Multiplying two logarithms (Solved)

As $\log2x=\log2+\log x,$

$$0>\log x\cdot\log2x=\log x(\log x+\log 2)$$

Now if $(y-a)(y-b)<0$ with $a<b,$ we can prove $$a<y<b$$

So, here we have $$-\log 2<\log x<0$$

$$2^{-1}<x<1$$

There is no particular rule for the product of logarithms, unlike for the sum.

Applying the latter, you can rewrite

$$\log(x)\log(2x)=\log(x)(\log(x)+\log(2))=t(t+\log(2))$$

and proceed as usual to find the domain of $t$. Then $x=e^t$.