$\frac{({\log }_{3}x-2)({\log }_{1/2}x-2)}{({\log }_{4}x-1)}>0$ , Then the value of x lies in the interval         

Let $S=\frac{({\log }_{3}x-2)({\log }_{1/2}x-2)}{({\log }_{4}x-1)}>0$

The factor ${\log }_{3}x-2$  becomes  zero at  $x=9.$
Similarly  the other two factors become zero at  $x=1/4$  and $x=4$  . Note that we must consider only positive values of  $x$.
   We now find the sign scheme of the expression by using some test points inside the       
    various sub-intervals.
We also use the fact that  $\log ax>0$  for $x>1$  if $a>1$ and  ${\log }_{a}x>0$ for $0<x<1$  if  $0<a<1$ .

$\Rightarrow 0<x<1/4$ Or $4<x<9$

$\Rightarrow x\in (0,1/4)\cup (4,9)$