Suppose that the doubling time of a bacteria is 18 hours and the initial amount is 10. How many do you have after 30 hrs? How long will it take to have 50?

Question

I need help I don't understand this problem. 

Answer 1

Let B(t) be the amount of bacteria at time (hour) t:

 

B(t) = 10*2^t/18

 

When t = 0, 2^t/18 = 2^0/18 = 2⁰ = 1, and B(t) = 10*1 = 10. which are the 10 bacteria we start with at time t = 0.  When t = 18, 2^t/18 = 2^18/18 = 2¹ = 2, and B(t) = 10*2 = 20, so the bacteria have doubled from 10 to 20 in 18 hours.

 

How many do you have in 30 hours?  Set t = 30:

 

B(t=30) = 10*2^30/18 = ____

 

Use your calculator to get the answer.

 

How long will it take to have 50 bacteria?  Set B(t) = 50 and solve for t:

 

50 = 10*2^t/18

5 = 2^t/18

log₂(5) = t/18

18*log₂(5) = t

 

Use your calculator to get the answer.  You know that the population doubles every 18 hours, so it starts at 10 and doubles to 20 after 18 hours, then doubles again to 40 after 2*18=36 hours, the doubles again to 80 after 3*18 = 54 hours.  So you know the answer will be more than 36 hours but less than 54 hours.