find x.. 70

First of all, we need x to be strictly positive (why?). Next, for x≥2, |x-2| = x-2, so for this region we get
(x-2)^{log₂x3 - 3 log_x4} = (x-2)³
which gives log₂x³ - 3 log_x4 = 3. The LHS simplifies to
3 log₂ x - 6 log_x 2 = 3.
Cancelling out 3 and using log_x 2 = 1 log₂ x, we get
log₂ x - 2 log₂ x = 1,
which finally gives
(log₂ x)² - log₂ x -2 =0
which solved as a quadratic equation in log₂ x give
log₂ x = 2 or -1
giving x = 4 or 1/2. Disregarding 1/2, we get x = 4.
For x<2, we see that though the LHS is always positive, the RHS becomes negative. So we cannot have any root less than 2. So the only root is x=4