If $$4^{\log_2 x} - 4x + 9^{\log_3 y} - 16y + 68 = 0$$, then $$y - x$$ equals:

$$4^{\log_2 x} - 4x + 9^{\log_3 y} - 16y + 68 = 0$$

Using the property of logarithms$$a^{\log_xy}=y^{\log_xa}$$ and $$\log_xx^2=2$$

We can rewrite $$4^{\log_2 x} - 4x + 9^{\log_3 y} - 16y + 68 = 0$$

=> $$x^{\log_24}-4x+y^{\log_39}-16y+68=0$$

=> $$x^2-4x+y^2-16y+68=0$$

We can rewrite the equation as sum of two perfect squares

=> $$\left(x-2\right)^2+\left(y-8\right)^2=0$$

We know that RHS is equal to zero so both perfect squares should be equal to zero.

Thus, x = 2 and y =8

Therefore, y-x = 8-2 = 6