SOLUTION : $1^n$
$f(x)=\sqrt{\log _{\frac{x}{2}} \log _2 \frac{2 x-1}{3 x}} \text {. For domain: } \log _{\frac{x+4}{2}}\left(\log _2 \frac{2 x-1}{3+x}\right) \leq 0$
Case-I $0<\frac{x+4}{2}<1 \Rightarrow-4<x<-2$ A
then $\log _{\frac{x+4}{2}}\left(\log _2 \frac{2 x-1}{3+x}\right) \leq 0 \Rightarrow \log _2 \frac{2 x-1}{3+x} \geq 1$
$\therefore \Rightarrow \Rightarrow \frac{2 x-1}{3+x} \geq 2 \Rightarrow \quad x<-3 \quad \ldots \ldots . . B \\$
$\therefore \Rightarrow \text { on } A \cap B x \in(-4,-3)$
Case-II $\frac{x+4}{2}>1$ or $\quad x>-2$
$\log _{\frac{x+4}{2}}\left(\log _2 \frac{2 x-1}{3+x}\right) \leq 0 \quad \Rightarrow \quad 0<\log _2 \frac{2 x-1}{3+x} \leq 1 \Rightarrow 1<\frac{2 x-1}{3+x} \leq 2 \\$
$\Rightarrow \quad x \in(4, \infty) \ldots \ldots \text { (ii) } \\$
$\therefore \quad \text { (i) } \cup \text { (ii) } \quad \text { Domain } x \in(-4,-3) \cup(4, \infty)$
