If `log_(3)2, log_(3)(2^(x)-5)` and `log_(3)(2^(x)-7//2)` are in AP then x is equal to:

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to determine the value of \( x \) such that the logarithmic expressions \( \log_3 2 \), \( \log_3 (2^x - 5) \), and \( \log_3 \left( \frac{2^x - 7}{2} \right) \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding AP Condition**:
For three terms \( A \), \( B \), and \( C \) to be in AP, the condition is:
\[ 2B = A + C \]
Here, let:
- \( A = \log_3 2 \)
- \( B = \log_3 (2^x - 5) \)
- \( C = \log_3 \left( \frac{2^x - 7}{2} \right) \)

2. **Applying the AP Condition**:
Substitute \( A \), \( B \), and \( C \) into the AP condition:
\[ 2 \log_3 (2^x - 5) = \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) \]

3. **Using Logarithmic Properties**:
The property of logarithms states that \( \log_a b + \log_a c = \log_a (bc) \). Thus, we can rewrite \( C \):
\[ \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) = \log_3 \left( 2 \cdot \frac{2^x - 7}{2} \right) = \log_3 (2^x - 7) \]
So, we have:
\[ 2 \log_3 (2^x - 5) = \log_3 (2^x - 7) \]

4. **Exponentiating Both Sides**:
To eliminate the logarithm, we exponentiate both sides:
\[ (2^x - 5)^2 = 2^x - 7 \]

5. **Expanding and Rearranging**:
Expanding the left side:
\[ 2^{2x} - 10 \cdot 2^x + 25 = 2^x - 7 \]
Rearranging gives:
\[ 2^{2x} - 11 \cdot 2^x + 32 = 0 \]

6. **Substituting for Simplicity**:
Let \( t = 2^x \). The equation becomes:
\[ t^2 - 11t + 32 = 0 \]

7. **Factoring the Quadratic**:
We can factor this quadratic:
\[ (t - 4)(t - 8) = 0 \]
Thus, \( t = 4 \) or \( t = 8 \).

8. **Finding Values of \( x \)**:
Recall \( t = 2^x \):
- If \( t = 4 \), then \( 2^x = 4 \) implies \( x = 2 \).
- If \( t = 8 \), then \( 2^x = 8 \) implies \( x = 3 \).

9. **Checking Validity**:
We need to ensure that both values of \( x \) satisfy the original logarithmic expressions:
- For \( x = 2 \):
\[ \log_3 (2^2 - 5) = \log_3 (-1) \quad \text{(not valid)} \]
- For \( x = 3 \):
\[ \log_3 (2^3 - 5) = \log_3 (3) \quad \text{(valid)} \]
\[ \log_3 \left( \frac{2^3 - 7}{2} \right) = \log_3 \left( \frac{1}{2} \right) \quad \text{(valid)} \]

Thus, the only valid solution is:
\[ \boxed{3} \]