Question

A bacteria culture grows with constant relative growth rate. The bacteria count was 2,560 after 4 hours and 2,621,440 after 12 hours.
(a) What is the relative growth rate? Express your answer as a percentage. (Round your answer to two decimal places.)
% per hr
(b) What was the initial size of the culture?
bacteria
(c) Find an expression for the exact number of bacteria after t hours, (t).
y(t) =
(d) Find the number of bacteria after 5.5 hours. (Round your answer to the nearest whole number.)
bacteria
(e) Find the rate of growth in bacteria per hour) after 5.5 hours. (Round your answer to the nearest whole number.)
bacteria per hour
(f) How many hours did it take for the population to reach 104,000? (Round your answer to two decimal places.)
hr

Answer 1

Answer:

A. 86.64%

B. 80

C. y(t)= 80$e^{0.8664t}$

D. y(5.5) = 9388

E. y(t) = 8134 bacteria for every hour

F.  8.28 hours

Step-by-step explanation:

a. The bacteria count after 4 hours y(4) = 2560

The bacteria count after 12 hours y(12) = 2621440

equation 1: ye^4k = 2560

equation 2: ye^12k= 2621440

divide (2) by (1),

e^8k = 1024

Taking ln both sides:

ln(e8k ) = ln(1024)

8k = 6.931

k=0.8664      // after dividing both sides by 8

rate of growth k% = 0.8664*100 = 86.64%

b. substituting for k in equation 2:

   ye^4*0.8664 = 2560

   Initial bacteria count y(4) = 80

c. y(t)= 80e^0.8664t

d. the number of bacteria after 5.5 hours => y(5.5) = 80$e^{0.8664(5.5)}$ ≈ 9388

e. y'(5.5) = 80*0.8664e^0.8664(5.5) = 69.31e^0.8664(5.5) = 8134

f. for bacteria population = 104000

where ye^0.8664t = 104000

   y = 80

   80e^0.8664t = 104000    //divide both sides by 80

   e^0.8664t = 1300

   ln(e^0.8664t = ln(1300)

   t = 7.17 / 0.8664 = 8.28 hours