The roots of the equation `log_2 (x^2 - 4 x + 5) = (x - 2)` are

Options

A. 4, 5
B. 2, -3
C. 2, 3
D. 3, 5

Answer

To solve the equation \( \log_2 (x^2 - 4x + 5) = x - 2 \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
Using the property of logarithms, we can rewrite the equation:
\[ \log_b(a) = c \implies a = b^c \]
Applying this to our equation:
\[ x^2 - 4x + 5 = 2^{(x - 2)} \]

### Step 2: Simplify the left-hand side
The left-hand side can be expressed as:
\[ x^2 - 4x + 5 = (x - 2)^2 + 1 \]
This is because:
\[ (x - 2)^2 = x^2 - 4x + 4 \quad \text{and adding 1 gives us} \quad x^2 - 4x + 5. \]
Thus, we rewrite the equation as:
\[ (x - 2)^2 + 1 = 2^{(x - 2)} \]

### Step 3: Substitute for simplicity
Let \( X = x - 2 \). Then the equation becomes:
\[ X^2 + 1 = 2^X \]

### Step 4: Analyze the equation
Now we need to find the values of \( X \) that satisfy:
\[ X^2 + 1 = 2^X \]
We can plot the graphs of \( y = X^2 + 1 \) and \( y = 2^X \) to find the intersection points.

### Step 5: Find intersection points
By examining the graphs or testing values:
- For \( X = 0 \):
\[ 0^2 + 1 = 1 \quad \text{and} \quad 2^0 = 1 \quad \text{(intersection)} \]
- For \( X = 1 \):
\[ 1^2 + 1 = 2 \quad \text{and} \quad 2^1 = 2 \quad \text{(intersection)} \]

### Step 6: Solve for \( x \)
Now we convert back to \( x \):
1. For \( X = 0 \):
\[ x - 2 = 0 \implies x = 2 \]
2. For \( X = 1 \):
\[ x - 2 = 1 \implies x = 3 \]

### Final Answer
The roots of the equation are:
\[ \boxed{2 \text{ and } 3} \]

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