# Population Growth question

1. Oct 21, 2009

### Slimsta

1. The problem statement, all variables and given/known data
http://img14.imageshack.us/img14/1286/99892583.jpg [Broken]

2. Relevant equations
in the picture

3. The attempt at a solution
so i know that
p(0)e^2k = 950
p(0)e^8k = 9500

how do i find k?
for p(0) i get 440.950939 is that right?

btw. ignore the answers in the picture

Last edited by a moderator: May 4, 2017
2. Oct 21, 2009

### lanedance

how do you find p(0) without knowing k?

try dividing you 2 equations together, then use logs

3. Oct 21, 2009

### Slimsta

what i did is made both equations equal to k
so
k1 = [ln(950/p(0))] / 2
k2 = [ln(9500/p(0))] / 8

k1 = k2
[ln(950/p(0))] / 2 = [ln(9500/p(0))] / 8
solving for p(0) i get 440.9509 which lets me plug it in the main equation and get the k value..
but looks like its wrong eh?

4. Oct 21, 2009

### lanedance

so dividing the 2 equations gives
$$e^{6k} = 10$$

can you solve for k?

you should get some constant a such that
$$k = \frac{1}{a} ln(10)$$

the equation then becomes
$$p(t) = p(0).e^{kt} = p(0).e^{(t/a) ln(10)} = p(0).e^{ln(10^{t/a}} = p(0) .10^{t/a}$$
$$p(0) = p(2).10^{-2/a}$$

which gives me the same p(0) value as you, so your way was fine

5. Oct 21, 2009

### Slimsta

oh so i just did it the long way... kk that makes sense.
now how do i find the growth rate after 5 hours?
k = 0.38376418 so growth rate = 38.376%

6. Oct 21, 2009

### lanedance

i would take the growth rate to mean p'(t)

7. Oct 22, 2009

### Slimsta

oh i got it!! tnx man

Last edited: Oct 22, 2009