The current population of a certain bacteria is 2373 organism. It is believed that bacteria's population is tripling every 10 minutes. Approximate the population of the bacteria 6 minutes from now

Question

The current population of a certain bacteria is 2373 organism. It is believed that bacteria's population is tripling every 10 minutes. Approximate the population of the bacteria 6 minutes from now

Answer 1

P(t) = Ae^kt

Need to solve for k

P/A = 3 = e^k10

ln(3) = 10k

k = ln(3)

P(t) = 2373e^ln(3)t/10 = 2373(3)^t/10 because e^ln(3) = 3

P(6) = 2373(3)^.6 ≈ 4587 bacteria