The set of all values of x satisfying `x^(logx(1-x)^(2))=9`

Options

A. a subset of R containing N
B. a subset of R containing Z ( set of all integers)
C. is a finite set containing at least two elements
D. a finite set

Answer

To solve the equation \( x^{\log_x(1-x)^2} = 9 \), we will follow these steps:

### Step 1: Rewrite the equation using logarithmic properties
We can rewrite the equation as:
\[ \log_x(1-x)^2 = \log_x(9) \]
This is because \( x^{\log_x(1-x)^2} = 9 \) implies that the exponent must equal the logarithm of 9 in base \( x \).

### Step 2: Apply the logarithmic identity
Using the property of logarithms \( \log_x(a^b) = b \cdot \log_x(a) \), we can simplify:
\[ \log_x(1-x)^2 = 2 \cdot \log_x(1-x) \]
Thus, the equation becomes:
\[ 2 \cdot \log_x(1-x) = \log_x(9) \]

### Step 3: Divide both sides by 2
Dividing both sides by 2 gives:
\[ \log_x(1-x) = \frac{1}{2} \log_x(9) \]

### Step 4: Rewrite the right-hand side
We can express \( \frac{1}{2} \log_x(9) \) as:
\[ \log_x(9^{1/2}) = \log_x(3) \]
So we have:
\[ \log_x(1-x) = \log_x(3) \]

### Step 5: Set the arguments equal to each other
Since the logarithm is a one-to-one function, we can set the arguments equal:
\[ 1-x = 3 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = 1 - 3 = -2 \]

### Step 7: Check the validity of \( x \)
Since \( x \) must be positive for the logarithm to be defined, \( x = -2 \) is not a valid solution.

### Step 8: Consider the case when \( 1-x \) is negative
We also consider the case where:
\[ 1-x = -3 \]
This gives:
\[ x = 1 + 3 = 4 \]

### Step 9: Check if \( x = 4 \) is valid
We need to ensure that \( x = 4 \) is valid:
- For \( x = 4 \), \( 1-x = 1-4 = -3 \), which is negative. The logarithm \( \log_x(1-x) \) is not defined for negative arguments.

### Conclusion
Since both potential solutions lead to invalid logarithmic arguments, we conclude that there are no valid solutions for the equation \( x^{\log_x(1-x)^2} = 9 \).