If `x^("log"_(x)(x^(2)-4x +5)) = (x-1)`, then x =

Options

A. 1
B. 2
C. 4
D. 5

Answer

To solve the equation \( x^{\log_{x}(x^2 - 4x + 5)} = x - 1 \), we can follow these steps:

### Step 1: Rewrite the equation using the property of logarithms
We know that \( a^{\log_b(c)} = c^{\log_b(a)} \). Therefore, we can rewrite the left-hand side of the equation:
\[ x^{\log_{x}(x^2 - 4x + 5)} = x^2 - 4x + 5 \]
So, we can rewrite the equation as:
\[ x^2 - 4x + 5 = x - 1 \]

### Step 2: Rearrange the equation
Now, we will move all terms to one side of the equation:
\[ x^2 - 4x + 5 - x + 1 = 0 \]
This simplifies to:
\[ x^2 - 5x + 6 = 0 \]

### Step 3: Factor the quadratic equation
Next, we will factor the quadratic equation \( x^2 - 5x + 6 \):
\[ (x - 2)(x - 3) = 0 \]

### Step 4: Solve for x
Setting each factor to zero gives us the possible solutions:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 5: Check the solutions
We need to check if both solutions satisfy the original equation.

**For \( x = 2 \)**:
\[ x^2 - 4x + 5 = 2^2 - 4 \cdot 2 + 5 = 4 - 8 + 5 = 1 \]
\[ x - 1 = 2 - 1 = 1 \]
Both sides are equal, so \( x = 2 \) is a valid solution.

**For \( x = 3 \)**:
\[ x^2 - 4x + 5 = 3^2 - 4 \cdot 3 + 5 = 9 - 12 + 5 = 2 \]
\[ x - 1 = 3 - 1 = 2 \]
Both sides are equal, so \( x = 3 \) is also a valid solution.

### Final Answer
The solutions to the equation are \( x = 2 \) and \( x = 3 \).