# 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 $\displaystyle P(t) = P_0e^{rt}$ and find $\displaystyle P_0$ and $\displaystyle r$ via substitution of $\displaystyle (4,400)$ and $\displaystyle (7,1600)$

After this find $\displaystyle P(10)$ 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 $\displaystyle 400 = P_0e^{4r}$

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

$\displaystyle 400 = P_0e^{4r}$

7. Solve it simultaneusly?

8. Originally Posted by lucky13
Solve it simultaneusly?
You have...

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

Divide the top equation by the bottom one.

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

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

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

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

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

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

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

Then solve...

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

You have...

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

Divide the top equation by the bottom one.

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

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

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

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

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

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

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

Then solve...

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

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

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

$\displaystyle =50 \cdot 2^7$

$\displaystyle =6400$