Here is another attempt to solve the equation:

1. On behalve of the summand $\displaystyle \log_{2}x$ x must be greater than zero.

2.

$\displaystyle \left| \log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x = 1$

$\displaystyle \left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| = \log_{2}\left(\frac2x\right)$

You have to consider 2 different cases:

$\displaystyle \log_{2}\left(\frac{4-x}{2x - 5}\right) > 0~\implies~ \frac{4-x}{2x - 5} > 1$

Then you must solve for x:

$\displaystyle \frac{4-x}{2x - 5} = \frac2x~\implies~ \boxed{x = \sqrt{10}}$

The negative solution isn't valid. See #1.

Or

$\displaystyle \log_{2}\left(\frac{4-x}{2x - 5}\right) < 0~\implies~ - \log_{2}\left(\frac{4-x}{2x - 5}\right) > 0 ~\implies~ \frac{2x - 5}{4-x} > 1$

Then you have to solve for x:

$\displaystyle \frac{2x - 5}{4-x} = \frac2x ~\implies~ \boxed{x \approx 2.886}$

The negative solution isn't valid. See #1.

Edit: To my great confusion both solutions don't satisfy the constraints to yield a fraction which is greater than 1.