SOLUTION: An experiment is carried out in which the number n of bacteria in a liquid, is given by the formula {{{n=650e^(kt)}}}, where t is the time in minutes after the beginning of the exp
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-> SOLUTION: An experiment is carried out in which the number n of bacteria in a liquid, is given by the formula {{{n=650e^(kt)}}}, where t is the time in minutes after the beginning of the exp
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Question 979620: An experiment is carried out in which the number n of bacteria in a liquid, is given by the formula , where t is the time in minutes after the beginning of the experiment and k is a constant. The number of bacteria doubles every 20 minutes. Find the exact value of k. Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! Hi there,
n = 650e^kt
The current value for the bacteria = 650
If the number doubles then n = 1300
So n = 650e^kt
=> 1300 = 650e^20k
=> 1300/650 = e^20k
=> 2 = e^20k
=> log(base e)2 = 20k log(base e) e
log(base e)e = 1
=> log(base e)2 = 20k
=> k = log(base e)2/20
=> k = 0.03466 (to 5 decimal places)
Hope this helps:-)
