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)   (Show Source): You can put this solution on YOUR website!
.
The doubling period of bacterial population 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,   =  = 94 (approximately).



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


    Therefore,   =  =  24576000.

Solved.

============

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