If `log_(3)2,log_(3)(2^(x)-5)` and `log_(3)(2^(x)-7//2)` are in arithmetic progression, then x=

Options

A. `2`
B. `3`
C. `4`
D. none

Answer

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

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression**:
For three numbers \( a \), \( b \), 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. **Setting Up the Equation**:
Using 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**:
We can combine the logs on the right side:
\[ \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) = \log_3 \left( 2 \cdot \frac{2^x - 7}{2} \right) = \log_3 (2^x - 7) \]
Thus, our equation becomes:
\[ 2 \log_3 (2^x - 5) = \log_3 (2^x - 7) \]

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:
\[ 4^x - 10 \cdot 2^x + 25 = 2^x - 7 \]
Rearranging gives:
\[ 4^x - 11 \cdot 2^x + 32 = 0 \]

6. **Substituting \( y = 2^x \)**:
Let \( y = 2^x \). Then we have:
\[ y^2 - 11y + 32 = 0 \]

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

8. **Finding \( x \)**:
- If \( y = 8 \), then \( 2^x = 8 \) implies \( x = 3 \).
- If \( y = 4 \), then \( 2^x = 4 \) implies \( x = 2 \).

9. **Checking Validity**:
We need to check if both values satisfy the original logarithmic conditions:
- For \( x = 3 \):
- \( 2^3 - 5 = 3 \) (valid)
- \( 2^3 - 7 = 1 \) (valid)
- For \( x = 2 \):
- \( 2^2 - 5 = -1 \) (invalid, as log of a negative number is not defined)

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