# Logarithm's and Such

1. Oct 10, 2006

### thomasrules

There are initially 200 bacteria in a culture. After five minutes, the population has grown to 4080 bacteria. Estimate the doubling period.

I DID THIS:

$$4080=200(k)^5\\\frac{4080}{200}=k^5\\k=1.82$$

IS THIS CORRECT?

Last edited: Oct 10, 2006
2. Oct 10, 2006

### Hootenanny

Staff Emeritus
I'm not quite sure what your doing here, but when solving exponential growth problems one should generally start with a differential equation such as;

$$\frac{dN}{dt} = kN$$

Where N is the number of bacteria. Do you know the solution to this differential?

3. Oct 10, 2006

### thomasrules

no your using derivative crap...not that way

4. Oct 10, 2006

### Hootenanny

Staff Emeritus
How do you propose to solve it?

5. Oct 10, 2006

### thomasrules

damnit how do u start a new line in latex

6. Oct 10, 2006

### Hootenanny

Staff Emeritus
Type a double backslash like this: \\

7. Oct 10, 2006

### thomasrules

I DID!!:

[.tex]4080=200(k)^5\\\frac{4080}{200}=k^5\\k=1.82[.tex]

i'm aware of the .

8. Oct 10, 2006

### thomasrules

ok:
4080=200(k)^5

4080/200}=k^5

k=1.82

9. Oct 10, 2006

### Hootenanny

Staff Emeritus
This is not correct.

10. Oct 10, 2006

$$\frac{dN}{dt} = kN$$. The solution of this equation is $$N_{0}e^{kt}$$. We know that $$N(0) = 200$$. So $$N= 200e^{kt}$$.

$$4080 = 200e^{5k}$$.

$$k = .6031$$

To find the doubling time, look at $$t = 1$$.

Last edited: Oct 10, 2006
11. Oct 11, 2006

### edavey8205

Y look at 1 minute. It's not answering the question for doubling time. If you are starting with 200, then double would be 400. and the question is what is the time for that. so it would be 400=200e^(0.6031*t). solve for t.

12. Oct 11, 2006

If T is the time (in minutes) to double, then the population would be multiplied by 2 every T minutes: in t minutes, you will have doubled t/T times: $P= P_02^{\frac{t}{T}}$. If, after 5 minutes the population has increased from 200 to 4080, you want to find T such that
$$2^{\frac5}{T}}= 4080/200= 20.4$$