Making $-\log(2-4x)\log(x-5)+\log(2x-1)=\log(3-4x)$ into a quadratic equation

Question

How we can work with below equation to get quadratic equation?

$$-\log(2-4x)\log(x-5)+\log(2x-1)=\log(3-4x)$$

No need to get the variable x

Answer 1

we know that $\log x$ is defined for $x>0$ so

$\log(2-4x) \Rightarrow 2-4x>0 ,\ \frac{1}{2}>x$

$\log(2x-1) \Rightarrow 2x-1 >0 ,\ x> \frac{1}{2}$

that is a contradiction which means there is not a solution