1. **State the problem:** We want to find the population of Bigtimeville in the year 2008 given that in 2025 the population is 63,881 and the net growth rate is $\frac{8}{1000}$ per year. 2. **Formula used:** The population growth can be modeled by the formula for exponential growth: $$ P(t) = P_0 e^{rt} $$ where: - $P(t)$ is the population at time $t$, - $P_0$ is the initial population, - $r$ is the growth rate per year, - $t$ is the time in years. 3. **Identify variables:** - Let $t=0$ correspond to the year 2008. - The year 2025 corresponds to $t=2025-2008=17$ years. - Given $P(17) = 63,881$ and $r = \frac{8}{1000} = 0.008$. 4. **Find $P_0$ (population in 2008):** Rearranging the formula: $$ P_0 = \frac{P(17)}{e^{r \times 17}} $$ 5. **Calculate:** $$ P_0 = \frac{63,881}{e^{0.008 \times 17}} = \frac{63,881}{e^{0.136}} $$ Calculate $e^{0.136}$: $$ e^{0.136} \approx 1.1457 $$ So, $$ P_0 = \frac{63,881}{1.1457} \approx 55,747 $$ 6. **Answer:** The population of Bigtimeville in 2008 was approximately 55,747 people.