b
(b) Let \({\log _{16}}x = y \Rightarrow {y^2} - y + {\log _{16}}k = 0\)

This quadratic equation will have exactly one solution if its discriminant vanishes.

\(\therefore {( - 1)^2} - 4.1.{\log _{16}}k = 0 \Rightarrow 1 = {\log _{16}}{k^4}\)

\( \Rightarrow \)\({k^4} = 16\) \( \Rightarrow \) \({k^2} = 4\) \( \Rightarrow \) \(k = \pm 2\).

But \({\log _{16}}k\) is not defined \(k < 0\), \(k = 2\).

\(\therefore\) Number of real values of \(k = 1\).