# Insect population growth problem

• Apr 23rd 2010, 01:03 PM
lucky13
Insect population growth problem
An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.
• Apr 23rd 2010, 01:52 PM
pickslides
Quote:

Originally Posted by lucky13
An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.

Use the model $P(t) = P_0e^{rt}$ and find $P_0$ and $r$ via substitution of $(4,400)$ and $(7,1600)$

After this find $P(10)$ and you will have the answer.
• Apr 23rd 2010, 02:09 PM
lucky13
Im not sure I understand, Where do I substitute those numbers to?
• Apr 23rd 2010, 02:29 PM
pickslides
Quote:

Originally Posted by lucky13
Im not sure I understand, Where do I substitute those numbers to?

In this case (4,400) = (t,P(t))

We get $400 = P_0e^{4r}$

Do the same for the other point.
What do you get?
• Apr 23rd 2010, 02:51 PM
lucky13
1600=Poe^7r?
• Apr 23rd 2010, 02:59 PM
pickslides
Quote:

Originally Posted by lucky13
1600=Poe^7r?

Yep, now solve your equation simultaneously with

$400 = P_0e^{4r}$
• Apr 23rd 2010, 03:14 PM
lucky13
Solve it simultaneusly?
• Apr 23rd 2010, 04:06 PM
Quote:

Originally Posted by lucky13
Solve it simultaneusly?

You have...

$P_0 e^{7r} = 1600$
$P_0 e^{4r} = 400$

Divide the top equation by the bottom one.

$\frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}$

=> $e^{(7-4)r} = 4$

=> $r = \frac{2}{3}\ln(2)$

Substitute that into $P_0 e^{4r} = 400$ to get...

$P_0 \cdot 4 \cdot 2^{2/3} = 400$

=> $P_0 = 50 \cdot 2^{1/3}$

You can simplify stuff if ya like, I've left it all in a kind of messy form.

Then solve...

$P_{10} = P_0 e^{10r}$
• Apr 23rd 2010, 05:14 PM
ione
Quote:

You have...

$P_0 e^{7r} = 1600$
$P_0 e^{4r} = 400$

Divide the top equation by the bottom one.

$\frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}$

=> $e^{(7-4)r} = 4$

=> $r = \frac{2}{3}\ln(2)$

Substitute that into $P_0 e^{4r} = 400$ to get...

$P_0 \cdot 4 \cdot 2^{2/3} = 400$

=> $P_0 = 50 \cdot 2^{1/3}$

You can simplify stuff if ya like, I've left it all in a kind of messy form.

Then solve...

$P_{10} = P_0 e^{10r}$

$P_{10} =50 \cdot 2^{1/3}e^{10(\frac{2}{3}\ln(2))}$

$=50 \cdot 2^{1/3}(e^{ln(2)})^{\frac{20}{3}}$

$=50 \cdot 2^{1/3}(2)^{\frac{20}{3}}$

$=50 \cdot 2^7$

$=6400$