Find the smallest integral value of `x` satisfying `(x-2)^(x^2-6x+8))>1`

Explanation

To solve the inequality \((x-2)^{(x^2 - 6x + 8)} > 1\), we can analyze the problem by considering different cases based on the value of \(x-2\).

### Step 1: Identify the cases based on \(x-2\)

1. **Case 1:** \(x - 2 > 1\) (i.e., \(x > 3\))
2. **Case 2:** \(x - 2 < 1\) (i.e., \(x < 3\))

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