If `7^(log 7(x^(2)-4x + 5))=x - 1`, x may have values

Options

A. 2, 3
B. 7
C. `-2, -3`
D. 2, -3

Answer

To solve the equation \( 7^{\log_7(x^2 - 4x + 5)} = x - 1 \), we will follow these steps:

### Step 1: Identify the condition for the logarithm
Since we have a logarithm in the equation, we need to ensure that the argument of the logarithm is greater than zero:
\[ x^2 - 4x + 5 > 0 \]

### Step 2: Analyze the quadratic expression
We can analyze the quadratic \( x^2 - 4x + 5 \) to determine if it is always positive. The discriminant \( D \) of the quadratic is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 \]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( x^2 - 4x + 5 > 0 \) for all real \( x \).

### Step 3: Apply the property of logarithms
Using the property of logarithms, we can simplify the equation:
\[ 7^{\log_7(x^2 - 4x + 5)} = x^2 - 4x + 5 \]
This leads us to:
\[ x^2 - 4x + 5 = x - 1 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ x^2 - 4x + 5 - x + 1 = 0 \]
This simplifies to:
\[ x^2 - 5x + 6 = 0 \]

### Step 5: Factor the quadratic
Next, we factor the quadratic:
\[ (x - 2)(x - 3) = 0 \]
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 6: Verify the solutions
We need to verify that both solutions satisfy the original condition \( x^2 - 4x + 5 > 0 \).

- For \( x = 2 \):
\[ 2^2 - 4 \cdot 2 + 5 = 4 - 8 + 5 = 1 > 0 \]

- For \( x = 3 \):
\[ 3^2 - 4 \cdot 3 + 5 = 9 - 12 + 5 = 2 > 0 \]

Both values satisfy the condition.

### Final Answer
Thus, the values of \( x \) that satisfy the equation are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

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