The set of real values of x for which $(\log_x 2) (\log_{2x} 2) (\log_2 4x) >1$, is

$(2^{-\sqrt{2}},1/2)∪(1,2^{\sqrt{2}})$

Various terms in the given inequation are meaningful if $x > 0$ and $x ≠ 1$.

Now, $(\log_x 2) (\log_{2x} 2) (\log_2 4x) >1$

$⇒\frac{\log_2 4x}{\log_2 x.\log_2 2x}>1$

$⇒\frac{(2+\log_2 x)}{\log_2 x(1+\log_2 x)}-1>1$

$⇒\frac{(\log_2 x+2)-\log_2 x(1+\log_2 x)}{\log_2 x(1+\log_2 x)}>1$

$⇒\frac{(\log_2 x)^2-2}{\log_2 x(\log_2 x+1)}<0$

$⇒\frac{(\log_2 x+\sqrt{2})(\log_2 x-\sqrt{2})}{\log_2 x(\log_2 x+1)}<0$

The signs of the expression

$\frac{(\log_2 x+\sqrt{2})(\log_2 x-\sqrt{2})}{\log_2 x(\log_2 x+1)}$ for different values of $\log_2 x$ are shown in Fig. 

$∴\frac{(\log_2 x+\sqrt{2})(\log_2 x-\sqrt{2})}{\log_2 x(\log_2 x+1)}<0$

$⇒-\sqrt{2}<\log_2x<-1$ or, $0<\log_2x<\sqrt{2}$

$⇒2^{-\sqrt{2}}<x<\frac{1}{2}$ or, $1<x<2^{\sqrt{2}}$

$⇒x∈(2^{-\sqrt{2}},1/2)∪(1,2^{\sqrt{2}})$