1. **State the problem:** We are given the population of Elmdale in 2010 and 2018 and need to find the continuous growth rate $r$ using the formula for continuous population growth: $$N(t) = N_0 e^{rt}$$ where: - $N(t)$ is the population at time $t$, - $N_0$ is the initial population, - $r$ is the growth rate, - $t$ is the time in years since the initial measurement. 2. **Identify given values:** - Initial population $N_0 = 34500$ (in 2010, so $t=0$) - Population in 2018 $N(8) = 48665$ (since 2018 is 8 years after 2010) - Time elapsed $t = 8$ years 3. **Write the formula with known values:** $$48665 = 34500 \times e^{8r}$$ 4. **Isolate the exponential term:** $$\frac{48665}{34500} = e^{8r}$$ 5. **Simplify the fraction:** $$\frac{48665}{34500} \approx 1.4107$$ 6. **Take the natural logarithm of both sides to solve for $r$:** $$\ln\left(1.4107\right) = \ln\left(e^{8r}\right)$$ $$\ln\left(1.4107\right) = 8r$$ 7. **Calculate the logarithm:** $$\ln\left(1.4107\right) \approx 0.344$$ 8. **Solve for $r$:** $$r = \frac{0.344}{8}$$ $$r = \frac{\cancel{0.344}}{\cancel{8}} = 0.043$$ 9. **Convert $r$ to a percentage and round to the nearest tenth:** $$r \approx 4.3\%$$ **Final answer:** The continuous growth rate $r$ for Elmdale is approximately **4.3%** per year.