If `7^((log_7(x^(2)-4x+5))) = x - 1 ` then x may have values

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 property of logarithms
Since the base of the logarithm and the base of the exponent are the same (both are 7), we can simplify the left side:
\[ \log_7(x^2 - 4x + 5) = x - 1 \]

### Step 2: Remove the logarithm
By the definition of logarithms, if \( \log_b(a) = c \), then \( a = b^c \). Therefore, we can rewrite the equation as:
\[ x^2 - 4x + 5 = 7^{(x - 1)} \]

### Step 3: Rearrange the equation
Now, we need to rearrange the equation to set it to zero:
\[ x^2 - 4x + 5 - 7^{(x - 1)} = 0 \]

### Step 4: Analyze the equation
This equation is a transcendental equation, which may not have a straightforward algebraic solution. However, we can analyze it by substituting values for \( x \) to find possible solutions.

### Step 5: Test integer values
Let's test some integer values for \( x \):

1. **For \( x = 2 \)**:
\[ 2^2 - 4(2) + 5 = 4 - 8 + 5 = 1 \]
\[ 7^{(2 - 1)} = 7^1 = 7 \]
Not equal.

2. **For \( x = 3 \)**:
\[ 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2 \]
\[ 7^{(3 - 1)} = 7^2 = 49 \]
Not equal.

3. **For \( x = 4 \)**:
\[ 4^2 - 4(4) + 5 = 16 - 16 + 5 = 5 \]
\[ 7^{(4 - 1)} = 7^3 = 343 \]
Not equal.

4. **For \( x = 1 \)**:
\[ 1^2 - 4(1) + 5 = 1 - 4 + 5 = 2 \]
\[ 7^{(1 - 1)} = 7^0 = 1 \]
Not equal.

### Step 6: Use the quadratic formula
The left side is a quadratic function, and we can find its roots:
\[ x^2 - 5x + 6 = 0 \]
Factoring gives:
\[ (x - 2)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Final Step: Verify solutions
1. **For \( x = 2 \)**:
\[ 7^{(\log_7(2^2 - 4(2) + 5))} = 7^{(\log_7(1))} = 1 \]
\[ 2 - 1 = 1 \quad \text{(Valid)} \]

2. **For \( x = 3 \)**:
\[ 7^{(\log_7(3^2 - 4(3) + 5))} = 7^{(\log_7(2))} = 2 \]
\[ 3 - 1 = 2 \quad \text{(Valid)} \]

Thus, the values of \( x \) that satisfy the equation are:
\[ \boxed{2 \text{ and } 3} \]