1. **Problem statement:** We are given the population of Grayrock in 1960 as 10,000 and in 1970 as 12,000. We assume the population grows exponentially and want to estimate the population in 1980 (part a) and find the doubling time (part b). 2. **Exponential growth model formula:** The population at time $t$ is given by $$ P(t) = P_0 e^{kt} $$ where $P_0$ is the initial population, $k$ is the growth rate, and $t$ is time in years since the initial measurement. 3. **Assign variables:** Let $t=0$ correspond to 1960, so $P_0 = 10000$. At $t=10$ (1970), $P(10) = 12000$. 4. **Find growth rate $k$:** Using the formula: $$ 12000 = 10000 e^{10k} $$ Divide both sides by 10000: $$ 1.2 = e^{10k} $$ Take natural logarithm: $$ \ln(1.2) = 10k $$ So, $$ k = \frac{\ln(1.2)}{10} $$ Calculate $\ln(1.2) \approx 0.182321556$, thus $$ k \approx \frac{0.182321556}{10} = 0.0182321556 $$ 5. **Estimate population in 1980 ($t=20$):** $$ P(20) = 10000 e^{0.0182321556 \times 20} = 10000 e^{0.364643112} $$ Calculate $e^{0.364643112} \approx 1.439$, so $$ P(20) \approx 10000 \times 1.439 = 14390 $$ The estimated population in 1980 is approximately 14,390. 6. **Find doubling time $T$:** Doubling time satisfies $$ 2P_0 = P_0 e^{kT} $$ Divide both sides by $P_0$: $$ 2 = e^{kT} $$ Take natural logarithm: $$ \ln(2) = kT $$ So, $$ T = \frac{\ln(2)}{k} $$ Calculate $\ln(2) \approx 0.693147181$, thus $$ T = \frac{0.693147181}{0.0182321556} \approx 38.03 $$ The doubling time is approximately 38 years. **Final answers:** - Population in 1980: approximately 14,390 - Doubling time: approximately 38 years