# Thread: Insect population growth problem

1. ## 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.

2. 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.

3. Im not sure I understand, Where do I substitute those numbers to?

4. 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?

5. 1600=Poe^7r?

6. Originally Posted by lucky13
1600=Poe^7r?
Yep, now solve your equation simultaneously with

$400 = P_0e^{4r}$

7. Solve it simultaneusly?

8. 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...

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

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...

$P_{10} = P_0 e^{10r}$
$P_{10} =50 \cdot 2^{1/3}e^{10(\frac{2}{3}\ln(2))}$

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

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

$=50 \cdot 2^7$

$=6400$