SOLUTION: I can't figure this out. Suppose that a colony of bacteria starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end o

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Question 481977: I can't figure this out.
Suppose that a colony of bacteria starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hours?
Thank you!

Found 2 solutions by solver91311, nerdybill:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


There are 48 half-hour periods in 24 hours, so there are bacteria at the end of 24 hours. Roughly 281 trillion, give or take a few second cousins twice removed.

John

My calculator said it, I believe it, that settles it
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Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose that a colony of bacteria starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hours?
.
Apply exponential growth formula:
y = ae^(rt)
where
a is initial amount (1)
y is amount after time t (2)
r is rate of growth/decline
t is time (.5)
.
so, now we have
2 = (1)e^(.5r)
solving for 'r':
2 = e^(.5r)
ln(2) = .5r
ln(2)/.5 = r
.
Now, our general formula is:
y = e^(t*ln(2)/.5)
so, when t = 24
y = e^(24*ln(2)/.5)
y = e^(48*ln(2))
y = 2.81 x 10^14 (A LOT!)