The count in a bacteria culture was 600 after 20 minutes and 1600 after 30 minut

Question

The count in a bacteria culture was 600 after 20 minutes and 1600 after 30 minutes. Assuming the count grows exponentially, What was the initial size of the culture? Find the doubling period. Find the population after 120 minutes. When will the population reach 13000.

Explanation / Answer

Dear Student Thank you for using Chegg ! Given that bacteria count grows exponentially P :- Population of bacteria at any time t P0 :- Initial population k :- Rate of increase t :- Time P = P0*e^(kt) Given count in a bateria culture was 600 after 20 minutes and 1600 after 30 minutes. => 600 = P0e^(k*(1/3)) Equation 1 1600 = P0e^(k*(1/2)) Equation 2 Dividing the above 2 equations 0.375 = e^(-k/6) Taking natural log both sides we get -0.98083 = -k/6 k = 5.88498 Substituting in equation 1 and calculating P0 600 = P0e^(5.88498*(1/3)) 600 = 7.111122P0 P0 = 84.37487 Initial Size = 84 b) Doubling period P = P0*e^(kt) P = 2P0 => 2P0 = P0*e^(5.88498*t) 2 = e^(5.88498*t) Taking natural log both sides we get 0.693147 = 5.88498t t = 0.117757 hrs c) Population after 120 mnutes i.e. 2 hrs P = P0*e^(kt) P = 84.37*e^(5.88498*2) P = 10909798 Solution d) When will the population reach 13000 13000 = 84.37e^(5.88498*t) 154.0832 = e^(5.88498*t) Taking natural log both sides 5.037493 = 5.88498*t t = 0.855991 hrs Solution

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