# Expononential Growth

1. Oct 2, 2007

### steven10137

1. The problem statement, all variables and given/known data
A bacterial population size N is known to be growing exponentially. If the population triples between noon and 2pm, at what time will N be 100 times the noon population.

2. Relevant equations
Firstly. Is this a distribution function??
If so; f(t)=ue^ut
where E(t) = 1/u

3. The attempt at a solution
I have no idea where to start ...
Perhaps;
t=2 -> 3 times initial population (N)
I have no idea ...

2. Oct 3, 2007

### Dick

This is an exponential growth problem. pop(t)=P0*exp(k*t) where P0 is the initial population and k is the growth constant with t=0 being noon. If t=2hr then pop(t)=3*P0. Can you find k? Once you've found k, can you say at what value of t is pop(t)=100*P0?

3. Oct 3, 2007

### steven10137

@ t=2 -> 3Po=Poe^kt
therefore ln3=2k and k=0.5493

P(t)=Poe^0.5493t
@ what time is P(t)=100Po
100Po=Poe^0.5493t
ln100=0.5493t
therefore t=8.38
or approxiamtely 8:23pm

Steven

4. Oct 3, 2007

### HallsofIvy

Staff Emeritus
Remember that all exponentials are equivalent: $a^x= e^{x ln(a)}$ so the only difference is a coefficient.
Since you are told that "the population triples between noon and 2pm", that is that it triples every 2 hours, it is much easier to use 3t/2 where t is in hours. Since there were initially N bacteria, P(t)= N(3t/2)= 100N. Solving 3t/2= 100, (t/2)ln(3)= ln(100) so t= 2ln(100)/ln(3) which gives exactly the answer you got. Of course, you could also use common logs to solve the equation.

5. Oct 3, 2007

### steven10137

good thinking HallsofIvy!

makes perfect sense
cheers
Steven