If `1, log_3sqrt(3^(1-x)+2), log_3 (4*3^x-1)` are in AP then x equals

Options

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

Answer

To solve the problem, we need to find the value of \( x \) such that the terms \( 1 \), \( \log_3 \sqrt{3^{1-x} + 2} \), and \( \log_3 (4 \cdot 3^x - 1) \) are in arithmetic progression (AP).

### Step 1: Set up the AP condition
For three numbers \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]
In our case:
\[ 2 \log_3 \sqrt{3^{1-x} + 2} = 1 + \log_3 (4 \cdot 3^x - 1) \]

### Step 2: Simplify the logarithmic expression
Using the property of logarithms, we can rewrite \( \log_3 \sqrt{3^{1-x} + 2} \):
\[ \log_3 \sqrt{3^{1-x} + 2} = \frac{1}{2} \log_3 (3^{1-x} + 2) \]
Substituting this into the AP condition gives:
\[ 2 \cdot \frac{1}{2} \log_3 (3^{1-x} + 2) = 1 + \log_3 (4 \cdot 3^x - 1) \]
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 \( \log_a b + \log_a c = \log_a (bc) \), we can rewrite the right-hand side:
\[ \log_3 (3^{1-x} + 2) = \log_3 (3) + \log_3 (4 \cdot 3^x - 1) \]
This simplifies to:
\[ \log_3 (3^{1-x} + 2) = \log_3 (3(4 \cdot 3^x - 1)) \]

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

### Step 5: Expand and simplify
Expanding the right-hand side:
\[ 3^{1-x} + 2 = 12 \cdot 3^x - 3 \]
Rearranging gives:
\[ 3^{1-x} + 5 = 12 \cdot 3^x \]

### Step 6: Substitute \( t = 3^x \)
Let \( t = 3^x \), then \( 3^{1-x} = \frac{3}{t} \):
\[ \frac{3}{t} + 5 = 12t \]
Multiplying through by \( t \) to eliminate the fraction:
\[ 3 + 5t = 12t^2 \]
Rearranging gives:
\[ 12t^2 - 5t - 3 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 12 \cdot (-3)}}{2 \cdot 12} \]
Calculating the discriminant:
\[ t = \frac{5 \pm \sqrt{25 + 144}}{24} = \frac{5 \pm \sqrt{169}}{24} = \frac{5 \pm 13}{24} \]
This gives two possible values for \( t \):
\[ t = \frac{18}{24} = \frac{3}{4} \quad \text{and} \quad t = \frac{-8}{24} = -\frac{1}{3} \]

### Step 8: Determine valid solutions
Since \( t = 3^x \) must be positive, we discard \( t = -\frac{1}{3} \). Thus, we have:
\[ 3^x = \frac{3}{4} \]

### Step 9: Take logarithm to solve for \( x \)
Taking logarithm base 3:
\[ x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4 \]

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