1. **Problem statement:** The population of gnomes grows exponentially. Initial population $P_0=7$. After 3 years, population $P(3)=12$. We need to find: a) The growth rate $r$. b) Population after 10 years. c) Time when population reaches 100. 2. **Formula for exponential growth:** $$P(t) = P_0 e^{rt}$$ where $P(t)$ is population at time $t$, $r$ is growth rate. 3. **Find growth rate $r$: ** Given $P(3)=12$, substitute: $$12 = 7 e^{3r}$$ Divide both sides by 7: $$\frac{12}{7} = e^{3r}$$ Take natural logarithm: $$\ln\left(\frac{12}{7}\right) = 3r$$ Solve for $r$: $$r = \frac{\ln\left(\frac{12}{7}\right)}{3}$$ Calculate: $$r \approx \frac{\ln(1.7143)}{3} \approx \frac{0.539}{3} = 0.1797$$ 4. **Population after 10 years:** Use formula: $$P(10) = 7 e^{0.1797 \times 10} = 7 e^{1.797}$$ Calculate: $$P(10) \approx 7 \times 6.03 = 42.21$$ So approximately 42 gnomes. 5. **Time when population reaches 100:** Set $P(t)=100$: $$100 = 7 e^{0.1797 t}$$ Divide both sides by 7: $$\frac{100}{7} = e^{0.1797 t}$$ Take natural logarithm: $$\ln\left(\frac{100}{7}\right) = 0.1797 t$$ Solve for $t$: $$t = \frac{\ln\left(\frac{100}{7}\right)}{0.1797}$$ Calculate: $$t \approx \frac{\ln(14.2857)}{0.1797} \approx \frac{2.659}{0.1797} = 14.8$$ So about 14.8 years. **Final answers:** - Growth rate $r \approx 0.1797$ per year - Population after 10 years $\approx 42$ gnomes - Time to reach 100 gnomes $\approx 14.8$ years