1. **State the problem:** Solve the system of equations: $$X + Y = 0$$ $$\log(X) + \log(Y) = 3$$ 2. **Recall logarithm properties:** The sum of logarithms can be rewritten as the logarithm of a product: $$\log(X) + \log(Y) = \log(XY)$$ So the second equation becomes: $$\log(XY) = 3$$ 3. **Rewrite the logarithmic equation:** Using the definition of logarithm (assuming base 10), $$XY = 10^3 = 1000$$ 4. **Use the first equation:** From $$X + Y = 0$$, we get: $$Y = -X$$ 5. **Substitute into the product equation:** $$X \cdot Y = X \cdot (-X) = -X^2 = 1000$$ 6. **Solve for $$X$$:** $$-X^2 = 1000 \implies X^2 = -1000$$ Since $$X^2$$ cannot be negative for real numbers, there is no real solution. 7. **Consider complex solutions:** $$X = \pm i\sqrt{1000} = \pm 10i\sqrt{10}$$ Then, $$Y = -X = \mp 10i\sqrt{10}$$ **Final answer:** The solutions are complex: $$\boxed{(X, Y) = \left(10i\sqrt{10}, -10i\sqrt{10}\right) \text{ or } \left(-10i\sqrt{10}, 10i\sqrt{10}\right)}$$