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

Explanation

To solve the problem, we need to determine the value of \( x \) such that the logarithmic expressions \( \log_{2}(5 \cdot 2^{x} + 1) \), \( \log_{4}(2^{1-x} + 1) \), and \( 1 \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
For three terms \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
Here, let:
\[ a = \log_{2}(5 \cdot 2^{x} + 1), \quad b = \log_{4}(2^{1-x} + 1), \quad c = 1 \]

2. **Express \( b \) in terms of base 2**:
We can convert \( b \) from base 4 to base 2:
\[ \log_{4}(2^{1-x} + 1) = \frac{\log_{2}(2^{1-x} + 1)}{\log_{2}(4)} = \frac{\log_{2}(2^{1-x} + 1)}{2} \]

3. **Substituting into the A.P. condition**:
Substitute \( a \), \( b \), and \( c \) into the A.P. condition:
\[ 2 \left( \frac{\log_{2}(2^{1-x} + 1)}{2} \right) = \log_{2}(5 \cdot 2^{x} + 1) + 1 \]
Simplifying gives:
\[ \log_{2}(2^{1-x} + 1) = \log_{2}(5 \cdot 2^{x} + 1) + 1 \]

4. **Using properties of logarithms**:
Recall that \( \log_{2}(a) + 1 = \log_{2}(2a) \). Thus, we can rewrite the equation:
\[ \log_{2}(2^{1-x} + 1) = \log_{2}(2(5 \cdot 2^{x} + 1)) \]

5. **Equating the arguments**:
Since the logarithms are equal, we can set the arguments equal to each other:
\[ 2^{1-x} + 1 = 2(5 \cdot 2^{x} + 1) \]

6. **Expanding and simplifying**:
Expanding the right-hand side:
\[ 2^{1-x} + 1 = 10 \cdot 2^{x} + 2 \]
Rearranging gives:
\[ 2^{1-x} - 10 \cdot 2^{x} - 1 = 0 \]

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

8. **Applying the quadratic formula**:
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 10, b = 1, c = -2 \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 10 \cdot (-2)}}{2 \cdot 10} = \frac{-1 \pm \sqrt{1 + 80}}{20} = \frac{-1 \pm \sqrt{81}}{20} = \frac{-1 \pm 9}{20} \]
This gives:
\[ y = \frac{8}{20} = \frac{2}{5} \quad \text{or} \quad y = \frac{-10}{20} = -\frac{1}{2} \quad (\text{not valid since } y > 0) \]

9. **Finding \( x \)**:
Since \( y = 2^{x} \), we have:
\[ 2^{x} = \frac{2}{5} \quad \Rightarrow \quad x = \log_{2}\left(\frac{2}{5}\right) = 1 - \log_{2}(5) \]

### Final Answer:
Thus, the value of \( x \) is:
\[ x = 1 - \log_{2}(5) \]