Question

The bacteria culture grows at a rate proportional to its size. After $2$ hours there are $600$ bacteria and after $8$ hours the count is $75000.$ Find the initial population and when the population reach $200000$ ?

Answer 1

P(t) = P(0) k^t

(OR P(t) = P(0) . exp[ln(k) . t] . )

P(2) = P(0) k^2 = 600
P(8) = P(0) k^8 = 75000

P(8)/P(2) = k^6 = 75000/600 = 125, so k = sqrt(5)

i) P(0) = 600/k^2 = 600/5 = 120
P(t) = 120 . 5^(t/2)

P(t) = 120 . 5^(t/2) = 200000

5^(t/2) = 200000 / 120
(t/2) = ln(200000 / 120) / ln(5)

t = 2 . ln(200000 / 120) / ln(5) = 9.2 hrs