To resolve the inequality \(\log_2(x^2 - 5x + 6) > 1\), the subsequent steps are required:
1. Determine the domain of the logarithm: The argument of the logarithm, \(x^2 - 5x + 6\), must be strictly positive. Therefore, we solve \(x^2 - 5x + 6 > 0\). Factoring the quadratic yields \((x-2)(x-3) > 0\), which is true for \(x \in (-\infty, 2) \cup (3, \infty)\).
2. Solve the logarithmic inequality: Given \(\log_2(x^2 - 5x + 6) > 1\), we convert this to exponential form: \(x^2 - 5x + 6 > 2^1\), simplifying to \(x^2 - 5x + 6 > 2\). Rearranging, we get \(x^2 - 5x + 4 > 0\). Factoring this quadratic gives \((x-1)(x-4) > 0\), which is satisfied when \(x \in (-\infty, 1) \cup (4, \infty)\).
3. Intersect the solution sets: The values of \(x\) must satisfy both domain restrictions and the inequality derived in step 2. The intersection of \(x \in (-\infty, 2) \cup (3, \infty)\) and \(x \in (-\infty, 1) \cup (4, \infty)\) is \(x \in (-\infty, 1) \cup (4, \infty)\).