Solve : `log_(5-x)(x^(2)-2x+65)=2`

Answer

To solve the equation \( \log_{(5-x)}(x^2 - 2x + 65) = 2 \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation in exponential form
Using the property of logarithms, we can rewrite the equation:
\[ (5 - x)^2 = x^2 - 2x + 65 \]

### Step 2: Expand the left side
Now, we will expand \( (5 - x)^2 \):
\[ (5 - x)^2 = 25 - 10x + x^2 \]
So, we can rewrite the equation as:
\[ 25 - 10x + x^2 = x^2 - 2x + 65 \]

### Step 3: Simplify the equation
Now, we will simplify the equation by subtracting \( x^2 \) from both sides:
\[ 25 - 10x = -2x + 65 \]

### Step 4: Rearrange the equation
Next, we will rearrange the equation to isolate the terms involving \( x \):
\[ -10x + 2x = 65 - 25 \]
This simplifies to:
\[ -8x = 40 \]

### Step 5: Solve for \( x \)
Now, we will solve for \( x \):
\[ x = \frac{40}{-8} = -5 \]

### Step 6: Check the validity of the solution
We need to check if \( x = -5 \) is valid in the context of the logarithm. The base of the logarithm \( 5 - x \) must be positive and not equal to 1:
\[ 5 - (-5) = 10 \quad (\text{valid, since } 10 > 0) \]
Now, check if \( x^2 - 2x + 65 \) is positive:
\[ (-5)^2 - 2(-5) + 65 = 25 + 10 + 65 = 100 \quad (\text{valid, since } 100 > 0) \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{-5} \]