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

Answer:

1) $P_o = \frac{125}{e^{2*0.5148097086}}= 44.643 \approx 45$

2) $P(t) = 44.643 e^{0.5148097086 t}$

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

4) $ln(559.998) =0.5148097086 t$

$t = \frac{ln(559.998)}{0.5148097086}=12.291 hours$

Step-by-step 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:

$P_o = \frac{125}{e^{2k}}$

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

$350 =\frac{125}{e^{2k}} e^{4k} = 125 e^{2k}$

And we can divide both sides by 125:

$\frac{14}{5} = e^{2k}$

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

$ln (\frac{14}{5}) = 2k$

$k = \frac{ln(14/5)}{2}=0.5148097086$

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

$P_o = \frac{125}{e^{2*0.5148097086}}= 44.643 \approx 45$

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:

$ln(559.998) =0.5148097086 t$

$t = \frac{ln(559.998)}{0.5148097086}=12.291 hours$