1. **State the problem:** We have a population of southern mountain caribou that decreases from 5393 in 1998 to 2691 in 2014. We assume the population changes exponentially with a constant per capita production rate $r$. We want to find: (a) The per capita production rate $r$. (b) The time when the population falls to the critical number 437. 2. **Formula used:** The exponential growth/decay model is $$P(t) = P_0 e^{rt}$$ where $P_0$ is the initial population, $P(t)$ is the population at time $t$, and $r$ is the per capita production rate. 3. **Given data:** - $P_0 = 5393$ (population in 1998, $t=0$) - $P(16) = 2691$ (population in 2014, $t=16$ years later) - Critical population $P_c = 437$ 4. **Find $r$:** Using the formula: $$2691 = 5393 e^{16r}$$ Divide both sides by 5393: $$\frac{2691}{5393} = e^{16r}$$ Intermediate step with cancellation: $$\frac{\cancel{2691}}{\cancel{5393}} = e^{16r}$$ Take natural logarithm on both sides: $$\ln\left(\frac{2691}{5393}\right) = 16r$$ Solve for $r$: $$r = \frac{\ln\left(\frac{2691}{5393}\right)}{16}$$ Calculate: $$r = \frac{\ln(0.4986)}{16} = \frac{-0.6963}{16} = -0.0435$$ Rounded to 4 decimal places: $$r = -0.0435$$ 5. **Find time $t$ when population reaches 437:** Use the formula: $$437 = 5393 e^{rt}$$ Divide both sides by 5393: $$\frac{437}{5393} = e^{rt}$$ Intermediate step with cancellation: $$\frac{\cancel{437}}{\cancel{5393}} = e^{rt}$$ Take natural logarithm: $$\ln\left(\frac{437}{5393}\right) = rt$$ Solve for $t$: $$t = \frac{\ln\left(\frac{437}{5393}\right)}{r}$$ Calculate numerator: $$\ln(0.0810) = -2.5140$$ Calculate $t$: $$t = \frac{-2.5140}{-0.0435} = 57.79$$ Rounded to 2 decimals: $$t = 57.79$$ **Final answers:** (a) $r = -0.0435$ (b) $t = 57.79$ years after 1998, i.e., around year 1998 + 57.79 = 2055.79