Sum of the roots of the equation `9^(log_(3)(log_(2)x))=log_(2)x-(log_(2)x)^(2)+1` is equal to

Options

A. `2`
B. `4`
C. `6`
D. `8`

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 left-hand side
We can rewrite \( 9^{\log_{3}(\log_{2}x)} \) as \( (3^2)^{\log_{3}(\log_{2}x)} \). By using the property of exponents, this simplifies to:
\[ 3^{2 \cdot \log_{3}(\log_{2}x)} = \log_{2}x \]
So, we have:
\[ \log_{2}x = \log_{2}(\log_{2}x)^2 \]

### Step 2: Let \( y = \log_{2}x \)
Substituting \( y \) for \( \log_{2}x \), we rewrite the equation:
\[ y^2 = y - y^2 + 1 \]
Rearranging gives us:
\[ 2y^2 - y - 1 = 0 \]

### Step 3: Solve the quadratic equation
Now we will solve the quadratic equation \( 2y^2 - y - 1 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = -1, c = -1 \):
\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ y = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ y = \frac{1 \pm 3}{4} \]
This gives us two solutions:
\[ y = 1 \quad \text{and} \quad y = -\frac{1}{2} \]

### Step 4: Find corresponding \( x \) values
Now we convert back to \( x \) using \( y = \log_{2}x \):
1. For \( y = 1 \):
\[ \log_{2}x = 1 \implies x = 2^1 = 2 \]
2. For \( y = -\frac{1}{2} \):
\[ \log_{2}x = -\frac{1}{2} \implies x = 2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}} \quad (\text{valid since } x > 0) \]

### Step 5: Sum of the roots
The roots we found are \( x = 2 \) and \( x = \frac{1}{\sqrt{2}} \). Now we calculate the sum of the roots:
\[ \text{Sum} = 2 + \frac{1}{\sqrt{2}} = 2 + \frac{\sqrt{2}}{2} = 2 + 0.707 \approx 2.707 \]

### Final Answer
Thus, the sum of the roots of the equation is:
\[ \text{Sum of the roots} = 2 + \frac{1}{\sqrt{2}} \]