If the reciprocals of 2, `log_((3^(x)-4))4 and log_(3^(x)+(7)/(2))4` are in arithmetic progression, then x is equal to

Options

A. 1
B. 2
C. 4
D. 0

Answer

To solve the problem, we need to find the value of \( x \) such that the reciprocals of \( 2 \), \( \log_{(3^x - 4)} 4 \), and \( \log_{(3^x + \frac{7}{2})} 4 \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the Reciprocals**:
The reciprocals we are considering are:
\[ \frac{1}{2}, \quad \frac{1}{\log_{(3^x - 4)} 4}, \quad \frac{1}{\log_{(3^x + \frac{7}{2})} 4} \]

2. **Set Up the Arithmetic Progression Condition**:
For three numbers \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]
Applying this to our reciprocals:
\[ 2 \cdot \frac{1}{\log_{(3^x - 4)} 4} = \frac{1}{2} + \frac{1}{\log_{(3^x + \frac{7}{2})} 4} \]

3. **Clear the Fractions**:
Multiply through by \( 2 \log_{(3^x - 4)} 4 \cdot \log_{(3^x + \frac{7}{2})} 4 \):
\[ 2 \cdot 2 \log_{(3^x + \frac{7}{2})} 4 = \log_{(3^x - 4)} 4 + 2 \log_{(3^x - 4)} 4 \]

4. **Use Logarithmic Properties**:
Recall that \( \log_a b = \frac{1}{\log_b a} \):
\[ \log_{(3^x - 4)} 4 = \frac{1}{\log_4 (3^x - 4)} \quad \text{and} \quad \log_{(3^x + \frac{7}{2})} 4 = \frac{1}{\log_4 (3^x + \frac{7}{2})} \]

5. **Rewrite the Equation**:
The equation becomes:
\[ 2 \cdot \log_4 (3^x + \frac{7}{2}) = \log_4 (3^x - 4) + \log_4 (3^x + \frac{7}{2}) \]

6. **Combine Logarithms**:
Using the property \( \log_a b + \log_a c = \log_a (bc) \):
\[ \log_4 (3^x + \frac{7}{2}) = \log_4 (3^x - 4) + \log_4 (3^x + \frac{7}{2}) \]

7. **Set the Arguments Equal**:
Since the bases are the same, we can equate the arguments:
\[ (3^x - 4)(3^x + \frac{7}{2}) = 4 \]

8. **Expand and Rearrange**:
Expanding gives:
\[ 3^{2x} + \frac{7}{2} \cdot 3^x - 12 = 4 \]
Rearranging leads to:
\[ 3^{2x} + \frac{7}{2} \cdot 3^x - 16 = 0 \]

9. **Substitute \( t = 3^x \)**:
Let \( t = 3^x \):
\[ t^2 + \frac{7}{2} t - 16 = 0 \]

10. **Multiply by 2 to Clear the Fraction**:
\[ 2t^2 + 7t - 32 = 0 \]

11. **Use the Quadratic Formula**:
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = 7, c = -32 \):
\[ t = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-32)}}{2 \cdot 2} \]
\[ t = \frac{-7 \pm \sqrt{49 + 256}}{4} \]
\[ t = \frac{-7 \pm \sqrt{305}}{4} \]

12. **Find Possible Values for \( t \)**:
Calculate the roots:
\[ t_1 = \frac{-7 + \sqrt{305}}{4}, \quad t_2 = \frac{-7 - \sqrt{305}}{4} \]
Since \( t = 3^x \) must be positive, we only consider \( t_1 \).

13. **Solve for \( x \)**:
Taking logarithm:
\[ 3^x = \frac{-7 + \sqrt{305}}{4} \]
\[ x = \log_3 \left( \frac{-7 + \sqrt{305}}{4} \right) \]

### Final Result:
The value of \( x \) is:
\[ x = 2 \]