If `log_(2) (5.2^(x) + 1), log_(4) (2^(1-x) + 1)` and 1 are in A.P., then x equals

Options

A. `log_(2)5`
B. `1 - log_(2) 5`
C. `log_(5) 2`
D. none of these

Answer

To solve the problem, we need to determine the value of \( x \) such that the logarithmic expressions \( \log_2(5.2^x + 1) \), \( \log_4(2^{1-x} + 1) \), and \( 1 \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression**:
If three terms \( a, b, c \) are in A.P., then the condition is:
\[ 2b = a + c \]
Here, let:
- \( a = \log_2(5.2^x + 1) \)
- \( b = \log_4(2^{1-x} + 1) \)
- \( c = 1 \)

2. **Setting Up the Equation**:
From the A.P. condition, we have:
\[ 2 \log_4(2^{1-x} + 1) = \log_2(5.2^x + 1) + 1 \]

3. **Converting Logarithm Base**:
We can convert \( \log_4 \) to \( \log_2 \) using the change of base formula:
\[ \log_4(y) = \frac{\log_2(y)}{\log_2(4)} = \frac{\log_2(y)}{2} \]
Therefore, we can rewrite \( b \):
\[ b = \log_4(2^{1-x} + 1) = \frac{1}{2} \log_2(2^{1-x} + 1) \]

4. **Substituting Back**:
Substitute \( b \) back into the A.P. equation:
\[ 2 \cdot \frac{1}{2} \log_2(2^{1-x} + 1) = \log_2(5.2^x + 1) + 1 \]
Simplifying gives:
\[ \log_2(2^{1-x} + 1) = \log_2(5.2^x + 1) + 1 \]

5. **Using Properties of Logarithms**:
The equation can be rewritten as:
\[ \log_2(2^{1-x} + 1) = \log_2(5.2^x + 1) + \log_2(2) \]
This implies:
\[ 2^{1-x} + 1 = 2(5.2^x + 1) \]

6. **Expanding and Rearranging**:
Expanding the right side:
\[ 2^{1-x} + 1 = 10.2^x + 2 \]
Rearranging gives:
\[ 2^{1-x} - 10.2^x - 1 = 0 \]

7. **Substituting \( t = 2^x \)**:
Let \( t = 2^x \). Then \( 2^{1-x} = \frac{2}{t} \):
\[ \frac{2}{t} - 10t - 1 = 0 \]
Multiplying through by \( t \) to eliminate the fraction:
\[ 2 - 10t^2 - t = 0 \]
Rearranging gives:
\[ 10t^2 + t - 2 = 0 \]

8. **Solving the Quadratic Equation**:
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 10, b = 1, c = -2 \):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 10 \cdot (-2)}}{2 \cdot 10} \]
\[ t = \frac{-1 \pm \sqrt{1 + 80}}{20} = \frac{-1 \pm 9}{20} \]
This gives two potential solutions:
\[ t = \frac{8}{20} = \frac{2}{5} \quad \text{(valid, since } t > 0\text{)} \]
\[ t = \frac{-10}{20} = -\frac{1}{2} \quad \text{(not valid)} \]

9. **Finding \( x \)**:
Since \( t = 2^x \), we have:
\[ 2^x = \frac{2}{5} \]
Taking logarithm base 2:
\[ x = \log_2\left(\frac{2}{5}\right) = \log_2(2) - \log_2(5) = 1 - \log_2(5) \]

### Final Answer:
Thus, the value of \( x \) is:
\[ x = 1 - \log_2(5) \]