Question

A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. Write the exponential equation representing this scenario modeled as a continuous growth model.

Answer 1

Answer:

$r = \frac{ln(\frac{25}{9})}{15}= 0.06811008317$

$A_o = \frac{360}{e^{5*0.06811008317}} = 256.0963179$

And our exponential model would be:

$A(t) = 256.0963179 e^{0.06811008317 t}$

Step-by-step explanation:

We want to adjust an exponential model given by this general expression:

$A(t)= A_o e^{rt}$

Where A(t) represent the number of bacteria after some t minuts

t represent the time in minutes

$A_o$ represent the initial amount of bacteria

$r$ represent the growth/decay rate

For this problem we know the following two conditions:

$A(5)= 360, A(20) = 1000$

Using the first condition we have this:

$360 = A_o e^{5r}$

We can solve for the initial amount $A_o$ and we got:

$A_o = \frac{360}{e^{5r}}$   (1)

Now using the second condition we have this:

$1000 = A_o e^{20r}$  (2)

Replacing equation (1) into (2) we have this:

$1000 =\frac{360}{e^{5r}} e^{20r} = 360 e^{15r}$   (3)

Now we can divide both sides by 360 and we got:

$\frac{1000}{360}=\frac{25}{9}= e^{15r}$

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

$ln(\frac{25}{9}) = 15r$

And solving for r we got:

$r = \frac{ln(\frac{25}{9})}{15}= 0.06811008317$

And replacing this value of r into equation (1) we got:

$A_o = \frac{360}{e^{5*0.06811008317}} = 256.0963179$

And our exponential model would be:

$A(t) = 256.0963179 e^{0.06811008317 t}$