MHB What is the Initial Population of Bacteria in a Culture?

[rayne1]

Assume the population of bacteria in a culture increases at a rate proportional to the current population. The population increased by 2455 from t = 2 to t = 3 and by 4314 from t = 4 to t =5. Find the initial population and how many times does the population increase each unit of time?

I don't really understand what I'm supposed to do.

 

[MarkFL]

We know the population will take the form:

$$P(t)=P_0e^{kt}$$

We are then given:

$$P(3)-P(2)=2455$$

$$P(5)-P(4)=4314$$

This will give you 2 equations and 2 unknowns...can you proceed to find the initial population $P_0$?

 

[rayne1]

We know the population will take the form:

$$P(t)=P_0e^{kt}$$

We are then given:

$$P(3)-P(2)=2455$$

$$P(5)-P(4)=4314$$

This will give you 2 equations and 2 unknowns...can you proceed to find the initial population $P_0$?

What would be the k value?

 

[MarkFL]

What would be the k value?

This would also have to be algebraically determined. Let's look at the first equation I gave:

$$P(3)-P(2)=2455$$

This then becomes (using the definition of $P(t)$):

$$P_0e^{3k}-P_0e^{2k}=2455$$

Now, if we factor on the left, we obtain:

$$P_0e^{2k}\left(e^k-1\right)=2455$$

And solving for $P_0$ we then get:

$$P_0=\frac{2455}{e^{2k}\left(e^k-1\right)}$$

Now, if we state the second equation I gave, we have (after factoring):

$$P_0e^{4k}\left(e^k-1\right)=4314$$

At this point we may substitute for $P_0$ that we obtained above:

$$\frac{2455}{e^{2k}\left(e^k-1\right)}e^{4k}\left(e^k-1\right)=4314$$

Now you may simplify, then solve for $e^k$, and then you can determine $P_0$.

 

[rayne1]

This would also have to be algebraically determined. Let's look at the first equation I gave:

$$P(3)-P(2)=2455$$

This then becomes (using the definition of $P(t)$):

$$P_0e^{3k}-P_0e^{2k}=2455$$

Now, if we factor on the left, we obtain:

$$P_0e^{2k}\left(e^k-1\right)=2455$$

And solving for $P_0$ we then get:

$$P_0=\frac{2455}{e^{2k}\left(e^k-1\right)}$$

Now, if we state the second equation I gave, we have (after factoring):

$$P_0e^{4k}\left(e^k-1\right)=4314$$

At this point we may substitute for $P_0$ that we obtained above:

$$\frac{2455}{e^{2k}\left(e^k-1\right)}e^{4k}\left(e^k-1\right)=4314$$

Now you may simplify, then solve for $e^k$, and then you can determine $P_0$.

Oh, thank you! I got a bit lazy to try everything out.