8^x-3•4^x-2^x+3=0

Solve for x:

• Let t = 2^x > 0. Then:

  • 8^x = t^3
  • 4^x = t^2
  • 2^x = t

• The equation becomes: t^3 − 3t^2 − t + 3 = 0

• Factor by grouping: (t^3 − 3t^2) + (−t + 3) = t^2(t − 3) − 1(t − 3) = (t − 3)(t^2 − 1)

• So (t − 3)(t − 1)(t + 1) = 0

Since t = 2^x > 0, t ∈ {1, 3} (t = −1 is invalid).

• 2^x = 1 ⇒ x = 0
• 2^x = 3 ⇒ x = log2(3)

Answer: x = 0 or x = log2(3) (approximately 1.585).