The number of real values of the parameter k | vedclass

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(B) Let $\log_{16} x = y$. The equation becomes $y^2 - y + \log_{16} k = 0$.This quadratic equation in $y$ will have exactly one solution for $x$ if t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $\log_{16} x = y$. The equation becomes $y^2 - y + \log_{16} k = 0$.This quadratic equation in $y$ will have exactly one solution for $x$ if the discriminant $D$ is equal to $0$.$D = (-1)^2 - 4(1)(\log_{16} k) = 0$.$1 - 4\log_{16} k = 0 \Rightarrow \log_{16} k = \frac{1}{4}$.$k = 16^{1/4} = (2^4)^{1/4} = 2$.Since $\log_{16} k$ must be defined,we require $k > 0$. Thus,$k = 2$ is the only valid real value.Therefore,the number of real values of $k$ is $1$.

The number of real values of the parameter $k$ for which the equation $(\log_{16} x)^2 - \log_{16} x + \log_{16} k = 0$ has exactly one real solution is:

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