Question

Solve for x:
2^{2x- 1} − 9 x 2^{x − 2} + 1 = 0

Answer 1

2^{2x − 1} − 9 x 2^{x − 2} + 1= 0
2^2x . 2⁻¹ − 9 x ^2x . 2⁻² + 1 = 0
Let 2^x = t, so 2^2x = t²
So, 2^2x . 2⁻¹ − 9 x 2^x . 2⁻² + 1 = 0 becomes `"t"^2/(2) - 9 xx "t"/(2^2) + 1` = 0
⇒ `"t"^2/(2) - (9"t")/(4) + 1`= 0
⇒ 2t² − 9t + 4 = 0
⇒ 2t² − 8t − t + 4 = 0
⇒ 2t(t − 4) − 1(t − 4) = 0
⇒ (t − 4)(2t − 1) = 0
⇒ t − 4 = 0 or 2t − 1 = 0
⇒ t = 4 or `"t" = (1)/(2)`
So, 2^x = 4 or 2^x = `(1)/(2)`
⇒ 2^x = 2² or 2^x = 2⁻¹
⇒ x = 2 or x = −1.