The roots of the equation `log_(2)(x^(2)-4x+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 in exponential form
Using the property of logarithms, we can rewrite the equation as:
\[ x^{2} - 4x + 5 = 2^{(x - 2)} \]

### Step 2: Simplify the left-hand side
The left-hand side simplifies to:
\[ x^{2} - 4x + 5 \]
This is a quadratic expression.

### Step 3: Check for specific values of \(x\)
We will check some integer values of \(x\) to see if they satisfy the equation.

#### Check \(x = 2\):
\[ LHS: 2^{2} - 4(2) + 5 = 4 - 8 + 5 = 1 \]
\[ RHS: 2^{(2 - 2)} = 2^{0} = 1 \]
Since \(LHS = RHS\), \(x = 2\) is a solution.

#### Check \(x = 3\):
\[ LHS: 3^{2} - 4(3) + 5 = 9 - 12 + 5 = 2 \]
\[ RHS: 2^{(3 - 2)} = 2^{1} = 2 \]
Since \(LHS = RHS\), \(x = 3\) is also a solution.

#### Check \(x = 5\):
\[ LHS: 5^{2} - 4(5) + 5 = 25 - 20 + 5 = 10 \]
\[ RHS: 2^{(5 - 2)} = 2^{3} = 8 \]
Since \(LHS \neq RHS\), \(x = 5\) is not a solution.

#### Check \(x = -3\):
\[ LHS: (-3)^{2} - 4(-3) + 5 = 9 + 12 + 5 = 26 \]
\[ RHS: 2^{(-3 - 2)} = 2^{-5} = \frac{1}{32} \]
Since \(LHS \neq RHS\), \(x = -3\) is not a solution.

### Step 4: Conclusion
The roots of the equation \( \log_{2}(x^{2} - 4x + 5) = x - 2 \) are:
\[ \boxed{2 \text{ and } 3} \]

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