Solve for x: `log_(2)x le 2/(log_(2)x-1)`

Answer

To solve the inequality \( \log_2 x \leq \frac{2}{\log_2 x - 1} \), we will follow these steps:

### Step 1: Rearranging the Inequality
Start by moving all terms to one side of the inequality:
\[ \log_2 x - \frac{2}{\log_2 x - 1} \leq 0 \]

### Step 2: Substituting Variables
Let \( t = \log_2 x \). Then the inequality becomes:
\[ t - \frac{2}{t - 1} \leq 0 \]

### Step 3: Finding a Common Denominator
To combine the terms, find a common denominator:
\[ \frac{t(t - 1) - 2}{t - 1} \leq 0 \]
This simplifies to:
\[ \frac{t^2 - t - 2}{t - 1} \leq 0 \]

### Step 4: Factoring the Numerator
Now, factor the quadratic in the numerator:
\[ \frac{(t - 2)(t + 1)}{t - 1} \leq 0 \]

### Step 5: Finding Critical Points
The critical points are found by setting the numerator and denominator to zero:
1. \( t - 2 = 0 \) gives \( t = 2 \)
2. \( t + 1 = 0 \) gives \( t = -1 \)
3. \( t - 1 = 0 \) gives \( t = 1 \)

### Step 6: Testing Intervals
Now we test the intervals determined by these critical points: \( (-\infty, -1) \), \( (-1, 1) \), \( (1, 2) \), and \( (2, \infty) \).

- **Interval \( (-\infty, -1) \)**: Choose \( t = -2 \):
\[ \frac{(-2 - 2)(-2 + 1)}{-2 - 1} = \frac{-4 \cdot -1}{-3} > 0 \]
- **Interval \( (-1, 1) \)**: Choose \( t = 0 \):
\[ \frac{(0 - 2)(0 + 1)}{0 - 1} = \frac{-2 \cdot 1}{-1} > 0 \]
- **Interval \( (1, 2) \)**: Choose \( t = 1.5 \):
\[ \frac{(1.5 - 2)(1.5 + 1)}{1.5 - 1} = \frac{-0.5 \cdot 2.5}{0.5} < 0 \]
- **Interval \( (2, \infty) \)**: Choose \( t = 3 \):
\[ \frac{(3 - 2)(3 + 1)}{3 - 1} = \frac{1 \cdot 4}{2} > 0 \]

### Step 7: Conclusion on Intervals
The inequality \( \frac{(t - 2)(t + 1)}{t - 1} \leq 0 \) holds in the interval \( (1, 2] \).

### Step 8: Back Substituting for \( x \)
Now, convert back to \( x \):
- From \( t = \log_2 x \):
- \( t = 1 \) gives \( x = 2^1 = 2 \)
- \( t = 2 \) gives \( x = 2^2 = 4 \)

### Step 9: Considering the Domain
Since \( \log_2 x \) is defined for \( x > 0 \), we also need to consider the domain:
- The solution must be \( x > 0 \).

### Final Solution
Thus, the solution set is:
\[ x \in (0, \frac{1}{2}) \cup (2, 4] \]