SOLUTION: A bacteria culture initially contains 1500 bacteria and doubles every half hour. Find the size of the baterial population after 100 minutes. Find the size of the bateri

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Question 1159087: A bacteria culture initially contains 1500 bacteria and doubles every half hour.
Find the size of the baterial population after 100 minutes.

Find the size of the baterial population after 4 hours.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
y=p%2A2%5E%282x%29
p, initial size
x, number of HOURS passed

If p is 1500 then y=15002%5E%282x%29.


100 minutes is 60 minutes plus 40 minutes, or 1%262%2F3 hours.
y=1500%2A2%5E%282%28cross%285%2F2%29%29%29
y=1500%2A2%5E%282%285%2F3%29%29
y=1500%2A2%5Ecross%285%29
y=1500%2A2%5E%2810%2F3%29
y=15120 after 100 minutes

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution and the answer  (of 48000 bacteria population)  by @josgarithmetic are  INCORRECT.

            I came to bring the correct solution/answer. 


N = 1500*2^(t/0.5) = 1500*2^(2t),


where  " t "  is the time in hours.


t = 100 minutes = 12%2F3 hours = 5%2F3 hours.



Therefore,  after 100 minute,  the bacteria population is

    N = 1500*2^(2*(5/3)) = 1500*2^(10/3) = 15119.    ANSWER




4 hours are 8 doubling periods.  Therefore, after 4 hours, the bacteria population is

    N = 1500*2^8  = 384000.    ANSWER

Solved.