Posted by George on Sunday, November 9, 2008 at 1:57pm.
I am a bit befuddled by this question.
According to your first part, we could say that the bacteria population is
N(t) = a(2)^(t/2) where N(t) is the number of bacteria and t is time in minutes after 6 am.
so at 6 am, t = 0, and N(0) = a(2)^0 = a
so we started with a bacteria
How many do we have at 6 pm?
that is 12 hours or 720 minutes, so t=720
N(720) = a(2)^36
= a(6.872 x 10^10) bacteria
then you ask "At what time does the syrup become half full?"
did you mean : At what time does the syrup can become half the capacity of the bacteria?
half of a(6.872 x 10^10) is a(3.436 x 10^10)
so we solve
a(3.436 x 10^10) = a(2)^(t/20)
notice the a drops out
I hope you know how to do logs.
log (3.436 x 10^10) = (t/20)log2
t/20 = log(3.436 x 10^10)/log2
t/20 = 35
t = 35(20) = 700 minutes, which would correspond to 5:40 pm
(At this point I said to myself, Duhhh!
since it doubled every 20 minutes, half the content must have been reached within the last doubling period, which of course is 20 minutes)
Give the last part a shot yourself, ok?
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