Calculating Doubling Time of Bacteria in Beaker

[stegz]

1. Homework Statement
A beaker contained 2000 bacteira. one hour later the beaker contained 2500 bacteria. What is the doubling time of the bacteria?


2. Homework 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!

 

[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

 

[HallsofIvy]

"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.