Find the product of the positive roots of the equation `sqrt2009(x)^(log_(2009)(x))=x^2`.

Answer

To solve the equation \( \sqrt{2009} \cdot x^{\log_{2009}(x)} = x^2 \), we will follow a systematic approach to simplify and find the roots.

### Step-by-Step Solution:

1. **Take logarithm on both sides**:
\[ \log_{2009}(\sqrt{2009} \cdot x^{\log_{2009}(x)}) = \log_{2009}(x^2) \]

2. **Apply the properties of logarithms**:
Using the product property of logarithms, we can split the left-hand side:
\[ \log_{2009}(\sqrt{2009}) + \log_{2009}(x^{\log_{2009}(x)}) = \log_{2009}(x^2) \]

3. **Simplify each term**:
- The first term can be simplified:
\[ \log_{2009}(\sqrt{2009}) = \log_{2009}(2009^{1/2}) = \frac{1}{2} \]
- The second term can be simplified using the power property:
\[ \log_{2009}(x^{\log_{2009}(x)}) = \log_{2009}(x) \cdot \log_{2009}(x) \]
- The right-hand side simplifies to:
\[ \log_{2009}(x^2) = 2 \log_{2009}(x) \]

4. **Combine the simplified terms**:
Now we can rewrite the equation:
\[ \frac{1}{2} + (\log_{2009}(x))^2 = 2 \log_{2009}(x) \]

5. **Rearrange into a standard quadratic form**:
Let \( y = \log_{2009}(x) \). The equation becomes:
\[ y^2 - 2y + \frac{1}{2} = 0 \]

6. **Use the quadratic formula to find \( y \)**:
The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = \frac{1}{2} \):
\[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{1}{2}}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 2}}{2} = \frac{2 \pm \sqrt{2}}{2} = 1 \pm \frac{\sqrt{2}}{2} \]

7. **Convert back to \( x \)**:
Since \( y = \log_{2009}(x) \):
\[ x_1 = 2009^{1 + \frac{\sqrt{2}}{2}}, \quad x_2 = 2009^{1 - \frac{\sqrt{2}}{2}} \]

8. **Find the product of the roots**:
The product of the roots \( x_1 \cdot x_2 \) is given by:
\[ x_1 \cdot x_2 = 2009^{(1 + \frac{\sqrt{2}}{2}) + (1 - \frac{\sqrt{2}}{2})} = 2009^{2} \]

### Final Answer:
The product of the positive roots of the equation is:
\[ \boxed{2009^2} \]