1. The problem is to find a strong logarithm question suitable for a 4th semester (S4) level. 2. A strong logarithm question often involves properties of logarithms, solving equations, or applying logarithms in real-world contexts. 3. Here is a challenging problem: Solve for $x$ in the equation $$\log_2(x^2 - 5x + 6) = 3.$$ 4. Recall the definition: $\log_b(a) = c$ means $b^c = a$. 5. Using this, rewrite the equation as $$x^2 - 5x + 6 = 2^3.$$ 6. Simplify the right side: $$2^3 = 8.$$ 7. So, $$x^2 - 5x + 6 = 8.$$ 8. Rearrange to form a quadratic equation: $$x^2 - 5x + 6 - 8 = 0 \Rightarrow x^2 - 5x - 2 = 0.$$ 9. Use the quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 \pm \sqrt{33}}{2}.$$ 10. Check the domain: The argument of the logarithm must be positive, so $$x^2 - 5x + 6 > 0.$$ 11. Factor the quadratic inside the log: $$(x-2)(x-3) > 0.$$ 12. This inequality holds when $x < 2$ or $x > 3$. 13. Evaluate the solutions: $$x_1 = \frac{5 + \sqrt{33}}{2} \approx 5.37 > 3,$$ valid. 14. $$x_2 = \frac{5 - \sqrt{33}}{2} \approx -0.37 < 2,$$ valid. 15. Both solutions satisfy the domain. 16. Final answer: $$x = \frac{5 \pm \sqrt{33}}{2}.$$