At 6am a few bacteria fall into a can of syrup in a broken garbage bag. The conditions of warmth, moisture, and food are perfect for growth and the population doubles every 20 minutes. By 6pm the bacteria are overcrowded and dry and their food is gone.At what time does the syrup become half full?

Question

A few bacteria search for more food and space. They find three more syrup cans how much of a time reprieve are they given by this find? When will the new cans be depleated?

I can't figure out how to do these problems without being given the number of bacteria that fell into the can to begin with...is there a way to do it that I'm not seeing?

Answer 1

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?