SOLUTION: The doubling period of bacterial population in 20 minutes. At time t = 120 minutes, the bacterial population was 6000. What was the initial population at time t = 0? Find the si

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Question 1115763: The doubling period of bacterial population in 20 minutes. At time t = 120 minutes, the bacterial population was 6000.
What was the initial population at time t = 0?
Find the size of the bacterial population after 4 hours?
Found 2 solutions by ikleyn, rothauserc:
Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
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The doubling period of bacterial population highlight%28cross%28in%29%29 is 20 minutes. At time t = 120 minutes, the bacterial population was 6000.
What was the initial population at time t = 0?
Find the size of the bacterial population after 4 hours?
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            * * * I read your condition as    "4 hours after  t= 120 minutes" * * *. 


1)  120 minutes = 6 times 20 minutes = 6 times doubling period.


    Therefore,  N%5Binitial%5D = 6000%2F2%5E6 = 94 (approximately).



2)  4 hours = 4*60 minutes = 4*3 = 12 doubling periods.


    Therefore,  N%5Bafter_4_hours%5D = 6000%2A2%5E12 =  24576000.

Solved.

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Be aware:   The solution by  @rothauserc  was totally   W R O N G !

                  His error was in that the  "linear rate",  as he defined it,  was irrelevant to the exponential rate.



Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
The population doubles every 20 minutes, in 120/20 there are 6 time periods
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Therefore initial population is (6000/2^6) = 93.75 approximately 94
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in 240 minutes there are 240/20 = 12 time periods
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we know the population is 6000 at time 120
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there are 120/20 = 6 additional time periods
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population after 240 minutes = 6000 * 2^6 = 384,000
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