1. Solve the equation: $$3^{2x} - 4 \cdot 3^x - 3 = 0$$ Step 1: Let $$y = 3^x$$. Then, the equation becomes $$y^2 - 4y - 3 = 0$$. Step 2: Solve the quadratic equation $$y^2 - 4y - 3 = 0$$ using the quadratic formula: $$y = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$$. Step 3: Since $$y = 3^x$$ and $$3^x > 0$$ for all real $$x$$, we discard the negative root: $$y = 2 + \sqrt{7}$$. Step 4: Solve for $$x$$: $$3^x = 2 + \sqrt{7} \implies x = \log_3(2 + \sqrt{7})$$. 2. Given sets: $$I = \{1,2,3,...,20\}$$ $$A = \{ x : 5 < x < 18 \} = \{6,7,8,9,10,11,12,13,14,15,16,17\}$$ $$B = \{\text{multiples of 3 between 2 and 20}\} = \{3,6,9,12,15,18\}$$ $$C = \{\text{prime numbers greater than 2 but less than 20}\} = \{3,5,7,11,13,17,19\}$$ Note: A Venn diagram would have three circles representing A, B, and C with overlaps indicating common elements. Step 1: Identify the intersections: - $$A \cap B = \{6,9,12,15\}$$ - $$A \cap C = \{7,11,13,17\}$$ (only primes within A) - $$B \cap C = \{3\}$$ (only 3 is both prime and multiple of 3) Step 2: Find the triple intersection: $$A \cap B \cap C = \{ x : x \in A \text{ and } x \in B \text{ and } x \in C \}$$ Check common elements of all three: - From above sets, the only candidate is 3 for $$B \cap C$$ but 3 is not in A. - No element is simultaneously in A, B, and C. Therefore, $$A \cap B \cap C = \emptyset$$. Final answers: (a) $$x = \log_3(2 + \sqrt{7})$$ (b) $$A \cap B \cap C = \emptyset$$