A bacteria culture grows with constant relative growth rate. After 2 hours there are 700 bacteria and after 8?

A bacteria culture grows with constant relative growth rate. After 2 hours there are 700 bacteria and after 8 hours the count is 70,000.





a) Find the initial population.


P(0) = ? bacteria





(b) Find an expression for the population after t hours.


P(t) = ?





(c) Find the number of cells after 7 hours.


P(7) = ? bacteria





(d) Find the rate of growth after 7 hours.


P'(7) = ? bacteria/hour





(e) When will the population reach 200,000?


t = ? hours

A bacteria culture grows with constant relative growth rate. After 2 hours there are 700 bacteria and after 8?
let A = initial count of bacteria





y = Ae^(kt)


700 = Ae^(2k)


7000 = Ae^(8k)





dividing...


10 = e^(6k)


e^k = 10^(1/6)





thus


700 = A 10^1/3


A = 700/[10^1/3] = 324.9112 or 324





y= A (10)^(t/6) .... this is the equation...





plug in t = 7... to get c


differentiate to get d.


y' = A/6 (10^(t/6)) * ln 10





when y = 200000 solve for t.


200000 = A (10)^(t/6)





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Reply:a) you only started with one bacteria, therefore the initial population is one





b) you have to find the average rate of change (ARC), i.e. slope of the expression. To do so, you make the given info coordinate points


(2,700)


(8,70000)





ARC: (70000-7000)/(8-2)=11550


P(t)=11550t+b


11550(2)=23100


23100-700=22400





Final expression:


P(t)=11550t - 22400





c) Pop 7 into t and solve





d) I would think you would take your answer from C and divide that by 7, but I don't know for sure.





e) put 200000 in as P and solve for t.