If `log_(3)2,log_(3)(2^(x)-5),log_(3)(2^(x)-7/2)` are in arithmetic progression, then the value of x is equal to

Options

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

Answer

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

### Step-by-step Solution:

1. **Understanding Arithmetic Progression**:
For three terms \( a, b, c \) to be in AP, the condition is:
\[ a + c = 2b \]
Here, we can assign:
- \( a = \log_3 2 \)
- \( b = \log_3 (2^x - 5) \)
- \( c = \log_3 \left(2^x - \frac{7}{2}\right) \)

2. **Setting Up the Equation**:
Using the AP condition, we write:
\[ \log_3 2 + \log_3 \left(2^x - \frac{7}{2}\right) = 2 \log_3 (2^x - 5) \]

3. **Using Logarithmic Properties**:
We can combine the left-hand side using the property of logarithms:
\[ \log_3 \left(2 \cdot \left(2^x - \frac{7}{2}\right)\right) = \log_3 \left((2^x - 5)^2\right) \]
This gives us:
\[ 2 \cdot \left(2^x - \frac{7}{2}\right) = (2^x - 5)^2 \]

4. **Expanding the Equation**:
Expanding both sides:
\[ 2 \cdot 2^x - 7 = (2^x - 5)(2^x - 5) \]
\[ 2 \cdot 2^x - 7 = (2^x)^2 - 10 \cdot 2^x + 25 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 0 = (2^x)^2 - 12 \cdot 2^x + 32 \]

6. **Substituting \( t = 2^x \)**:
Let \( t = 2^x \). The equation becomes:
\[ t^2 - 12t + 32 = 0 \]

7. **Solving the Quadratic Equation**:
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \]
\[ t = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2} \]
This gives:
\[ t = \frac{16}{2} = 8 \quad \text{or} \quad t = \frac{8}{2} = 4 \]

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

9. **Checking Validity**:
We need to check which value satisfies the logarithmic conditions:
- For \( x = 2 \): \( \log_3 (2^2 - 5) = \log_3 (-1) \) (not valid).
- For \( x = 3 \): \( \log_3 (2^3 - 5) = \log_3 (3) \) and \( \log_3 (2^3 - \frac{7}{2}) = \log_3 \left(\frac{1}{2}\right) \) (valid).

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