# Population growth using logarithims

1. Jan 25, 2010

### trials

1. The problem statement, all variables and given/known data

A culture begins with 100,000 bacteria and grows to 125,000 bacteria after 20 min. What is the doubling period to the nearest minute?

2. Relevant equations

Current=Original(rate)^time

3. The attempt at a solution

I can get make the first part out. 125000=100000(rate)^2 I have a feeling its wrong but its as far as i can get.

2. Jan 25, 2010

### Char. Limit

Well, yeah. The doubling period is time, and it's supposed to be the thing you're solving for.

3. Jan 25, 2010

### trials

So would my rate be 2? if so then would this be right 120000=100000(2)^t
1.25=2^t
log1.25/log2=t?

What confuses me is when it says doubling period, is that time or the rate?

4. Jan 25, 2010

### Char. Limit

The original is 100,000 and it changes to 125,000 in 20 minutes. So, first solve for the rate, then put in the rate to the equation 200,000=100,000(rate)^(time)

5. Jan 25, 2010

### trials

Well i can get this far but i cant get farther sorry.
125000=100000(rate)^20
125000/100000=r^20
log (125000/100000)=20log r

I don't know how to go farther, if r had a value and i was solving for time i would have no problem with this.

6. Jan 25, 2010

### trials

alrighhhht worked it out on my own :)

I figured, why log to find the rate, 20th root it.

125000/100000=r^20
(20th root) 1.25=r
r=1.01

200000=100000(1.01)^t
2=1.01^t
log 2/log1.01=t
t=62 minutes which is the answer in the back of my textbook. Thanks for setting me on the right path Char.Limit :D

7. Jan 25, 2010

### Char. Limit

No problem. And it's true... roots are almost always easier than logs.