The smallest integral x satisfying the inequality `(1-log_(4)x)/(1+log_(2)x)le (1)/(2)x is.

Options

A. `sqrt(2)`
B. 2
C. 3
D. 4

Explanation

To solve the inequality \(\frac{1 - \log_{4} x}{1 + \log_{2} x} \leq \frac{1}{2} x\), we will follow these steps:

### Step 1: Rewrite the logarithms
We start with the inequality:
\[ \frac{1 - \log_{4} x}{1 + \log_{2} x} \leq \frac{1}{2} x \]
We know that \(\log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{\log_{2} x}{2}\). Thus, we can rewrite the inequality as:
\[ \frac{1 - \frac{\log_{2} x}{2}}{1 + \log_{2} x} \leq \frac{1}{2} x \]

### Step 2: Substitute \(y = \log_{2} x\)
Let \(y = \log_{2} x\). Then, the inequality becomes:
\[ \frac{1 - \frac{y}{2}}{1 + y} \leq \frac{1}{2} \cdot 2^{y} \]
This simplifies to:
\[ \frac{1 - \frac{y}{2}}{1 + y} \leq 2^{y-1} \]

### Step 3: Clear the fraction
Multiply both sides by \(2(1 + y)\) (noting that \(1 + y > 0\) for \(y > -1\)):
\[ 2(1 - \frac{y}{2}) \leq 2^{y} (1 + y) \]
This simplifies to:
\[ 2 - y \leq 2^{y} + 2^{y} y \]

### Step 4: Rearranging the inequality
Rearranging gives:
\[ 2 - y - 2^{y} - 2^{y} y \leq 0 \]

### Step 5: Analyze the function
Let \(f(y) = 2 - y - 2^{y} - 2^{y} y\). We need to find the values of \(y\) for which \(f(y) \leq 0\).

### Step 6: Find critical points
We can check values of \(y\):
- For \(y = -1\):
\[ f(-1) = 2 + 1 - \frac{1}{2} - \frac{1}{2} = 2 \]
- For \(y = 0\):
\[ f(0) = 2 - 0 - 1 - 0 = 1 \]
- For \(y = 1\):
\[ f(1) = 2 - 1 - 2 - 2 = -3 \]
- For \(y = 2\):
\[ f(2) = 2 - 2 - 4 - 8 = -12 \]

### Step 7: Determine intervals
From the evaluations, we see that \(f(y)\) changes sign between \(y = 0\) and \(y = 1\). Thus, we can conclude:
- \(y < -1\) or \(y \geq 1\) gives us the solutions.

### Step 8: Convert back to \(x\)
Since \(y = \log_{2} x\):
- If \(y < -1\), then \(x < 2^{-1} = \frac{1}{2}\).
- If \(y \geq 1\), then \(x \geq 2^{1} = 2\).

### Step 9: Find the smallest integer
The smallest integer satisfying the inequality is:
\[ \boxed{2} \]