A cell of some bacteria divides into two cells every 30 minuntes. The initial population is 5 bacteria?

Hint: The calculation will be easier if you let the unit of time be 20 minutes, so that one hour is (60/30) units of time.





Find the population after t hours


FInd the population after 2 hours .


When will the population reach 10?

A cell of some bacteria divides into two cells every 30 minuntes. The initial population is 5 bacteria?
q1


P(t) = 5*2^(t/(1/2) = 5*2^(2t)





q2


P(2) = 5*2^(2*2) = 5*16 = 80





q3


10 = 5*2^(2t)


2^(2t) = 10/5 = 2^1


2t = 1


t = 1/2 hour = 30 mins
Reply:Let T be a time period in which the population doubles (in this case 30 mins but don’t worry about that right now)


At T = 0 y = 5


At T= 1 y = 5 x 2^1 (= 5 x 2 = 10 as a check for the pattern)


at T = 2 y = 5 x 2^2 (= 5 x 4 = 20)


At T = 3 y = 5 x 2^3 (= 5 x 8 = 40)


So when T = n y = 5 x 2^n


But T = 30t, where t is the time in minutes


So when t = 30 y = 5 x 2^1


when t = 60 y = 5 x 2^2


when t = 90 y = 5 x 2^3


So to find what power of 2 we need, can you see that you take the time and divide it by 30?


So instead of T = 5 x 2^n


you have t = 5 X 2 ^ ……..?








PS to find when the population reaches 10, you takes logs of both sides - that is the only way to get n on its own