A bacteria culture grows with a constant relative growth rate. After 2 hours there are 600 bacteria and after 8 hours the count is 75,000.

Question

(a) Find the initial population.

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

(c) Find the number of cells after 7 hours. (Round your answer to the nearest integer.)

(d) Find the rate of growth after 7 hours. (Round your answer to the nearest integer.)

(e) When will the population reach 200,000? (Round your answer to one decimal place.)

Answer 1

This is an exponential growth model.

The number of bacteria P after t hours: P(t) = P₀ e^kt

Find the constants from the given information:

P(2) = P₀ e^2k = 600

P(8) = P₀ e^8k = 75000

Divide the second equation by the first, to cancel out P₀:

e^8k/e^2k = 75000/600

e^6k = 125

k = (1/6)ln125 = 0.804719

Find P₀ from the first equation:

P₀ = 600/e^2k = 600/e^2*0.804719 = 120

So your exponential growth model is: P(t) = 120e^0.804719t

(c) Evaluate P(7)

(d) Evaluate P ' (7)

(e) Solve the exponential equation:

P(t) = 120e^0.804719t = 200,000

The answer t will be in hours.

Answer 2

2 hours 600

8 hours 75000

A= Pe^rt

600=Pe^2r

75000=Pe^8r

75000/600 = e^6r

750/6 = e^6r

125 = e^6r

ln125 = 6r

r = (ln125)/6

600=Pe^(ln125)/3

P = 600/e^(ln125)/3 = 120

P = 120= initial population

A after t hours

= 120e^rt

= 120e^t(ln125)/3

after 7 hours

A = 120e^7(ln125)/3

= 9375

r= (ln125)/6 = about 0.8045

= about 2 rounded to nearest integer

or = about 161%

A = 200,000 when

200,000 = Pe^rt = 120e^.8045t solve for t

200000/120 = e^.8045t

20,000/12 = 10,000/6 = 5000/3 = e^.8045t

ln(5000/3) = .8045t

t = [ln(5000/3)]/.8045 =9.22 hours to reach 200,000