The number of ordered pairs (x, y) satisfying `4("log"_(2) x^(2))^(2) + 1 = 2 "log"_(2)y " and log"_(2)x^(2) ge "log"_(2) y`, is

Options

A. 1
B. 2
C. more than 2 but finite
D. infinite

Explanation

To solve the problem step by step, we need to analyze the given equation and the conditions provided.

### Step 1: Rewrite the given equation
The equation we need to solve is:
\[ 4(\log_2 x^2)^2 + 1 = 2 \log_2 y \]

### Step 2: Simplify the logarithmic expressions
We can rewrite \(\log_2 x^2\) as \(2 \log_2 x\). Thus, we have:
\[ 4(2 \log_2 x)^2 + 1 = 2 \log_2 y \]
This simplifies to:
\[ 16 (\log_2 x)^2 + 1 = 2 \log_2 y \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 2 \log_2 y = 16 (\log_2 x)^2 + 1 \]
Dividing both sides by 2:
\[ \log_2 y = 8 (\log_2 x)^2 + \frac{1}{2} \]

### Step 4: Express \(y\) in terms of \(x\)
Exponentiating both sides to eliminate the logarithm:
\[ y = 2^{8 (\log_2 x)^2 + \frac{1}{2}} \]
This can be rewritten as:
\[ y = 2^{\frac{1}{2}} \cdot 2^{8 (\log_2 x)^2} = \sqrt{2} \cdot (x^8)^{\log_2 2} = \sqrt{2} \cdot x^8 \]

### Step 5: Analyze the second condition
The second condition given is:
\[ \log_2 x^2 \geq \log_2 y \]
This implies:
\[ x^2 \geq y \]

### Step 6: Substitute \(y\) into the inequality
Substituting \(y\) from Step 4 into the inequality:
\[ x^2 \geq \sqrt{2} \cdot x^8 \]

### Step 7: Rearranging the inequality
Rearranging gives us:
\[ x^2 - \sqrt{2} x^8 \geq 0 \]
Factoring out \(x^2\):
\[ x^2 (1 - \sqrt{2} x^6) \geq 0 \]

### Step 8: Finding the critical points
The critical points occur when:
1. \(x^2 = 0\) (which gives \(x = 0\), but \(x\) must be positive)
2. \(1 - \sqrt{2} x^6 = 0\) leading to:
\[ x^6 = \frac{1}{\sqrt{2}} \]
\[ x = \left(\frac{1}{\sqrt{2}}\right)^{\frac{1}{6}} = 2^{-\frac{1}{12}} \]

### Step 9: Determine the intervals
The inequality \(x^2 (1 - \sqrt{2} x^6) \geq 0\) holds true when:
- \(x^2 \geq 0\) (always true for \(x > 0\))
- \(1 - \sqrt{2} x^6 \geq 0\) which gives:
\[ x^6 \leq \frac{1}{\sqrt{2}} \]
Thus:
\[ x \leq \left(\frac{1}{\sqrt{2}}\right)^{\frac{1}{6}} = 2^{-\frac{1}{12}} \]

### Step 10: Finding the range for \(y\)
From the earlier steps, we have:
\[ y = \sqrt{2} \cdot x^8 \]
Substituting \(x = 2^{-\frac{1}{12}}\):
\[ y = \sqrt{2} \cdot \left(2^{-\frac{1}{12}}\right)^8 = \sqrt{2} \cdot 2^{-\frac{2}{3}} = 2^{\frac{1}{2} - \frac{2}{3}} = 2^{-\frac{1}{6}} \]

### Conclusion
The ordered pairs \((x, y)\) satisfy the conditions with \(y\) ranging from \(0\) to \(2^{-\frac{1}{6}}\) as \(x\) varies from \(0\) to \(2^{-\frac{1}{12}}\). Since \(y\) can take infinitely many values in the range, the number of ordered pairs \((x, y)\) is infinite.

### Final Answer
The number of ordered pairs \((x, y)\) satisfying the conditions is infinite.

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