Solve :

8 × 2^2x + 4 × 2^{x + 1} = 1 + 2^x

Given,

⇒ 8 × 2^2x + 4 × 2^{x + 1} = 1 + 2^x

⇒ 8 × 2^(x)(2) + 4 × 2^x.2¹ = 1 + 2^x

Substituting 2^x = a, in above equation, we get :

⇒ 8 × a² + 4a × 2 = 1 + a

⇒ 8a² + 8a = 1 + a

⇒ 8a² + 8a - a - 1 = 0

⇒ 8a(a + 1) - 1(a + 1) = 0

⇒ (8a - 1)(a + 1) = 0

⇒ 8a - 1 = 0 or a + 1 = 0

⇒ 8a = 1 or a = -1

a cannot be negative as 2^x, for any value of x is greater than 0.

⇒ 8(2^x) = 1

⇒ 2^x = $\dfrac{1}{8}$

⇒ $2^x = \dfrac{1}{2^3}$

⇒ 2^x = 2^-3

⇒ x = -3.

Hence, x = -3.