A bacterial population grows at a rate proportional to the population size at time t. Let y(t) be the population size at time t. By experiment it is determined that the population at t=15 min is 16000 and at t=45 min it is 22000.

Question

(a) What was the initial population?
(b)What will the population be at time t=65 min?

Pls help! I know how to solve this kind of question if the initial condition is given. How do I approach it this time?

Answer 1

780000

Answer 2

y(t) = a e^(kt) , where a is the intitial count, and k is a constant, and t is in minutes

You have two ordered pairs, (15,16000) and (45,22000)

for (15,16000)
16000 = a e(15k)
for (45,22000)
22000 = a e^(45k)
divide the 2nd by the 1st
1.375 =e^(30k)
ln both sides , that is : (log base e)
ln 1.375 = 30k
k = ln 1.375/30 = .010615.. (store it in your calculator )
now plut that into your first equation of
16000 = a e^(15(.0106..) )
I got a = 13645

now that you have both a and k in the equation, plug in t = 65
(here is where storing the value of k is important)

Answer 3

btw, jorge's answer is bogus.
I got 27203

Answer 4

The population of a southern city follows the exponential law. If the population doubled in size over 2929 months and the current population is 30 comma 00030,000​, what will the population be 33 years from​ now

Answer 5

The population of a southern city follows the exponential law. If the population doubled in size over 29 months and the current population is 30,000​, what will the population be 33 years from​ now