Solve `log_(1-x)(x-2) ge-1`.

Explanation

To solve the inequality \( \log_{(1-x)}(x-2) \geq -1 \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \log_{(1-x)}(x-2) \geq -1 \]
This can be rewritten using the property of logarithms:
\[ \log_{(1-x)}(x-2) + 1 \geq 0 \]
This implies:
\[ \log_{(1-x)}(x-2) \geq -1 \]

### Step 2: Convert the Logarithmic Inequality
Using the definition of logarithms, we can convert the logarithmic inequality:
\[ \log_{(1-x)}(x-2) \geq -1 \implies x-2 \geq (1-x)^{-1} \]
This means:
\[ x-2 \geq \frac{1}{1-x} \]

### Step 3: Multiply Both Sides by \( (1-x) \)
To eliminate the fraction, we multiply both sides by \( (1-x) \). However, we must consider the sign of \( (1-x) \):
1. If \( 1-x > 0 \) (i.e., \( x < 1 \)), the inequality remains the same.
2. If \( 1-x < 0 \) (i.e., \( x > 1 \)), the inequality reverses.

Thus, we have two cases to consider.

### Case 1: \( x < 1 \)
\[ (x-2)(1-x) \geq 1 \]
Expanding this gives:
\[ -x^2 + 3x - 2 \geq 1 \implies -x^2 + 3x - 3 \geq 0 \]
Rearranging gives:
\[ x^2 - 3x + 3 \leq 0 \]
Now we find the roots of the quadratic equation \( x^2 - 3x + 3 = 0 \) using the discriminant:
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 3 = 9 - 12 = -3 \]
Since the discriminant is negative, there are no real roots, and the quadratic is always positive. Therefore, there are no solutions in this case.

### Case 2: \( x > 1 \)
In this case, we reverse the inequality:
\[ (x-2)(1-x) \leq 1 \]
Expanding gives:
\[ -x^2 + 3x - 2 \leq 1 \implies -x^2 + 3x - 3 \leq 0 \]
Rearranging gives:
\[ x^2 - 3x + 3 \geq 0 \]
Again, the discriminant is negative, indicating that this quadratic is always positive. Therefore, there are no solutions in this case either.

### Step 4: Check the Domain of the Logarithm
For the logarithm to be defined, we need:
1. \( x - 2 > 0 \) which gives \( x > 2 \)
2. \( 1 - x > 0 \) which gives \( x < 1 \)
3. \( 1 - x \neq 1 \) which gives \( x \neq 0 \)

### Conclusion
Combining these conditions, we find that:
- \( x > 2 \) contradicts \( x < 1 \).
Thus, there are no values of \( x \) that satisfy all conditions.

The solution set is empty:
\[ \text{Solution: } \emptyset \]