If `9^("log"3("log"_(2) x)) = "log"_(2)x - ("log"_(2)x)^(2) + 1,` then x =

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To solve the equation \( 9^{\log_3(\log_2 x)} = \log_2 x - (\log_2 x)^2 + 1 \), we will follow these steps:

### Step 1: Rewrite the base
We can rewrite \( 9 \) as \( 3^2 \):
\[ 9^{\log_3(\log_2 x)} = (3^2)^{\log_3(\log_2 x)} = 3^{2 \log_3(\log_2 x)} \]
Using the property \( a^{\log_a b} = b \), we have:
\[ 3^{2 \log_3(\log_2 x)} = \log_2 x^2 \]

### Step 2: Set the equation
Now we can set the equation:
\[ \log_2 x^2 = \log_2 x - (\log_2 x)^2 + 1 \]

### Step 3: Let \( t = \log_2 x \)
Let \( t = \log_2 x \). Then the equation becomes:
\[ t^2 = t - t^2 + 1 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 2t^2 - t - 1 = 0 \]

### Step 5: Solve the quadratic equation
Now we can solve the quadratic equation \( 2t^2 - t - 1 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = -1, c = -1 \):
\[ t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ t = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]
Calculating the two possible values:
1. \( t = \frac{4}{4} = 1 \)
2. \( t = \frac{-2}{4} = -\frac{1}{2} \)

### Step 6: Convert back to \( x \)
Now we convert back to \( x \):
1. If \( t = 1 \):
\[ \log_2 x = 1 \implies x = 2^1 = 2 \]
2. If \( t = -\frac{1}{2} \):
\[ \log_2 x = -\frac{1}{2} \implies x = 2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \]

### Step 7: Determine valid solutions
Since we need \( x > 1 \), the only valid solution is:
\[ x = 2 \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{2} \]