Question

the number of bacteria in a culture is increasing according to the law of exponential growth.there are 125 bacteria in the culture after 2 hours and 350 bacteria after 4 hours.1)find the initial population2)write an exponential growth model for the bacteria population.Let t represent time in hours3)use the model to determine the number of bacteria after 8 hours4)after how many hours will the bacteria count be 25,000?

Answer 1

*Answers below*

Explanation:

Answer 2

1)
`P_o = (125)/(e^(2*0.5148097086))= 44.643 \approx 45`

2)
`P(t) = 44.643 e^(0.5148097086 t)`

3)
`P(t=8) = 44.643 e^(0.5148097086*8)=2744.009 \approx 2744`

4)
`ln(559.998) =0.5148097086 t`

`t = (ln(559.998))/(0.5148097086)=12.291 hours`

Explanation:

The exponential model on this case is given by the following formula:

`P(t) = P_o e^(kt)`

Where P represent the population, k the growth/decay constant and t the time in hours for this case.

`P_o` represent the initial population

We have some initial conditions given:

`P(2) = 125, P(4) = 350`

Part 1

From the initial conditions we have the following equations:

`125 = P_o e^(2k)` (1)

`350 = P_o e^(4k)` (2)

We can solve for
`P_o` from equation (1) like this:

`P_o = (125)/(e^(2k))`

And we can replace this into equation (2) and we got:

`350 =(125)/(e^(2k)) e^(4k) = 125 e^(2k)`

And we can divide both sides by 125:

`(14)/(5) = e^(2k)`

Now we can apply natural log on both sides and we got:

`ln ((14)/(5)) = 2k`

`k = (ln(14/5))/(2)=0.5148097086`

And now since we have the value of k we can solve for
`P_o` like this:

`P_o = (125)/(e^(2*0.5148097086))= 44.643 \approx 45`

Part 2

For this case the exponential model is given by:

`P(t) = 44.643 e^(0.5148097086 t)`

Part 3

For this case we just need to replace t=8 and see what we got:

`P(t=8) = 44.643 e^(0.5148097086*8)=2744.009 \approx 2744`

Part 4

For this case we want to solve this:

`25000 = 44.643 e^(0.5148097086 t)`

We can divide both sides by 44.643 and we got:

`559.998 = e^(0.5148097086 t)`

Now we can apply natural logs on both sides:

`ln(559.998) =0.5148097086 t`

`t = (ln(559.998))/(0.5148097086)=12.291 hours`