1. **Problem statement:** The population of a town decreased by 8% in 2077 B.S., then decreased by 10% in 2078 B.S. and onward. At the beginning of 2080 B.S., the population is 556600. We need to answer the following: a) Formula for population increase condition. b) Population at the beginning of 2078 B.S. c) Population at the beginning of 2077 B.S. d) Compare populations at 2077 B.S. and 2080 B.S. 2. **a) Formula for population increase condition:** The general formula for population increase or decrease is: $$P_{n} = P_{0} (1 + r)^n$$ where $P_0$ is the initial population, $r$ is the rate of increase (positive for increase, negative for decrease), and $n$ is the number of years. 3. **b) Population at the beginning of 2078 B.S.:** Let $P_{2077}$ be the population at the beginning of 2077 B.S. The population decreased by 8% in 2077, so at the beginning of 2078: $$P_{2078} = P_{2077} \times (1 - 0.08) = 0.92 P_{2077}$$ 4. **c) Population at the beginning of 2077 B.S.:** From 2078 to 2080, the population decreased by 10% each year for 2 years. So, $$P_{2080} = P_{2078} \times (1 - 0.10)^2 = P_{2078} \times 0.9^2 = P_{2078} \times 0.81$$ Given $P_{2080} = 556600$, we find $P_{2078}$: $$P_{2078} = \frac{556600}{0.81} = 687654.32$$ 5. Using $P_{2078} = 0.92 P_{2077}$, we find $P_{2077}$: $$P_{2077} = \frac{P_{2078}}{0.92} = \frac{687654.32}{0.92} = 747009.27$$ 6. **d) Comparison of populations at 2077 and 2080:** Population at 2077: 747009 (approx) Population at 2080: 556600 The population decreased by: $$747009 - 556600 = 190409$$ Percentage decrease over 3 years: $$\frac{190409}{747009} \times 100 \approx 25.5\%$$ **Final answers:** a) Formula: $$P_n = P_0 (1 + r)^n$$ b) Population at 2078: approximately 687654 c) Population at 2077: approximately 747009 d) Population decreased by approximately 25.5% from 2077 to 2080.