Answer:
They will have the same population after 6 hours
Step-by-step explanation:
The general exponential equation format for this is;
P = P_o(e^(kt))
For Culture A, we are told that it had an initial population of 200 and doubles every hour.
Thus;
After 1 hour, P = 400
After 2 hours, P = 800
After 3 hours, P = 1600
Thus;
At t = 1, we have;
400 = 200(e^(k × 1))
400/200 = e^(k)
e^(k) = 2 - - - (eq 1)
At t = 2, we have;
800 = 200(e^(k × 2))
800/200 = e^(2k)
e^(2k) = 4 - - - (eq 2)
To find k, let's divide eq 1 by eq 2.
(e^(2k))/e^(k) = 4/2
e^(2k - k) = 2
e^(k) = 2
k = In 2
k = 0.6931
Thus;
P = 200e^(0.6931t)
For Culture B, we are told that it had an initial population of 819200 but has been contaminated. Its population is now decreasing by half every hour.
Thus;
After 1 hour, P = 409600
After 2 hours, P = 204800
After 3 hours, P = 102400
Thus;
At t = 1, we have;
409600 = 819200(e^(k × 1))
409600/819200 = e^(k)
e^(k) = 0.5 - - - (eq 1)
At t = 2, we have;
204800 = 819200(e^(k × 2))
204800/819200 = e^(2k)
e^(2k) = 0.25 - - - (eq 2)
To find k, let's divide eq 1 by eq 2.
(e^(2k))/e^(k) = 0.25/0.5
e^(k) = 0.5
k = In 0.5
k = -0.6931
Thus;
P = 819200(e^(-0.6931t))
We want to find the time when the 2 cultures will have the same population. Thus;
200e^(0.6931t) = 819200(e^(-0.6931t))
Arranging, we have;
(e^(0.6931t))/(e^(-0.6931t)) = 819200/200
e^(0.6931t - (-0.6931t) = 4096
e^(1.3862t) = 4096
1.3862t = In 4096
1.3862t = 8.3178
t = 8.3178/1.3862
t ≈ 6 hours
