The value of x for which `log_(3) (2^(1-x)+3), log_(9) 4 and log_(27) (2^(x)-1)^(3)` form an A.P. is

Options

A. `11//6`
B. `6//11`
C. `log_(2) (11//6)`
D. 1

Answer

To find the value of \( x \) for which the expressions \( \log_3(2^{1-x} + 3) \), \( \log_9(4) \), and \( \log_{27}((2^x - 1)^3) \) form an arithmetic progression (A.P.), we can follow these steps:

### Step 1: Set up the A.P. condition
For three numbers \( a \), \( b \), and \( c \) to be in A.P., the condition is:
\[ 2b = a + c \]
Here, we can assign:
- \( a = \log_3(2^{1-x} + 3) \)
- \( b = \log_9(4) \)
- \( c = \log_{27}((2^x - 1)^3) \)

### Step 2: Express all logarithms in terms of base 3
We need to convert all logarithms to the same base. We know:
\[ \log_9(4) = \frac{\log_3(4)}{\log_3(9)} = \frac{\log_3(4)}{2} \quad \text{(since } \log_3(9) = 2\text{)} \]
\[ \log_{27}((2^x - 1)^3) = \frac{\log_3((2^x - 1)^3)}{\log_3(27)} = \frac{3\log_3(2^x - 1)}{3} = \log_3(2^x - 1) \]

### Step 3: Substitute into the A.P. condition
Now substituting these into the A.P. condition:
\[ 2 \cdot \frac{\log_3(4)}{2} = \log_3(2^{1-x} + 3) + \log_3(2^x - 1) \]
This simplifies to:
\[ \log_3(4) = \log_3(2^{1-x} + 3) + \log_3(2^x - 1) \]

### Step 4: Use the property of logarithms
Using the property of logarithms that states \( \log_a(m) + \log_a(n) = \log_a(m \cdot n) \), we rewrite the equation:
\[ \log_3(4) = \log_3((2^{1-x} + 3)(2^x - 1)) \]

### Step 5: Remove the logarithm
Since the logarithms are equal, we can equate the arguments:
\[ 4 = (2^{1-x} + 3)(2^x - 1) \]

### Step 6: Expand and simplify
Expanding the right-hand side:
\[ 4 = 2^{1-x} \cdot 2^x - 2^{1-x} + 3 \cdot 2^x - 3 \]
This simplifies to:
\[ 4 = 2^1 - 2^{1-x} + 3 \cdot 2^x - 3 \]
\[ 4 = 2 - 2^{1-x} + 3 \cdot 2^x - 3 \]
\[ 4 = -2^{1-x} + 3 \cdot 2^x - 1 \]
\[ 5 = 3 \cdot 2^x - 2^{1-x} \]

### Step 7: Substitute \( 2^{1-x} \)
Let \( y = 2^x \). Then \( 2^{1-x} = \frac{2}{y} \). Substitute this into the equation:
\[ 5 = 3y - \frac{2}{y} \]
Multiplying through by \( y \) to eliminate the fraction:
\[ 5y = 3y^2 - 2 \]
Rearranging gives:
\[ 3y^2 - 5y - 2 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
\[ y = \frac{5 \pm \sqrt{25 + 24}}{6} \]
\[ y = \frac{5 \pm \sqrt{49}}{6} \]
\[ y = \frac{5 \pm 7}{6} \]
Calculating the two possible values:
1. \( y = \frac{12}{6} = 2 \)
2. \( y = \frac{-2}{6} = -\frac{1}{3} \) (not valid since \( y = 2^x \) must be positive)

Thus, \( y = 2 \) implies:
\[ 2^x = 2 \implies x = 1 \]

### Final Answer
The value of \( x \) for which the given logarithmic expressions form an A.P. is:
\[ \boxed{1} \]