Solve the following inequation .
(iii) `log_(2)(2-x)ltlog_(1//2)(x+1)`

Answer

To solve the inequation \( \log_2(2-x) < \log_{1/2}(x+1) \), we will follow these steps:

### Step 1: Rewrite the logarithm with base \( \frac{1}{2} \)
Using the property of logarithms, we can rewrite \( \log_{1/2}(x+1) \) as:
\[ \log_{1/2}(x+1) = -\log_2(x+1) \]
Thus, the inequation becomes:
\[ \log_2(2-x) < -\log_2(x+1) \]

### Step 2: Combine the logarithms
We can rewrite the inequation as:
\[ \log_2(2-x) + \log_2(x+1) < 0 \]
Using the property of logarithms \( \log_a(b) + \log_a(c) = \log_a(bc) \), we have:
\[ \log_2((2-x)(x+1)) < 0 \]

### Step 3: Exponentiate both sides
Exponentiating both sides gives:
\[ (2-x)(x+1) < 1 \]

### Step 4: Expand and rearrange the inequality
Expanding the left side:
\[ 2 - x + 2x - x^2 < 1 \implies -x^2 + x + 2 < 1 \]
Rearranging gives:
\[ -x^2 + x + 1 < 0 \implies x^2 - x - 1 > 0 \]

### Step 5: Solve the quadratic inequality
To solve \( x^2 - x - 1 > 0 \), we first find the roots of the equation \( x^2 - x - 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ x_1 = \frac{1 - \sqrt{5}}{2}, \quad x_2 = \frac{1 + \sqrt{5}}{2} \]

### Step 6: Test intervals
The roots divide the number line into three intervals:
1. \( (-\infty, \frac{1 - \sqrt{5}}{2}) \)
2. \( \left(\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right) \)
3. \( \left(\frac{1 + \sqrt{5}}{2}, \infty\right) \)

We will test a point from each interval to determine where the inequality \( x^2 - x - 1 > 0 \) holds.

- For \( x < \frac{1 - \sqrt{5}}{2} \) (e.g., \( x = -1 \)):
\[ (-1)^2 - (-1) - 1 = 1 + 1 - 1 = 1 > 0 \quad \text{(True)} \]

- For \( \frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2} \) (e.g., \( x = 0 \)):
\[ 0^2 - 0 - 1 = -1 < 0 \quad \text{(False)} \]

- For \( x > \frac{1 + \sqrt{5}}{2} \) (e.g., \( x = 2 \)):
\[ 2^2 - 2 - 1 = 4 - 2 - 1 = 1 > 0 \quad \text{(True)} \]

### Step 7: Conclusion
The solution to the inequality \( x^2 - x - 1 > 0 \) is:
\[ x \in \left(-\infty, \frac{1 - \sqrt{5}}{2}\right) \cup \left(\frac{1 + \sqrt{5}}{2}, \infty\right) \]