A bacteria culture started with 5500 bacteria and increased

A bacteria culture started with 5500 bacteria and increased by 35% in the first 2 hours. (a) Find an expression for the number of bacteria t hours after it was first observed. Q(t) = 5500e( ln(1.35) 2)t

b) Find the number of bacteria that will be present after 8 hours. (Round your answer to the nearest whole number.)

(c) When will the population reach 35,000? (Round your answer to one decimal place.) hr (d) How long does it take the population to triple in size? (Round your answer to one decimal place.) hr

#2

A bacteria culture started with 5500 bacteria and increased by 35% in the first 2 hours.
(a) Find an expression for the number of bacteria t hours after it was first observed.

(b) Find the number of bacteria that will be present after 8 hours. (Round your answer to the nearest whole number.)

(c) When will the population reach 35,000? (Round your answer to one decimal place.)
hr

(d) How long does it take the population to triple in size? (Round your answer to one decimal place.)
hr

Solution

Initial population of bacteria=5500

After 2 hours it increased by 35%

And 35% of 5500= 1925

Therefore after 2 hours,polulation= 7425

We have to use the following formula here

N(t)= N(0) e^kt

7425=5500 e^2t

7425/5500= e^2t

1.35= e^2tTaking ln on both sides

t=ln1.35/2

Therefore the required function is N(t) = 5500e^{(ln(1.35)/2)t}

b. t= 8 hrs

N(8)= 5500e^{(ln(1.35)/2)8} = 5500 e^4ln(1.35) = 18268

c. 35000= 5500 e^{(ln(1.35)/2)t}

   35000/5500= e^{(ln(1.35)/2)t}

Taking ln on both sides

ln(35000/5500)=t* ln(1.35)/2

2 ln(35000/5500)= t ln 1.35

3.7 = t* ln 1.35

t= 12.3 hrs

d. Initial population =5500

Triple of this population = 3*5500= 16500

16500=5500e^{(ln(1.35)/2)t}

3= e^{(ln(1.35)/2)t}

Taking ln on both sides

ln 3= t* ln(1.35)/2

2 ln 3/ln1.35 = t

t=7.3 hrs