Logging inequalities 2

Calm down! In general, $\log_b a > 0\ \ \Leftrightarrow\ \ \begin{cases} b > 1\ \text{and}\ a > 1 & \text{or} \\ 0 < b < 1\ \text{and}\ 0 < a < 1. & \end{cases}$

We must therefore determine the set $(A \cap C) \cup (B \cap D)$ , with the sets $A, B, C, D$ defined as the solution sets: $(A):\ \frac{x^2 - 12x + 30}{10} > 1 \\ (B):\ 0 < \frac{x^2 - 12x + 30}{10} < 1 \\ (C):\ \log_2 \frac{2x}{5} > 1\ \ \therefore\ \ \frac{2x}5 > 2 \\ (D):\ 0 < \log_2 \frac{2x}{5} < 1\ \ \therefore\ \ 1 < \frac{2x}5 < 2$

The solutions sets are $A = \langle \leftarrow, 2 \rangle \cup \langle 10, \rightarrow \rangle; \\ B = \langle 2, 6 - \sqrt 6 \rangle \cup \langle 6 + \sqrt 6, 10 \rangle; \\ C = \langle 5, \rightarrow \rangle; \\ D = \langle \tfrac52, 5 \rangle.$

The solution set of the inequality is therefore $(A \cap C) \cup (B \cap D) = \langle 10, \rightarrow \rangle \cup \langle \tfrac52, 6-\sqrt 6\rangle.$