Solve : `log_1- x (3-x)=log_(3-x)(1-x)`

Answer

To solve the equation \( \log_{1-x}(3-x) = \log_{3-x}(1-x) \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
Using the property of logarithms that states \( \log_a b = \frac{1}{\log_b a} \), we can rewrite the equation:
\[ \log_{1-x}(3-x) = \frac{1}{\log_{3-x}(1-x)} \]
This implies:
\[ \log_{1-x}(3-x) = \log_{3-x}(1-x) \]

### Step 2: Let \( t = \log_{1-x}(3-x) \)
This gives us:
\[ t = \frac{1}{t} \]
Multiplying both sides by \( t \) (assuming \( t \neq 0 \)):
\[ t^2 = 1 \]
Thus, we have two possible solutions:
\[ t = 1 \quad \text{or} \quad t = -1 \]

### Step 3: Solve for \( t = 1 \)
If \( t = 1 \):
\[ \log_{1-x}(3-x) = 1 \]
This implies:
\[ 3-x = 1-x \]
Solving this gives:
\[ 3 = 1 \quad \text{(which is not possible)} \]

### Step 4: Solve for \( t = -1 \)
If \( t = -1 \):
\[ \log_{1-x}(3-x) = -1 \]
This implies:
\[ 3-x = \frac{1}{1-x} \]
Multiplying both sides by \( 1-x \):
\[ (3-x)(1-x) = 1 \]
Expanding the left side:
\[ 3 - 3x - x + x^2 = 1 \]
This simplifies to:
\[ x^2 - 4x + 3 = 1 \]
Rearranging gives:
\[ x^2 - 4x + 2 = 0 \]

### Step 5: Use the quadratic formula
To find the roots of the quadratic equation \( x^2 - 4x + 2 = 0 \), we apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = 2 \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2} \]

### Step 6: Check the validity of the solutions
1. **For \( x = 2 + \sqrt{2} \)**:
- \( 1 - x = 1 - (2 + \sqrt{2}) = -1 - \sqrt{2} \) (negative base, invalid)

2. **For \( x = 2 - \sqrt{2} \)**:
- \( 1 - x = 1 - (2 - \sqrt{2}) = -1 + \sqrt{2} \) (positive base, valid)

### Conclusion
The only valid solution is:
\[ \boxed{2 - \sqrt{2}} \]