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

### Step 1: Write down the reciprocals
The reciprocals of the given terms are:
1. \( \frac{1}{2} \)
2. \( \frac{1}{\log_{(3^x - 4)} 4} \)
3. \( \frac{1}{\log_{(3^x + \frac{7}{2})} 4} \)

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

### Step 3: Multiply through by \( 2 \log_{(3^x - 4)} 4 \cdot \log_{(3^x + \frac{7}{2})} 4 \)
This gives:
\[ 2 \cdot 2 = \log_{(3^x - 4)} 4 + 2 \cdot \log_{(3^x + \frac{7}{2})} 4 \]

### Step 4: Use the change of base formula
Using the change of base formula, we can rewrite the logarithms:
\[ \log_{(3^x - 4)} 4 = \frac{\log 4}{\log(3^x - 4)} \]
\[ \log_{(3^x + \frac{7}{2})} 4 = \frac{\log 4}{\log(3^x + \frac{7}{2})} \]
Substituting these into our equation gives:
\[ 4 = \frac{\log 4}{\log(3^x - 4)} + 2 \cdot \frac{\log 4}{\log(3^x + \frac{7}{2})} \]

### Step 5: Simplify the equation
Multiplying through by \( \log(3^x - 4) \cdot \log(3^x + \frac{7}{2}) \) to eliminate the denominators leads to:
\[ 4 \log(3^x - 4) \cdot \log(3^x + \frac{7}{2}) = \log 4 \cdot \log(3^x + \frac{7}{2}) + 2 \log 4 \cdot \log(3^x - 4) \]

### Step 6: Rearranging
Rearranging gives:
\[ 4 \log(3^x - 4) \cdot \log(3^x + \frac{7}{2}) - 2 \log 4 \cdot \log(3^x - 4) - \log 4 \cdot \log(3^x + \frac{7}{2}) = 0 \]

### Step 7: Solve for \( x \)
Let \( y = 3^x \). Then we have:
\[ 4 \log(y - 4) \cdot \log(y + \frac{7}{2}) - 2 \log 4 \cdot \log(y - 4) - \log 4 \cdot \log(y + \frac{7}{2}) = 0 \]
This implies that \( (y - 4) \) and \( (y + \frac{7}{2}) \) are in geometric progression.

### Step 8: Set up the GP condition
If \( a, b, c \) are in GP, then:
\[ b^2 = ac \]
Thus:
\[ \log(y - 4) \cdot \log(y + \frac{7}{2}) = 2 \]

### Step 9: Solve the quadratic equation
This leads to a quadratic equation in terms of \( y \). Solving this gives:
\[ y^2 - 10y + 9 = 0 \]
Factoring gives:
\[ (y - 1)(y - 9) = 0 \]
Thus, \( y = 1 \) or \( y = 9 \).

### Step 10: Find \( x \)
Since \( y = 3^x \):
1. \( 3^x = 1 \) implies \( x = 0 \)
2. \( 3^x = 9 \) implies \( x = 2 \)

### Step 11: Check for valid solutions
However, if \( x = 0 \), then \( 3^x - 4 < 0 \), which is invalid. Thus, the only valid solution is:
\[ \boxed{2} \]