-1)<2$
$\Rightarrow x^2-x-2<0$
$\Rightarrow (x-2)(x+1)<0$
$\Rightarrow x\in (-1,2)...\left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$x\in (1,2)$

$\because (1,2)\equiv (a,b)$
$\therefore a+b=3$