1. **State the problem:** We have a population model given by the formula $$P = 71.2e^{0.0329t}$$ where $t=1$ corresponds to the year 1971. We want to find the year when the population reaches 360,000. 2. **Understand the variables:** Here, $P$ is the population and $t$ is the time in years since 1970 (since $t=1$ is 1971, $t=0$ is 1970). 3. **Set up the equation:** We want to find $t$ such that $$P = 360000 = 71.2e^{0.0329t}$$ 4. **Solve for $t$:** $$360000 = 71.2e^{0.0329t}$$ Divide both sides by 71.2: $$\frac{360000}{71.2} = \cancel{\frac{71.2}{71.2}} e^{0.0329t}$$ $$5056.18 = e^{0.0329t}$$ 5. **Take the natural logarithm of both sides:** $$\ln(5056.18) = \ln\left(e^{0.0329t}\right)$$ $$\ln(5056.18) = 0.0329t$$ 6. **Calculate the logarithm and solve for $t$:** $$t = \frac{\ln(5056.18)}{0.0329}$$ $$t = \frac{8.528}{0.0329} \approx 259.3$$ 7. **Interpret the result:** Since $t=0$ corresponds to 1970, the year when the population reaches 360,000 is approximately: $$1970 + 259.3 \approx 2229$$ **Final answer:** The population will reach 360,000 around the year **2229** according to the model.