1. **State the problem:** Given the equation $\log_3 x + \log_3 y = 2$, find the minimum value of $2x + 3y$. 2. **Use logarithm properties:** Recall that $\log_a b + \log_a c = \log_a (bc)$. So, $$\log_3 x + \log_3 y = \log_3 (xy) = 2.$$ This implies $$xy = 3^2 = 9.$$ 3. **Rewrite the problem:** We want to minimize $$2x + 3y$$ subject to the constraint $$xy = 9.$$ 4. **Express one variable in terms of the other:** From $xy=9$, we get $$y = \frac{9}{x}.$$ 5. **Substitute into the expression to minimize:** $$2x + 3y = 2x + 3 \cdot \frac{9}{x} = 2x + \frac{27}{x}.$$ 6. **Minimize the function:** Let $$f(x) = 2x + \frac{27}{x},$$ where $x > 0$ (since $\log_3 x$ is defined only for $x>0$). 7. **Find critical points by differentiation:** $$f'(x) = 2 - \frac{27}{x^2}.$$ Set $f'(x) = 0$: $$2 - \frac{27}{x^2} = 0 \implies 2 = \frac{27}{x^2} \implies x^2 = \frac{27}{2} \implies x = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2}.$$ 8. **Find corresponding $y$:** $$y = \frac{9}{x} = \frac{9}{\frac{3\sqrt{6}}{2}} = \frac{9 \cdot 2}{3\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}.$$ 9. **Calculate minimum value:** $$2x + 3y = 2 \cdot \frac{3\sqrt{6}}{2} + 3 \cdot \sqrt{6} = 3\sqrt{6} + 3\sqrt{6} = 6\sqrt{6}.$$ **Final answer:** The minimum value of $2x + 3y$ given $\log_3 x + \log_3 y = 2$ is $$\boxed{6\sqrt{6}}.$$