1. **State the problem:** Solve the equation $$\log_3(x + 2) + \log_3(x - 1) = 2$$ for $x$. 2. **Recall the logarithm property:** The sum of logarithms with the same base can be combined as the logarithm of the product: $$\log_b A + \log_b B = \log_b (A \times B)$$ 3. **Apply the property:** $$\log_3(x + 2) + \log_3(x - 1) = \log_3\big((x + 2)(x - 1)\big)$$ So the equation becomes: $$\log_3\big((x + 2)(x - 1)\big) = 2$$ 4. **Rewrite the logarithmic equation in exponential form:** $$\log_3 y = 2 \implies y = 3^2 = 9$$ Therefore: $$(x + 2)(x - 1) = 9$$ 5. **Expand the left side:** $$x^2 - x + 2x - 2 = 9$$ $$x^2 + x - 2 = 9$$ 6. **Bring all terms to one side:** $$x^2 + x - 2 - 9 = 0$$ $$x^2 + x - 11 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, and $c=-11$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 1 \times (-11) = 1 + 44 = 45$$ So: $$x = \frac{-1 \pm \sqrt{45}}{2} = \frac{-1 \pm 3\sqrt{5}}{2}$$ 8. **Check domain restrictions:** Since the logarithms require their arguments to be positive: $$x + 2 > 0 \implies x > -2$$ $$x - 1 > 0 \implies x > 1$$ The stricter condition is $x > 1$. 9. **Evaluate the solutions:** $$x_1 = \frac{-1 + 3\sqrt{5}}{2} \approx \frac{-1 + 6.708}{2} = \frac{5.708}{2} = 2.854 > 1$$ (valid) $$x_2 = \frac{-1 - 3\sqrt{5}}{2} \approx \frac{-1 - 6.708}{2} = \frac{-7.708}{2} = -3.854 < 1$$ (not valid) 10. **Final answer:** $$\boxed{x = \frac{-1 + 3\sqrt{5}}{2}}$$