Question

The number of real values of the parameter k for which

(log₁₆x)² – (log₁₆x) + (log₁₆ k) = 0 with real coefficients will

have exactly one solution is –

• A. (a) 2
• A. (b) 1
• C. 4
• D. None

Answer 1

log₁₆x = $\frac{1 \pm \sqrt{1 - 4\log_{16}k}}{2}$

⇒ for exactly one solution log₁₆k = $\frac{1}{4}$

⇒ k = (16)^1/4

⇒ k = 2 only one real positive value.