Calculus Exponential Growth Function Problem. How can I start this?

Question

A bacteria culture initially contains 581 bacteria and doubles every half hour.

Find the size of the baterial population after 40 minutes.

Find the size of the baterial population after 8 hours.

Can anyone provide a step-by-step solution to this problem?

Answer 1

suppose that no.of bacteria initially is x0=581when t=0and it is x at time t(in hours),then

             dx/dt is proportional to x

             dx/dt=kx .........(1)

on integrating,

              log x=kt+c

              x=e^kt+c

              x=ce^kt(take e^c as c) ........(2)

              when t=0,x=x0=581

              thus,xo=c=581

          (2) becomes,

                       x=581e^kt ...........(3)

given: it doubles every half hour

           thus,x=2x0,when t = 1/2

            so,x=4x0, when t=1

    (3) becomes,  

                     4*581=581e^k

                  so,e^k=4

now,let x=x1,when t=40 min=2/3 hrs

                   so,x1=581e^2/3k

                      x1=581e^k^2/3

                      x1=581(4)^2/3     (since e^k=4)

thus,size of bacterial population after 40 min is (4)^2/3 times 581,i.e,

                                x1=1643.068

similarly.now ,x=x2,when t=8

                     x2=581e^8k

                      x2=581(e^k)^8

                      x2=581(4)^8    (since e^k=4)

                    thus,x2=65536

Answer 2

A=Åe^(kt)

where Å = initial value of A

t in hrs

thus from the problems, we are given "initially contains 581 bacteria"

Å = 581

k will be positive since the bacteria doubles

thus find k:

2=e^(k(0.5)) - [this is due to the particle double, so we know the left hand side will be double Å]

k = (ln2) / 0.5

k ~ 1.3863

thus the overall eqn will be A=581e^(1.3863t)

(i) after 40 min = 2/3hrs: A=581e^(1.3863(2/3))

A = 1464.0283

(i) after 8hrs

A=581e^(1.3863(8))

A = 3.8076* 10^(7)