The number of real values of the parameter `k` for which `(log_(16)x)^2-(log)_(16)x+(log)_(16)k=0` with real coefficients will have exactly one solution is 2 (b) 1 (c) 4 (d) none of these

Options

A. 2
B. 1
C. 4
D. none of these

Explanation

To solve the problem, we need to find 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 solution. Let's denote \( t = \log_{16} x \). The equation can then be rewritten as:

\[ t^2 - t + \log_{16} k = 0 \]

This is a quadratic equation in \( t \). For a quadratic equation \( at^2 + bt + c = 0 \) to have exactly one solution, the discriminant must be zero. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = 1 \), \( b = -1 \), and \( c = \log_{16} k \). Thus, the discriminant becomes:

\[ D = (-1)^2 - 4 \cdot 1 \cdot \log_{16} k = 1 - 4 \log_{16} k \]

Setting the discriminant equal to zero for the quadratic to have exactly one solution:

\[ 1 - 4 \log_{16} k = 0 \]

Solving for \( \log_{16} k \):

\[ 4 \log_{16} k = 1 \]

\[ \log_{16} k = \frac{1}{4} \]

Now, we can convert this logarithmic equation into its exponential form:

\[ k = 16^{\frac{1}{4}} \]

Calculating \( 16^{\frac{1}{4}} \):

\[ 16 = 2^4 \implies 16^{\frac{1}{4}} = (2^4)^{\frac{1}{4}} = 2^{4 \cdot \frac{1}{4}} = 2^1 = 2 \]

Thus, we have found one value of \( k \):

\[ k = 2 \]

Next, we should check if there are any other possible values for \( k \). The equation \( \log_{16} k = \frac{1}{4} \) gives us a unique solution for \( k \) since the logarithmic function is one-to-one.

Therefore, the only real value of \( k \) that satisfies the condition for the quadratic equation to have exactly one solution is:

\[ k = 2 \]

So, the number of real values of the parameter \( k \) is:

\[ \boxed{1} \]