Population growth using logarithims

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Homework Help Overview

The problem involves a bacterial culture that starts with 100,000 bacteria and grows to 125,000 in 20 minutes. The objective is to determine the doubling period of the bacteria using logarithmic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the growth of the bacteria and the equations used to model this growth. There are attempts to express the growth in terms of a rate and time, with some questioning whether the doubling period refers to time or rate.

Discussion Status

Participants are exploring different methods to solve for the doubling period, with some providing guidance on how to approach the problem. There is an ongoing examination of the equations and assumptions involved, with no explicit consensus reached on the best method yet.

Contextual Notes

There is some confusion regarding the definitions of terms such as "doubling period" and "rate," which may affect the interpretation of the problem. Participants are working within the constraints of the homework assignment, which may limit the methods they can use.

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Homework Statement



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?

Homework Equations



Current=Original(rate)^time

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.
 
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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?
 
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 can't 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
 
No problem. And it's true... roots are almost always easier than logs.
 

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