If log 2, $log(2^{x}-1)$, $log(2^{x}+3)$ are in arithmetic progression, then x is equal to:

Question

Accepted Answer

To solve the problem, we need to verify why Option 2, $log_{2}5$, is the correct answer for the value of $ x $ when $ \log 2 $, $ \log(2^{x}-1) $, and $ \log(2^{x}+3) $ are in arithmetic progression.

### Explanation:

1. **Understanding Arithmetic Progression (AP):**
   - In an arithmetic progression, the difference between consecutive terms is constant. If $ a $, $ b $, and $ c $ are in AP, then $ 2b = a + c $.

2. **Applying AP to the Given Logs:**
   - Let $ a = \log 2 $, $ b = \log(2^{x}-1) $, and $ c = \log(2^{x}+3) $.
   - According to the AP property: 
     $$
     2\log(2^{x}-1) = \log 2 + \log(2^{x}+3)
     $$

3. **Using Logarithmic Properties:**
   - The equation becomes:
     $$
     \log((2^{x}-1)^2) = \log(2 \cdot (2^{x}+3))
     $$
   - By the property of logarithms, if $ \log A = \log B $, then $ A = B $.
   - Therefore:
     $$
     (2^{x}-1)^2 = 2 \cdot (2^{x}+3)
     $$

4. **Solving the Equation:**
   - Expanding both sides:
     $$
     (2^{x}-1)^2 = 2^{2x} - 2 \cdot 2^{x} + 1
     $$
     $$
     2 \cdot (2^{x}+3) = 2^{x+1} + 6
     $$
   - Equating:
     $$
     2^{2x} - 2 \cdot 2^{x} + 1 = 2^{x+1} + 6
     $$
   - Simplifying:
     $$
     2^{2x} - 2 \cdot 2^{x} - 2^{x+1} + 1 - 6 = 0
     $$
     $$
     2^{2x} - 4 \cdot 2^{x} - 5 = 0
     $$
   - Let $ y = 2^{x} $, then the equation becomes:
     $$
     y^2 - 4y - 5 = 0
     $$
   - Solving this quadratic equation:
     $$
     (y - 5)(y + 1) = 0
     $$
   - Thus, $ y = 5 $ or $ y = -1 $. Since $ y = 2^{x} $ and must be positive, $ y = 5 $.

5. **Finding $ x $:**
   - Since $ y = 2^{x} = 5 $, we have:
     $$
     x = \log_{2}5
     $$

### Conclusion:

- **Option 2: $ \log_{2}5 $ is the correct answer** because it satisfies the condition that the logarithms are in arithmetic progression.
- **Other Options:**
  - **Option 1: $ 5/2 $** does not satisfy the equation derived from the AP condition.
  - **Option 3: $ \log_{3}2 $** and **Option 4: $ \log_{5}3 $** similarly do not satisfy the derived equation, as they do not equate $ 2^{x} $ to 5.

Thus, the correct answer is indeed Option 2, $ \log_{2}5 $.