1. **State the problem:** We need to solve the system of equations: $$x + y = 69$$ $$\log(x) + \log(y) = 3$$ 2. **Recall the logarithm property:** The sum of logarithms is the logarithm of the product: $$\log(x) + \log(y) = \log(xy)$$ So the second equation becomes: $$\log(xy) = 3$$ 3. **Rewrite the logarithmic equation:** Assuming the logarithm is base 10, $$xy = 10^3 = 1000$$ 4. **Use substitution:** From the first equation, $$y = 69 - x$$ Substitute into the product equation: $$x(69 - x) = 1000$$ 5. **Form a quadratic equation:** $$69x - x^2 = 1000$$ Rearranged: $$x^2 - 69x + 1000 = 0$$ 6. **Solve the quadratic equation:** Using the quadratic formula, $$x = \frac{69 \pm \sqrt{69^2 - 4 \times 1 \times 1000}}{2}$$ Calculate the discriminant: $$69^2 = 4761$$ $$4 \times 1000 = 4000$$ $$\sqrt{4761 - 4000} = \sqrt{761} \approx 27.58$$ 7. **Find the roots:** $$x_1 = \frac{69 + 27.58}{2} = \frac{96.58}{2} = 48.29$$ $$x_2 = \frac{69 - 27.58}{2} = \frac{41.42}{2} = 20.71$$ 8. **Find corresponding y values:** For $x_1 = 48.29$, $$y_1 = 69 - 48.29 = 20.71$$ For $x_2 = 20.71$, $$y_2 = 69 - 20.71 = 48.29$$ 9. **Check solutions:** Both pairs $(48.29, 20.71)$ and $(20.71, 48.29)$ satisfy the equations. **Final answer:** $$\boxed{(x,y) = (48.29, 20.71) \text{ or } (20.71, 48.29)}$$