Question

Let u = (log₂x)² – 6(log₂x) + 12, where x is a real number. Then the equation x^u = 256 has :

(a) No solution for x 

(b) Exactly one solution for x 

(c) Exactly two distinct solutions for x 

(d) Exactly three distinct solutions for x

Answer 1

Given, u = (log₂x)² – 6(log₂x) + 12 = p2 – 6p + 12      (where p = log₂x)  ...(i) 

Also, given, x^u = 256 

Taking log to the base 2 of both the sides, we have 

u log₂x = log₂256 = log₂2⁸ = 8 log₂2 ⇒ u log₂x = 8 ⇒ u = \(\frac{8}{log_2x}\) = \(\frac{8}{p}\)

From (i) and (ii) \(\frac{8}{p}\) = p² - 6p + 12

⇒ 8 = p³ – 6p² + 12 ⇒ p³ – 6p² + 12p – 8 = 0 

⇒ (p – 2)³ = 0 ⇒ p = 2.                                      [Using (a – b)³ = a³ – 3a²b + 3ab² – b³] 

∴ log₂x = 2 ⇒ x = 2² = 4 

Hence the equation u⁴ = 256 has exactly one solution.