1. **State the problem:** Solve the equation $$\log_3(x) + \log_3(2x + 3) = 3$$ for $x$ and check for extraneous solutions. 2. **Use the logarithm property:** Recall that $$\log_b(a) + \log_b(c) = \log_b(ac)$$ when $a, c > 0$. 3. **Apply the property:** $$\log_3(x) + \log_3(2x + 3) = \log_3\big(x(2x + 3)\big) = 3$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\log_3\big(x(2x + 3)\big) = 3 \implies x(2x + 3) = 3^3 = 27$$ 5. **Expand and form a quadratic equation:** $$2x^2 + 3x = 27$$ $$2x^2 + 3x - 27 = 0$$ 6. **Factor the quadratic:** $$ (2x + 9)(x - 3) = 0 $$ 7. **Solve for $x$:** $$2x + 9 = 0 \implies x = -\frac{9}{2}$$ $$x - 3 = 0 \implies x = 3$$ 8. **Check for extraneous solutions:** - The domain of $\log_3(x)$ requires $x > 0$. - The domain of $\log_3(2x + 3)$ requires $2x + 3 > 0 \implies x > -\frac{3}{2}$. Since $x = -\frac{9}{2}$ is less than zero, it is **extraneous** and must be discarded. 9. **Final solution:** $$\boxed{x = 3}$$