# 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:

$$4080=200(k)^5 4080/200=k^5 k=1.82$$

IS THIS CORRECT?

2. Oct 10, 2006

### edavey8205

what equation are you using? Exponential growth is N=Ni*e^(k*t). It doen't look like you did that. And simply solving the equation is not answering estimate the doubling period. once you solve for k you will need to then solve for t when the bacteria population is 400.

3. Oct 10, 2006

### thomasrules

ok:
4080=200(k)^5

4080/200}=k^5

k=1.82

I"M USING N=k(a)^x

4. Oct 11, 2006

### OlderDan

You can do it your way, but you actually used (k) in place of (a), or N = a(k)^x. Your k is just a bit off. Check it again. Now you need to find x that satisfies

400 = 200(k)^x

It doesn't have to be 400 and 200. All that is required is that the ratio be 2.

5. Oct 11, 2006

$$N = N_{0}e^{kt}$$, $$N(0) = 200$$.

$$N = 200e^{kt}$$

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

$$k \doteq 0.603$$

doubling time = 1.149 mins

Last edited: Oct 11, 2006