If `7^(log_(7)(x^(2)-4x+5))=(x-1)`, then `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 can follow these steps:

### Step 1: Simplify the left side using the properties of logarithms
Using the property \( a^{\log_{a}(b)} = b \), we can simplify the left side:
\[ 7^{\log_{7}(x^{2}-4x+5)} = x^{2} - 4x + 5 \]
Thus, we have:
\[ x^{2} - 4x + 5 = x - 1 \]

### Step 2: Rearrange the equation
Now, we can rearrange the equation to bring all terms to one side:
\[ x^{2} - 4x + 5 - x + 1 = 0 \]
This simplifies to:
\[ x^{2} - 5x + 6 = 0 \]

### Step 3: Factor the quadratic equation
Next, we can 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 values of \( x \):
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 5: Check the validity of the solutions
We need to ensure that the solutions satisfy the original logarithmic condition. The argument of the logarithm \( x^{2} - 4x + 5 \) must be greater than 0:
1. For \( x = 2 \):
\[ 2^{2} - 4(2) + 5 = 4 - 8 + 5 = 1 \quad (\text{valid, since } 1 > 0) \]
2. For \( x = 3 \):
\[ 3^{2} - 4(3) + 5 = 9 - 12 + 5 = 2 \quad (\text{valid, since } 2 > 0) \]

Both values of \( x \) are valid.

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

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