Let `x_(1)` and `x_(2)` be two solutions of the equalition `log_(x)(3x^(log_(5)x)+4) = 2log_(5)x` , then the product `x_(1)x_(2)` is equal to

Options

A. 2
B. 4
C. 3
D. 1

Explanation

To solve the equation \( \log_{x}(3x^{\log_{5}x}+4) = 2\log_{5}x \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \log_{x}(3x^{\log_{5}x}+4) = 2\log_{5}x \]

### Step 2: Change of Base
Let \( u = \log_{5}x \). Then, we can express \( x \) as:
\[ x = 5^{u} \]
Now, substituting \( x \) into the equation, we have:
\[ \log_{5^{u}}(3(5^{u})^{u}+4) = 2u \]

### Step 3: Simplifying the Left Side
Using the change of base formula, we can rewrite the left side:
\[ \frac{\log_{5}(3(5^{u})^{u}+4)}{\log_{5}(5^{u})} = \frac{\log_{5}(3(5^{u})^{u}+4)}{u} \]
Thus, the equation becomes:
\[ \frac{\log_{5}(3(5^{u})^{u}+4)}{u} = 2u \]

### Step 4: Multiply Both Sides by \( u \)
Multiplying both sides by \( u \) gives:
\[ \log_{5}(3(5^{u})^{u}+4) = 2u^{2} \]

### Step 5: Expressing \( (5^{u})^{u} \)
We know that \( (5^{u})^{u} = 5^{u^2} \), so we rewrite the equation:
\[ \log_{5}(3 \cdot 5^{u^2} + 4) = 2u^{2} \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm gives:
\[ 3 \cdot 5^{u^2} + 4 = 5^{2u^{2}} \]

### Step 7: Rearranging the Equation
Rearranging the equation leads to:
\[ 5^{2u^{2}} - 3 \cdot 5^{u^2} - 4 = 0 \]

### Step 8: Letting \( p = 5^{u^2} \)
Let \( p = 5^{u^2} \). The equation then becomes:
\[ p^2 - 3p - 4 = 0 \]

### Step 9: Factoring the Quadratic
Factoring gives:
\[ (p - 4)(p + 1) = 0 \]
Thus, we have:
\[ p - 4 = 0 \quad \text{or} \quad p + 1 = 0 \]
This leads to:
\[ p = 4 \quad \text{or} \quad p = -1 \]
Since \( p = 5^{u^2} \) cannot be negative, we discard \( p = -1 \).

### Step 10: Solving for \( u \)
From \( p = 4 \):
\[ 5^{u^2} = 4 \]
Taking logarithm base 5:
\[ u^2 = \log_{5}4 \]
Thus:
\[ u = \pm \sqrt{\log_{5}4} \]

### Step 11: Finding \( x_1 \) and \( x_2 \)
Recall that \( u = \log_{5}x \):
\[ \log_{5}x_1 = \sqrt{\log_{5}4} \quad \text{and} \quad \log_{5}x_2 = -\sqrt{\log_{5}4} \]
This gives:
\[ x_1 = 5^{\sqrt{\log_{5}4}} \quad \text{and} \quad x_2 = 5^{-\sqrt{\log_{5}4}} \]

### Step 12: Finding the Product \( x_1x_2 \)
Now, we calculate the product:
\[ x_1 x_2 = 5^{\sqrt{\log_{5}4}} \cdot 5^{-\sqrt{\log_{5}4}} = 5^{0} = 1 \]

Thus, the product \( x_1 x_2 \) is equal to \( 1 \).

### Final Answer
The product \( x_1 x_2 \) is \( \boxed{1} \).