Difference between minimum and maximum

$\begin{aligned} 16^x - \frac 52 \left(2^{2x+1}\right) + 4 & = 0 \\ 16^x - 5 \left(2^{2x}\right) + 4 & = 0 \\ 4^{2x} - 5 \left(4^{x}\right) + 4 & = 0 & \small \color{#3D99F6} \text{Note that it is a quadratic equation of }4^x. \\ \left(4^x - 1\right)\left(4^x - 4\right) & = 0 \end{aligned}$

$\implies 4^x = \begin{cases} 1 & \implies x = 0 \\ 4 & \implies x = 1 \end{cases}$

Therefore, the different of the maximum and minimum real roots is $1-0 = \boxed{1}$ .