# Growth Rate

1. Dec 6, 2008

### stegz

1. The problem statement, all variables and given/known data
A beaker contained 2000 bacteira. one hour later the beaker contained 2500 bacteria. What is the doubling time of the bacteria?

2. Relevant equations
rate = (distance)/(time)
Time to double = .693/((ln(1+r))^t)

3. The attempt at a solution
rate = 2500/2000
My biggest problem is trying to find the rate, I used this at first, but think it is giving me the wrong answer. I know how to finish the problem, i just need help finding the rate. Thanks!

2. Dec 6, 2008

### jmarcian

i would assume exponential growth rate here... use:

P=Ae^(rt)
A is initial quantity, P is quantity at time t, t is time of course, and r is rate

in the first part of the equation they give you enough info to solve for r, (using ln's which i assume you know how to do), now that you know r you have enough information to some for your solution. (it may not look like at first, but just think what happens after the initial population DOUBLES......)

good luck,
jared

3. Dec 7, 2008

### HallsofIvy

Staff Emeritus
"Rate= distance/time"? There is no "distance" here- that's not what "rate" means here!

If the bacteria double in T hours, you multiply by 2 for every "T" in the time: that is, for t hours, you multiply by 2 t/T times: $B(t)= B_0 2^{t/T}$. you are told that B(0)= 2000 and B(1)= 2500. That gives you two equations to solve for $B_0$ and T.