If $log_{10}2$, $log_{10}(2^{x}-1)$ and $log_{10}(2^{x}+3)$ be three consecutive terms of an arithmetic progression, then :

Question

Accepted Answer

To solve the problem, we need to verify why Option 2 ($x = \log_{2}5$) is the correct answer, given that $log_{10}2$, $log_{10}(2^{x}-1)$, and $log_{10}(2^{x}+3)$ are consecutive terms of an arithmetic progression (AP).

### Explanation:

1. **Understanding the Arithmetic Progression:**
   - In an arithmetic progression, the difference between consecutive terms is constant. Therefore, if $a$, $b$, and $c$ are consecutive terms, then $b - a = c - b$.
   - Applying this to our logarithmic terms:
     $$
     \log_{10}(2^{x} - 1) - \log_{10}2 = \log_{10}(2^{x} + 3) - \log_{10}(2^{x} - 1)
     $$

2. **Simplifying the Equation:**
   - Using the properties of logarithms, we can rewrite the differences:
     $$
     \log_{10}\left(\frac{2^{x} - 1}{2}\right) = \log_{10}\left(\frac{2^{x} + 3}{2^{x} - 1}\right)
     $$
   - Equating the arguments of the logarithms (since the bases are the same and the logarithm function is one-to-one):
     $$
     \frac{2^{x} - 1}{2} = \frac{2^{x} + 3}{2^{x} - 1}
     $$

3. **Solving the Equation:**
   - Cross-multiplying gives:
     $$
     (2^{x} - 1)^2 = 2(2^{x} + 3)
     $$
   - Expanding and simplifying:
     $$
     2^{2x} - 2 \cdot 2^{x} + 1 = 2 \cdot 2^{x} + 6
     $$
     $$
     2^{2x} - 4 \cdot 2^{x} - 5 = 0
     $$
   - Let $y = 2^{x}$. The equation becomes:
     $$
     y^2 - 4y - 5 = 0
     $$
   - Solving this quadratic equation using the quadratic formula:
     $$
     y = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}
     $$
   - This gives $y = 5$ or $y = -1$. Since $y = 2^{x}$ must be positive, we have $y = 5$.

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

### Conclusion:

- **Option 2 ($x = \log_{2}5$) is correct** because it satisfies the condition that the logarithmic terms form an arithmetic progression.
- **Why Other Options are Incorrect:**
  - **Option 1 ($x = \log_{2}3$):** If $x = \log_{2}3$, then $2^{x} = 3$, which does not satisfy the derived equation $y^2 - 4y - 5 = 0$.
  - **Option 3 ($x = \log_{5}3$):** This implies $2^{x} = 3^{\log_{5}2}$, which again does not satisfy the quadratic equation.
  - **Option 4 ($x = \log_{5}2$):** This implies $2^{x} = 2^{\log_{5}2}$, which is not a solution to the quadratic equation derived.

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