If `1,log_9(3^(1-x)+2), log_3(4*3^x-1)` are in A.P then x equals to

Options

A. `1 log_(3) 4`
B. `1 - log_(3) 4 `
C. `log_(3) 0.25`
D. `log_(4) 3 `

Answer

To solve the problem, we need to find the value of \( x \) such that the terms \( 1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1) \) are in Arithmetic Progression (A.P).

### Step 1: Set up the A.P condition
For three terms \( a, b, c \) to be in A.P, the condition is:
\[ 2b = a + c \]
Here, let:
- \( a = 1 \)
- \( b = \log_9(3^{1-x} + 2) \)
- \( c = \log_3(4 \cdot 3^x - 1) \)

Thus, we have:
\[ 2 \log_9(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1) \]

### Step 2: Convert \( \log_9 \) to \( \log_3 \)
Using the change of base formula, we can convert \( \log_9 \) to \( \log_3 \):
\[ \log_9(3^{1-x} + 2) = \frac{\log_3(3^{1-x} + 2)}{\log_3(9)} = \frac{\log_3(3^{1-x} + 2)}{2} \]
Substituting this back into the A.P condition gives:
\[ 2 \cdot \frac{\log_3(3^{1-x} + 2)}{2} = 1 + \log_3(4 \cdot 3^x - 1) \]
This simplifies to:
\[ \log_3(3^{1-x} + 2) = 1 + \log_3(4 \cdot 3^x - 1) \]

### Step 3: Use properties of logarithms
Using the property \( \log_a b + \log_a c = \log_a(b \cdot c) \), we can rewrite the equation:
\[ \log_3(3^{1-x} + 2) = \log_3(3) + \log_3(4 \cdot 3^x - 1) \]
This leads to:
\[ \log_3(3^{1-x} + 2) = \log_3(3(4 \cdot 3^x - 1)) \]

### Step 4: Remove the logarithm
Since the logarithm is one-to-one, we can equate the arguments:
\[ 3^{1-x} + 2 = 3(4 \cdot 3^x - 1) \]

### Step 5: Simplify the equation
Expanding the right side gives:
\[ 3^{1-x} + 2 = 12 \cdot 3^x - 3 \]
Rearranging this leads to:
\[ 3^{1-x} - 12 \cdot 3^x + 5 = 0 \]

### Step 6: Substitute \( y = 3^x \)
Let \( y = 3^x \). Then \( 3^{1-x} = \frac{3}{y} \). Substituting this into the equation gives:
\[ \frac{3}{y} - 12y + 5 = 0 \]
Multiplying through by \( y \) to eliminate the fraction results in:
\[ 3 - 12y^2 + 5y = 0 \]
Rearranging gives:
\[ 12y^2 - 5y - 3 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 12 \cdot (-3)}}{2 \cdot 12} \]
Calculating the discriminant:
\[ 25 + 144 = 169 \]
Thus:
\[ y = \frac{5 \pm 13}{24} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{18}{24} = \frac{3}{4} \)
2. \( y = \frac{-8}{24} \) (not valid since \( y \) must be positive)

### Step 8: Find \( x \)
Since \( y = 3^x \), we have:
\[ 3^x = \frac{3}{4} \]
Taking logarithm base 3:
\[ x = \log_3\left(\frac{3}{4}\right) = 1 - \log_3(4) \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{1 - \log_3(4)} \]