The ordered pair `(x,y)` satisfying the equation
`x^(2)=1+6 log_(4)y and y^(2)=2^(x)y+2^(2x+1)`
and `(x_(1), y_(1)) and (x_(2), y_(2))`, then find the value of `log_(2)|x_(1)x_(2)y_(1)y_(2)|`.

Answer

To solve the given equations and find the value of \( \log_{2}|x_{1}x_{2}y_{1}y_{2}| \), we will follow these steps:

### Step 1: Analyze the equations
We are given two equations:
1. \( x^2 = 1 + 6 \log_{4} y \)
2. \( y^2 = 2^x y + 2^{2x + 1} \)

### Step 2: Rewrite the second equation
From the second equation, we can rearrange it as follows:
\[ y^2 - 2^x y - 2^{2x + 1} = 0 \]
This is a quadratic equation in \( y \).

### Step 3: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2^x, c = -2^{2x + 1} \):
\[ y = \frac{2^x \pm \sqrt{(2^x)^2 - 4 \cdot 1 \cdot (-2^{2x + 1})}}{2 \cdot 1} \]
\[ y = \frac{2^x \pm \sqrt{2^{2x} + 4 \cdot 2^{2x + 1}}}{2} \]
\[ y = \frac{2^x \pm \sqrt{2^{2x} + 4 \cdot 2^{2x} \cdot 2}}{2} \]
\[ y = \frac{2^x \pm \sqrt{2^{2x}(1 + 4)}}{2} \]
\[ y = \frac{2^x \pm \sqrt{5 \cdot 2^{2x}}}{2} \]
\[ y = \frac{2^x \pm \sqrt{5} \cdot 2^x}{2} \]
\[ y = \frac{(1 \pm \sqrt{5}) \cdot 2^x}{2} \]

### Step 4: Determine valid values for \( y \)
Thus, we have two possible values for \( y \):
\[ y_1 = \frac{(1 + \sqrt{5}) \cdot 2^x}{2}, \quad y_2 = \frac{(1 - \sqrt{5}) \cdot 2^x}{2} \]
Since \( y \) must be positive, we will only consider \( y_1 \).

### Step 5: Substitute \( y_1 \) into the first equation
Now substitute \( y_1 \) into the first equation:
\[ x^2 = 1 + 6 \log_{4} \left( \frac{(1 + \sqrt{5}) \cdot 2^x}{2} \right) \]
Using the change of base formula:
\[ \log_{4} a = \frac{\log_{2} a}{\log_{2} 4} = \frac{\log_{2} a}{2} \]
Thus,
\[ x^2 = 1 + 6 \cdot \frac{1}{2} \log_{2} \left( \frac{(1 + \sqrt{5}) \cdot 2^x}{2} \right) \]
\[ x^2 = 1 + 3 \log_{2} \left( (1 + \sqrt{5}) \cdot 2^{x-1} \right) \]
\[ x^2 = 1 + 3 \left( \log_{2}(1 + \sqrt{5}) + (x - 1) \right) \]
\[ x^2 = 1 + 3 \log_{2}(1 + \sqrt{5}) + 3x - 3 \]
\[ x^2 - 3x + 2 + 3 \log_{2}(1 + \sqrt{5}) = 0 \]

### Step 6: Solve the quadratic equation for \( x \)
Using the quadratic formula:
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (2 + 3 \log_{2}(1 + \sqrt{5}))}}{2 \cdot 1} \]
\[ x = \frac{3 \pm \sqrt{9 - 8 - 12 \log_{2}(1 + \sqrt{5})}}{2} \]
\[ x = \frac{3 \pm \sqrt{1 - 12 \log_{2}(1 + \sqrt{5})}}{2} \]

### Step 7: Find \( y_1 \) and \( y_2 \)
Substituting the values of \( x \) back into the equation for \( y \):
\[ y_1 = \frac{(1 + \sqrt{5}) \cdot 2^{x_1}}{2}, \quad y_2 = \frac{(1 - \sqrt{5}) \cdot 2^{x_2}}{2} \]

### Step 8: Calculate \( \log_{2}|x_{1}x_{2}y_{1}y_{2}| \)
Now we need to find:
\[ \log_{2}|x_{1}x_{2}y_{1}y_{2}| \]
Substituting the values:
\[ \log_{2}|x_1 x_2 y_1 y_2| = \log_{2}|x_1 x_2 \cdot \frac{(1 + \sqrt{5}) \cdot 2^{x_1}}{2} \cdot \frac{(1 - \sqrt{5}) \cdot 2^{x_2}}{2}| \]
\[ = \log_{2}|x_1 x_2 \cdot \frac{(1 + \sqrt{5})(1 - \sqrt{5})}{4} \cdot 2^{x_1 + x_2}| \]
\[ = \log_{2}|x_1 x_2| + \log_{2}\left(\frac{(1 + \sqrt{5})(1 - \sqrt{5})}{4}\right) + \log_{2}(2^{x_1 + x_2}) \]
\[ = \log_{2}|x_1 x_2| + \log_{2}(-1) + (x_1 + x_2) \]

### Final Step: Conclusion
The final answer will depend on the specific values of \( x_1 \) and \( x_2 \) derived from the quadratic solutions.