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_5 2`
C. `log_(5)2`
D. `1-log_(2)5`

Explanation

To solve the problem, we need to find 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 Arithmetic Progression**:
For three terms \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
Here, let:
- \( a = \log_2(5 \cdot 2^x + 1) \)
- \( b = \log_4(2^{1-x} + 1) \)
- \( c = 1 \)

Thus, we have:
\[ 2 \log_4(2^{1-x} + 1) = \log_2(5 \cdot 2^x + 1) + 1 \]

2. **Converting \( \log_4 \) to \( \log_2 \)**:
We know that:
\[ \log_4(y) = \frac{\log_2(y)}{\log_2(4)} = \frac{\log_2(y)}{2} \]
Therefore, we can rewrite \( b \):
\[ 2 \log_4(2^{1-x} + 1) = \log_2(2^{1-x} + 1) \]

3. **Substituting back into the A.P. condition**:
Now substituting back into our A.P. condition:
\[ \log_2(2^{1-x} + 1) = \log_2(5 \cdot 2^x + 1) + 1 \]

4. **Using properties of logarithms**:
We can express \( 1 \) in terms of logarithm:
\[ 1 = \log_2(2) \]
Thus, we have:
\[ \log_2(2^{1-x} + 1) = \log_2(5 \cdot 2^x + 2) \]

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

6. **Rearranging the equation**:
Rearranging gives:
\[ 2^{1-x} - 5 \cdot 2^x + 1 - 2 = 0 \]
Simplifying:
\[ 2^{1-x} - 5 \cdot 2^x - 1 = 0 \]

7. **Substituting \( y = 2^x \)**:
Let \( y = 2^x \), then \( 2^{1-x} = \frac{2}{y} \):
\[ \frac{2}{y} - 5y - 1 = 0 \]
Multiplying through by \( y \) (assuming \( y \neq 0 \)):
\[ 2 - 5y^2 - y = 0 \]
Rearranging gives:
\[ 5y^2 + y - 2 = 0 \]

8. **Using the quadratic formula**:
We can solve this quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5, b = 1, c = -2 \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} \]
\[ = \frac{-1 \pm \sqrt{1 + 40}}{10} = \frac{-1 \pm \sqrt{41}}{10} \]

9. **Finding valid solutions**:
We have two potential solutions for \( y \):
\[ y = \frac{-1 + \sqrt{41}}{10} \quad \text{(valid since it's positive)} \]
\[ y = \frac{-1 - \sqrt{41}}{10} \quad \text{(not valid since it's negative)} \]

10. **Finding \( x \)**:
Since \( y = 2^x \):
\[ 2^x = \frac{-1 + \sqrt{41}}{10} \]
Taking logarithm base 2:
\[ x = \log_2\left(\frac{-1 + \sqrt{41}}{10}\right) \]

### Final Answer:
Thus, the value of \( x \) is:
\[ x = \log_2\left(\frac{-1 + \sqrt{41}}{10}\right) \]