Solve the problem.
-The size P of a small herbivore population at time t (in years) obeys the function $\mathrm { P } ( \mathrm { t } ) = 800 \mathrm { e } ^ { 0.24 \mathrm { t } }$ if they have enough food and the predator population stays constant. After how many years will the population reach 1,600?
A) 8.66 yrs
B) 27.85 yrs
C) 7.05 yrs
D) 2.89 yrs

Question

Quizplus · Accepted Answer

To find the time t when the population reaches 1,600, solve the equation $800 \mathrm{e}^{0.24t} = 1600$. Dividing both sides by 800 gives $\mathrm{e}^{0.24t} = 2$. Taking the natural logarithm of both sides, we get $0.24t = \ln(2)$. Solving for t gives $t = \frac{\ln(2)}{0.24} \approx 2.89$ years.

Solve the Problem P(t)=800e0.24t\mathrm { P } ( \mathrm { t } ) = 800 \mathrm { e } ^ { 0.24 \mathrm { t } }P(t)=800e0.24t

Solve the Problem $\mathrm { P } ( \mathrm { t } ) = 800 \mathrm { e } ^ { 0.24 \mathrm { t } }$