1. **State the problem:** Solve the equation $\log_3(x-2) + 2\log_9(x+3) = \log_3 14$ for $x$. 2. **Recall the properties of logarithms:** - Change of base: $\log_9 a = \frac{\log_3 a}{\log_3 9} = \frac{\log_3 a}{2}$ since $9 = 3^2$. - Power rule: $a \log_b c = \log_b c^a$. 3. **Rewrite the equation using base 3 logs:** $$\log_3(x-2) + 2 \cdot \log_9(x+3) = \log_3(x-2) + 2 \cdot \frac{\log_3(x+3)}{2} = \log_3(x-2) + \log_3(x+3)$$ 4. **Combine the logarithms:** $$\log_3(x-2) + \log_3(x+3) = \log_3[(x-2)(x+3)]$$ 5. **Set equal to the right side:** $$\log_3[(x-2)(x+3)] = \log_3 14$$ 6. **Since $\log_3 A = \log_3 B$, then $A = B$ (domain permitting):** $$(x-2)(x+3) = 14$$ 7. **Expand and simplify:** $$x^2 + 3x - 2x - 6 = 14$$ $$x^2 + x - 6 = 14$$ $$x^2 + x - 20 = 0$$ 8. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=1$, $c=-20$. $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-20)}}{2} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$$ 9. **Calculate the two roots:** - $x = \frac{-1 + 9}{2} = \frac{8}{2} = 4$ - $x = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$ 10. **Check domain restrictions:** - $x-2 > 0 \Rightarrow x > 2$ - $x+3 > 0 \Rightarrow x > -3$ Only $x=4$ satisfies both. **Final answer:** $$\boxed{4}$$