Why is $x$ in there?

$log_{x}(2x)+log_{2}(x)\geqslant 3$

$log_{x}(2)+log_{x}(x)+log_{2}(x)\geqslant 3$

$log_{x}(2)+1+log_{2}(x)\geqslant 3$

$log_{x}(2)+log_{2}(x)\geqslant 2$

$\frac{log\, 2}{log\,x}+\frac{log\, x}{log\,2}\geqslant 2$

$\frac{(log\, 2)^{2}+(log\,x)^{2}}{log\,x\,log\, 2}\geqslant 2$

now a=log x and b=log 2

$\frac{a^{2}+b^{2}}{ab}\geqslant 2$

if a>0 then x > 1 and

$a^{2}+b^{2}\geqslant 2{ab}$

if a<0 then 0<x<1 and

$a^{2}+b^{2}\leqslant 2{ab}$

continuing with a>0

$a^{2}+b^{2}-2{ab}\geqslant 0$

$(a-b)^{2}\geqslant 0$

therefore for a>0 a can be anything (and requires x>1)

but for a<0 a cannot be real

therefore the domain of x is (1, +inf)