SOLUTION: The logistic growth model P(t)=1270/1+27.22e^-0.348t represents the population of a bacterium in a culture tube after "t" hours. A) What was the initial amount of bacteria in the

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The logistic growth model P(t)=1270/1+27.22e^-0.348t represents the population of a bacterium in a culture tube after "t" hours. A) What was the initial amount of bacteria in the       Log On


   

Question 87056: The logistic growth model P(t)=1270/1+27.22e^-0.348t represents the population of a bacterium in a culture tube after "t" hours.
A) What was the initial amount of bacteria in the population?
B)After how many hours is the population of bacteria 1000? Round to the nearest hour.
C) What is the limitig size of P(t), the poluation of bacterium?
* For B) I put 1000 in for "P" and C) I think the answer would be 1270...right?
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
A) What was the initial amount of bacteria in the population?
Initial amount would be determined after 0 hours ... t = 0
P%28t%29+=+1270%2F%281+%2B+27.22e%5E%28-0.348t%29%29
P%280%29+=+1270%2F%281+%2B+27.22e%5E%280%29%29
P%280%29+=+1270%2F%281+%2B+27.22%29
P%280%29+=+1270%2F28.22
Initial amount would be about 45
B)After how many hours is the population of bacteria 1000? Round to the nearest hour.
P%28t%29+=+1270%2F%281+%2B+27.22e%5E%28-0.348t%29%29
1000+=+1270%2F%281+%2B+27.22e%5E%28-0.348t%29%29
1+%2B+27.22e%5E%28-0.348t%29+=+1270%2F1000
1+%2B+27.22e%5E%28-0.348t%29+=+1.27
27.22e%5E%28-0.348t%29+=+0.27
e%5E%28-0.348t%29+=+0.27%2F27.22
log%28e%2Ce%5E%28-0.348t%29%29+=+log%28e%2C0.27%2F27.22%29
-0.348t+=+log%28e%2C0.27%2F27.22%29
t+=+-log%28e%2C0.27%2F27.22%29%2F0.348
C) What is the limitig size of P(t), the poluation of bacterium?
P%28t%29+=+1270%2F%281+%2B+27.22e%5E%28-0.348t%29%29
Firstly, we would have to map out 1 + 27.22e^(-0.348t). Since time is positive, we must see how 1 + 27.22e^(-0.348t) reacts as t increases positively.
Red: 1 + 27.22/e^(0.348t)
Green: 1270/(1 + 27.22e^(-0.348t))

1270 is being divided by an exponentially smaller number every hour
as 1 + 27.22/e^(0.348t) approaches infinity for time ... it equals 1
1270 / 1 is the limit