Question

population growth model. Can anybody please help me out in trying to solve this problem? It's my homework and I don't seem to understand what I am getting.

The count in a culture of bacteria was 400 after 2 hours and 25,600 after 6 hours.
(a) What is the relative rate of growth of the bacteria population? Express your answer as a percentage. (Round your answer to the nearest whole number.)
= ? %

(b) What was the initial size of the culture? (Round your answer to the nearest whole number.)
=# ? bacteria

(c) Find a function that models the number of bacteria n(t) after t hours. (Round your r value to two decimal places.)
n(t) =

(d) Find the number of bacteria after 4.5 hours. (Round your answer to the nearest hundred.)
=#? bacteria

(e) After how many hours will the number of bacteria reach 50,000? (Round your answer to two decimal places.)
=#? hr

Answer 1

general equation:
number = a e^(kt) , where a is the initial amount, t is the time in hours and k is a constant.

case1: when t = 2, number = 400
400 = a e^(2k)

case2: when t = 6 , number = 25600
25600 = a e^(6k)

divide the 2nd equation by the 1st

64 = e^(4k)
4k = ln64
k = ln64/4 = 1.039721

back in 1st:
400 = a(e^(2(1.039721))
400 = a(8)
a = 50

b) 50
c) see above
d) number = 50 e^(4.5(1.039721) = 9050.9..
= 9100 to the nearest 100

e) 50000 = 50 e^(1.039721 t)
1000 = e^(1.039721t)
ln 1000 = 1.039721t
t = appr 6.64 hours

or

looking at the "exact" value of a = 50 , I would guess that this is "doubling" type of question, and we should find the doubling time "d"

let number = a(2)^(t/d)

400 = a 2^(2/d)
25600 = a 2^(6/d)
divide as before, ...
64 = 2^(4/d)
2^6 = 2^(4/d)
4/d = 6
d = 2/3

in 1st:
400 = a 2^(2/(2/3))
400 = a 2^(3)
a = 50 ---> same as before

so number = 50 2^(t/(2/3))
n(t) = 50 2^(3t/2)

I will do the last one using this new equation

50000 = 50 2^(3t/2)
1000 = 2^(3t/2)
log both sides
log 1000 = (3t/2) log2
3/log2 = 3t/2
3t = 6/log2
t = 2/log2 = 6.64 , just as before