Question

The number of real values of the parameter k for which $(lo{g}_{16}x{)}^2-lo{g}_{16}x+lo{g}_{16}k=0$ with real coefficients will have exactly one solution is

• A. 2
• B. 1
• C. 4
• D. None of these

Answer 1

The correct option is B 1

Let $lo{g}_{16}x=y\Rightarrow y^2-y+lo{g}_{16}k=0$
This quadratic equation will have exactly one solution if its discriminant vanishes.
$\therefore (-1{)}^2-\mathrm{4.1.}lo{g}_{16}k=0\Rightarrow 1=lo{g}_{16}{k}^4$
$\Rightarrow k^4=16\Rightarrow k^2=4\Rightarrow k=\pm 2.$
But $lo{g}_{16}k$ is not defined if $k<0,$
$\therefore k=2.$
$\therefore$ Number of real values of k = 1.