The equation has
- A.atleast one real solution
- B.exactly three real solutions
- C.exactly one irrational solution
- D.complex roots
Show answer and solution
Correct option: B
Solution
Taking log base 2 of both sides: [(3/4)(log₂x)² + log₂x - 5/4] · log₂x = 1/2.
Let t = log₂x. Then (3/4)t³ + t² - (5/4)t - 1/2 = 0, or multiplying by 4: 3t³ + 4t² - 5t - 2 = 0.
Factor: try t = -2: 3(-8) + 4(4) - 5(-2) - 2 = -24 + 16 + 10 - 2 = 0 ✓. So (t+2) is a factor. Dividing: 3t³ + 4t² - 5t - 2 = (t+2)(3t² - 2t - 1) = (t+2)(3t+1)(t-1).
So t = -2, t = -1/3, t = 1, giving x = 2⁻² = 1/4, x = 2^(-1/3) (irrational), x = 2¹ = 2. All three are distinct positive reals (x > 0 required for the original expression), so there are exactly three real solutions — two rational (1/4 and 2) and one irrational (2^(1/3) in denominator).
Let t = log₂x. Then (3/4)t³ + t² - (5/4)t - 1/2 = 0, or multiplying by 4: 3t³ + 4t² - 5t - 2 = 0.
Factor: try t = -2: 3(-8) + 4(4) - 5(-2) - 2 = -24 + 16 + 10 - 2 = 0 ✓. So (t+2) is a factor. Dividing: 3t³ + 4t² - 5t - 2 = (t+2)(3t² - 2t - 1) = (t+2)(3t+1)(t-1).
So t = -2, t = -1/3, t = 1, giving x = 2⁻² = 1/4, x = 2^(-1/3) (irrational), x = 2¹ = 2. All three are distinct positive reals (x > 0 required for the original expression), so there are exactly three real solutions — two rational (1/4 and 2) and one irrational (2^(1/3) in denominator).
[1989]