View Full Version : Population growth using logarithims
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.
Char. Limit
Jan25-10, 04:07 PM
Well, yeah. The doubling period is time, and it's supposed to be the thing you're solving for.
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?
Char. Limit
Jan25-10, 04:13 PM
Well, start with this:
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)
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.
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
Char. Limit
Jan25-10, 09:57 PM
No problem. And it's true... roots are almost always easier than logs.
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