The solution set of `(x-2)^(x^(2)-6x+8)gt1` is

Options

A. `(2,oo)`
B. `(2,3)uu(4,oo)`
C. `(4,5)uu(5,oo)`
D. `(2,3)uu(4,5)`

Answer

To solve the inequality \((x-2)^{(x^2 - 6x + 8)} > 1\), we will analyze the expression in two cases based on the properties of exponents.

### Step 1: Identify the cases
The expression \((x-2)^{(x^2 - 6x + 8)} > 1\) can be true under two conditions:
1. \(x - 2 > 1\) and \(x^2 - 6x + 8 > 0\)
2. \(x - 2 < 1\) and \(x^2 - 6x + 8 < 0\)

### Step 2: Solve Case 1
**Case 1: \(x - 2 > 1\)**

This simplifies to:
\[ x > 3 \]

Next, we need to solve \(x^2 - 6x + 8 > 0\). We can factor the quadratic:
\[ x^2 - 6x + 8 = (x - 2)(x - 4) \]

To find where this expression is greater than zero, we determine the roots:
- The roots are \(x = 2\) and \(x = 4\).

Now, we analyze the sign of the quadratic:
- The quadratic is positive when \(x < 2\) or \(x > 4\).

Since we already have \(x > 3\) from the first condition, we combine this with the quadratic's positive region:
- The valid range from Case 1 is \(x > 4\).

### Step 3: Solve Case 2
**Case 2: \(x - 2 < 1\)**

This simplifies to:
\[ x < 3 \]

Next, we need to solve \(x^2 - 6x + 8 < 0\). As established, the quadratic factors to:
\[ (x - 2)(x - 4) < 0 \]

The quadratic is negative between its roots:
- The valid range is \(2 < x < 4\).

Since we have \(x < 3\) from the first condition, we combine this with the quadratic's negative region:
- The valid range from Case 2 is \(2 < x < 3\).

### Step 4: Combine the results
From Case 1, we found \(x > 4\), and from Case 2, we found \(2 < x < 3\). Therefore, the solution set for the inequality \((x-2)^{(x^2 - 6x + 8)} > 1\) is:
\[ x \in (2, 3) \cup (4, \infty) \]

### Final Solution
The solution set of the inequality \((x-2)^{(x^2 - 6x + 8)} > 1\) is:
\[ \boxed{(2, 3) \cup (4, \infty)} \]