Homework Help: 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

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

Show that $\lim_{n\to\infty}(1+\frac{r}{n})^n=e^r$