At the start of an experiment, 3000 bacteria are present in a colony. Two hours later, the population is 4900.

Assume that the population grows exponentially, that is, according to the law
N(t) = N0ekt.

please help. thanks
At the start of an experiment, 3000 bacteria are present in a colony. Two hours later, the population is 4900. (Assume t is in hours.)
(a) Determine the growth constant k. (Round your answer to four decimal places.)
k = .2453 i managed to get the right answer here.

(b) Determine the population five hours after the start of the experiment. (Round your answer to the nearest whole number.)
bacteria
in this part i am confused on how one finds out what k equals....
(c) When will the population reach 10,000? (Round your answer to the nearest whole number.)
hr

qb
you use the same k, so
N(5) = 3000*e^(.2453*5) = 10,228 <------

qc
10000 = 3000*e^(.2453t)
e^(.2453t) = (10/3)
.2453t = ln(10/3)
t = ln(10/3)/.2453 = 5 hrs <------

k would be the same, because you're trying to find the population after a time t given a growth constant k. You solved for k, so you just plug that same one back in.

3000e^.2453(5)

and whatever that comes out to be is your answer.

For the next part, solve as such -


10,000 = 3000e^.2453t

Divide the 3000 out

10000/3000 = e^.2453t

take the natural log of both sides to eliminate the x and you get

ln 10000/3000 = .2453t

Thus, t = the natural log of 10000/3000 divided by .2453 or

(ln 10000/3000)/.2453

The k value is correct. Now you need to solve for what it would be for 5 hours using the k value you found.

B)
N(t) = N0e^kt.
N(t) = 3000e^(0.2453*5) = 10227.8

C)
N(t) = N0e^kt.
10000 = 3000e^(0.2453*t)
ln (10000/3000) = ln (e^(0.2453*t)
1.204 = 0.2453*t
t = 4.908 hours

This checks out because the amount is just shy of why we obtained in part B.