SOLUTION: three yeast cells are placed in a laboratory dish at 9:00 am. the number of yeast cells doubles in every 5 minutes period. the number of cells in the dish at 9:30 am the same day i

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Question 237581: three yeast cells are placed in a laboratory dish at 9:00 am. the number of yeast cells doubles in every 5 minutes period. the number of cells in the dish at 9:30 am the same day is
a) 30 b) 64 c) 192 d) 256 e) 300
Answer by nerdybill(7384) About Me  (Show Source):
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three yeast cells are placed in a laboratory dish at 9:00 am. the number of yeast cells doubles in every 5 minutes period. the number of cells in the dish at 9:30 am the same day is
a) 30 b) 64 c) 192 d) 256 e) 300
.
This is an "exponential growth" problem...apply :
A(t) = Pe^(rt)
.
For your problem, they give:
Let P (initial amount) = 3
then A(t) (amount at time t) = 2(3) = 6
r is growth ratio -- this is what we're looking for
t is 5 minutes
.
So,
A(t) = Pe^(rt)
6 = 3e^(5r)
Dividing both sides by 3:
2 = e^(5r)
ln 2 = 5r
ln(2)/5 = r
.
Finally, for t = 30 (9:30 - 9:00)
we have
A(t) = 3e^(ln(2)/5t)
A(30) = 3e^(ln(2)/5(30))
A(30) = 3e^(ln(2)(6))
A(30) = 3(64)
A(30) = 192 cells