If `x_(1)` and `x_(2)` are the roots of the equation `sqrt(2010)x^(log_(2010^(x))) = x^(2)`, then find the Zeroes at the end of the product `(x_(1)x_(2))`.

Answer

To solve the equation \(\sqrt{2010} \cdot x^{\log_{2010}(x)} = x^2\) and find the number of zeros at the end of the product \(x_1 x_2\) (where \(x_1\) and \(x_2\) are the roots), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sqrt{2010} \cdot x^{\log_{2010}(x)} = x^2 \]
This can be rewritten as:
\[ 2010^{1/2} \cdot x^{\log_{2010}(x)} = x^2 \]

### Step 2: Take logarithm on both sides
Taking logarithm (base 2010) on both sides gives:
\[ \log_{2010}\left(2010^{1/2} \cdot x^{\log_{2010}(x)}\right) = \log_{2010}(x^2) \]

### Step 3: Apply logarithm properties
Using the properties of logarithms, we can expand the left side:
\[ \log_{2010}(2010^{1/2}) + \log_{2010}(x^{\log_{2010}(x)}) = 2 \log_{2010}(x) \]
This simplifies to:
\[ \frac{1}{2} + \log_{2010}(x) \cdot \log_{2010}(x) = 2 \log_{2010}(x) \]

### Step 4: Let \(t = \log_{2010}(x)\)
Letting \(t = \log_{2010}(x)\), we rewrite the equation:
\[ \frac{1}{2} + t^2 = 2t \]

### Step 5: Rearranging the equation
Rearranging gives us a standard quadratic equation:
\[ t^2 - 2t + \frac{1}{2} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = -2\), and \(c = \frac{1}{2}\):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{1}{2}}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 - 2}}{2} = \frac{2 \pm \sqrt{2}}{2} = 1 \pm \frac{1}{\sqrt{2}} \]

### Step 7: Finding \(x_1\) and \(x_2\)
Thus, the roots \(t_1\) and \(t_2\) are:
\[ t_1 = 1 + \frac{1}{\sqrt{2}}, \quad t_2 = 1 - \frac{1}{\sqrt{2}} \]
Converting back to \(x\):
\[ x_1 = 2010^{t_1}, \quad x_2 = 2010^{t_2} \]

### Step 8: Product of the roots
The product \(x_1 x_2\) is:
\[ x_1 x_2 = 2010^{t_1} \cdot 2010^{t_2} = 2010^{t_1 + t_2} \]
Calculating \(t_1 + t_2\):
\[ t_1 + t_2 = \left(1 + \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) = 2 \]
Thus,
\[ x_1 x_2 = 2010^2 \]

### Step 9: Finding the number of zeros at the end
To find the number of zeros at the end of \(2010^2\), we factor \(2010\):
\[ 2010 = 2 \times 3 \times 5 \times 67 \]
Thus,
\[ 2010^2 = 2^2 \times 3^2 \times 5^2 \times 67^2 \]
The number of zeros at the end of a number is determined by the minimum power of 10 in its prime factorization, which is \(2^1 \times 5^1\). Therefore, the number of zeros is:
\[ \min(2, 2) = 2 \]

### Final Answer
The number of zeros at the end of the product \(x_1 x_2\) is \(2\).

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