Question 1115763
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The doubling period of bacterial population {{{highlight(cross(in))}}} <U>is</U> 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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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;* * * I read your condition as  &nbsp;&nbsp; "4 hours after &nbsp;t= 120 minutes" * * *.



<pre>
1)  120 minutes = 6 times 20 minutes = 6 times doubling period.


    Therefore,  {{{N[initial]}}} = {{{6000/2^6}}} = 94 (approximately).



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


    Therefore,  {{{N[after_4_hours]}}} = {{{6000*2^12}}} =  24576000.
</pre>

Solved.


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<U>Be aware</U>: &nbsp;&nbsp;The solution by &nbsp;@rothauserc &nbsp;was totally &nbsp;&nbsp;<U>W R O N G</U> ! 


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