1. **State the problem:** A water extraction machine starts with a maximum capacity of 80 gallons per hour but loses 10% efficiency every 2 hours. We need to find how many hours it takes to extract 640 gallons. 2. **Define variables and formula:** Let $h$ be the total hours of operation. The machine operates in 2-hour intervals, and at the start of each interval, efficiency decreases by 10%. The capacity during the $n^{th}$ 2-hour interval is: $$C_n = 80 \times (0.9)^{n-1}$$ 3. **Calculate water extracted per interval:** Each interval lasts 2 hours, so water extracted in the $n^{th}$ interval is: $$W_n = C_n \times 2 = 80 \times (0.9)^{n-1} \times 2 = 160 \times (0.9)^{n-1}$$ 4. **Sum water extracted until total reaches 640 gallons:** We want the smallest integer $k$ such that: $$\sum_{n=1}^k W_n \geq 640$$ 5. **Sum of geometric series:** The sum is: $$S_k = 160 \sum_{n=0}^{k-1} (0.9)^n = 160 \times \frac{1 - (0.9)^k}{1 - 0.9} = 160 \times \frac{1 - (0.9)^k}{0.1} = 1600 \times (1 - (0.9)^k)$$ 6. **Solve inequality:** $$1600 \times (1 - (0.9)^k) \geq 640$$ $$1 - (0.9)^k \geq \frac{640}{1600} = 0.4$$ $$ (0.9)^k \leq 0.6$$ 7. **Take natural logarithm:** $$k \ln(0.9) \leq \ln(0.6)$$ $$k \geq \frac{\ln(0.6)}{\ln(0.9)}$$ 8. **Calculate $k$:** $$k \geq \frac{-0.5108}{-0.1054} \approx 4.85$$ Since $k$ must be an integer number of intervals, $k=5$ intervals. 9. **Calculate total hours:** Each interval is 2 hours, so total hours: $$h = 2 \times 5 = 10$$ **Final answer:** It will take 10 hours to extract 640 gallons under the given conditions.