1. **Solve the equation:** $$3^{2x} - 4(3^x) - 3 = 0$$ Step 1: Let $$y = 3^x$$, then the equation becomes: $$y^2 - 4y - 3 = 0$$ Step 2: Solve the quadratic equation: $$y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$$ Step 3: Since $$y = 3^x > 0$$, both roots are positive: $$y_1 = 2 + \sqrt{7}, \quad y_2 = 2 - \sqrt{7}$$ Step 4: Solve for $$x$$: $$x = \log_3 y$$ So, $$x_1 = \log_3 (2 + \sqrt{7}), \quad x_2 = \log_3 (2 - \sqrt{7})$$ Note that $$2 - \sqrt{7} \approx -0.645$$ is negative, so only $$x_1$$ is a valid solution. **Final solution:** $$x = \log_3 (2 + \sqrt{7})$$ --- 2. **Define the sets:** - Universal set $$U = \{1,2,3,\ldots,20\}$$ - $$A = \{ x : 5 < x < 18 \} = \{6,7,8,\ldots,17\}$$ - $$B = \{\text{multiples of } 3 \text{ between } 2 \text{ and } 20\} = \{3,6,9,12,15,18\}$$ - $$C = \{\text{prime numbers } > 2 \text{ but } < 20\} = \{3,5,7,11,13,17,19\}$$ 3. **Find the intersection $$A \cap B \cap C$$:** - First find $$A \cap B$$: numbers in both $$A$$ and $$B$$ $$A \cap B = \{6,9,12,15\}$$ - Now find intersection with $$C$$: $$A \cap B \cap C = \{6,9,12,15\} \cap \{3,5,7,11,13,17,19\} = \varnothing$$ **Answer:** $$A \cap B \cap C = \emptyset$$ --- Note: The Venn diagram would show sets $$A$$, $$B$$, and $$C$$ inside $$U$$ with overlaps, but here the triple intersection is empty.