If the equation `x^(log_(a)x^(2))=(x^(k-2))/a^(k),a ne 0`has exactly one solution for x, then the value of k is/are

Options

A. `6+4sqrt2`
B. `2+6sqrt3)`
C. `6-4sqrt2`
D. `2-6sqrt3`

Explanation

To solve the equation \( x^{\log_a{x^2}} = \frac{x^{k-2}}{a^k} \) for the value of \( k \) such that there is exactly one solution for \( x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^{\log_a{x^2}} = \frac{x^{k-2}}{a^k} \]

### Step 2: Take logarithm on both sides
Taking logarithm to the base \( a \) on both sides, we have:
\[ \log_a{x^{\log_a{x^2}}} = \log_a{\left(\frac{x^{k-2}}{a^k}\right)} \]

### Step 3: Apply logarithmic properties
Using the property \( \log_b{(m^n)} = n \log_b{m} \) and \( \log_b{\left(\frac{m}{n}\right)} = \log_b{m} - \log_b{n} \), we can rewrite both sides:
\[ \log_a{x^2} \cdot \log_a{x} = \log_a{x^{k-2}} - \log_a{a^k} \]
This simplifies to:
\[ 2 \log_a{x} \cdot \log_a{x} = (k-2) \log_a{x} - k \]
or
\[ 2 (\log_a{x})^2 = (k-2) \log_a{x} - k \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ 2 (\log_a{x})^2 - (k-2) \log_a{x} + k = 0 \]

### Step 5: Let \( t = \log_a{x} \)
Let \( t = \log_a{x} \). The equation becomes:
\[ 2t^2 - (k-2)t + k = 0 \]

### Step 6: Condition for exactly one solution
For this quadratic equation to have exactly one solution, the discriminant must be zero:
\[ b^2 - 4ac = 0 \]
Here, \( a = 2 \), \( b = -(k-2) \), and \( c = k \). Thus, we have:
\[ (k-2)^2 - 4 \cdot 2 \cdot k = 0 \]

### Step 7: Expand and simplify the discriminant
Expanding gives:
\[ (k-2)^2 - 8k = 0 \]
This simplifies to:
\[ k^2 - 4k + 4 - 8k = 0 \]
or
\[ k^2 - 12k + 4 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ k = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
Calculating the discriminant:
\[ k = \frac{12 \pm \sqrt{144 - 16}}{2} = \frac{12 \pm \sqrt{128}}{2} = \frac{12 \pm 8\sqrt{2}}{2} = 6 \pm 4\sqrt{2} \]

### Final Result
Thus, the values of \( k \) are:
\[ k = 6 + 4\sqrt{2} \quad \text{and} \quad k = 6 - 4\sqrt{2} \]

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