4^x - 3 * 2^(x+1) + 8 <= 0

(2^x)^2 - 6 * 2^x + 8 <= 0

Let y = 2^x. Then y^2 - 6y + 8 <= 0.

(y-2)(y-4) <= 0, so 2 <= y <= 4.

2 <= 2^x <= 4, which means 1 <= x <= 2.