The roots of $$2^{2x}-10×2^x+16=0$$​ are:

Correct option is D

​Given:

$$2^{2x} - 10 \times 2^x + 16 = 0$$​

Solution:

Let $$y = 2^x$$​, so $$2^{2x} = y^2$$​​

Rewritten Equation:

​$$y^2 - 10y + 16 = 0$$​​

Solve quadratic equation using factorization:

$$y^2 - 10y + 16 = (y - 2)(y - 8) = 0$$​​

Roots: $$y = 2 \, \text{or} \, y = 8.$$​​

Since y = 2^x​

​$$2^x = 2 \implies x = 1, \\ \ \\ 2^x = 8 \implies x = 3$$​​