# Solving the Logistic Model: Find P(66) with P(0)=3

• glid02
In summary, the conversation is about solving a logistic model equation to find the population at a given time. The conversation includes discussions about integrating the equation, separating variables, and using logarithmic functions. The expert summarizer provides a step-by-step summary of the conversation, highlighting key points such as finding the constant and collecting logarithmic terms before taking the exponential.

#### glid02

Here is the first question:
A population P obeys the logistic model. It satisfies the equation
https://webwork.math.uga.edu/webwork2_files/tmp/equations/11/02d0a645c053f1f6002d746c78143f1.png

Assume that P(0)=3. Find P(66)

First I multiplied both sides by dt and integrated, giving:
P=6/700Pt(7-P)+c
If P(0)=3 then c=3
P=6/700Pt(7-P)+3

Then I divided everything by P and had
1=6/700t(7/P-1)+3/P

Now to find P(66)
1=6/700*66(7/P-1)+3/P
1=396/700(7/P-1)+3/P
1=2772/700P-396/700+3/P
4872/700P=1096/700
P=4.445

That's not right, what am I missing?
Thanks.

Last edited by a moderator:
Your first equation is not visible. Perhaps you could rewrite it?

You have to accept some sort of web certificate to view the equation. It seems to be located on some university's website. Heres the equation that I get:

$$\frac{dP}{dt} = \frac {6}{700}P(7-P)$$

ranger said:
You have to accept some sort of web certificate to view the equation. It seems to be located on some university's website. Heres the equation that I get:

$$\frac{dP}{dt} = \frac {6}{700}P(7-P)$$

Ahh, ok, thanks for that, ranger. I must have clicked no automatically!

glid02 said:
Here is the first question:
A population P obeys the logistic model. It satisfies the equation
https://webwork.math.uga.edu/webwork2_files/tmp/equations/11/02d0a645c053f1f6002d746c78143f1.png

Assume that P(0)=3. Find P(66)

First I multiplied both sides by dt and integrated, giving:
P=6/700Pt(7-P)+c
If P(0)=3 then c=3
P=6/700Pt(7-P)+3

Then I divided everything by P and had
1=6/700t(7/P-1)+3/P

Now to find P(66)
1=6/700*66(7/P-1)+3/P
1=396/700(7/P-1)+3/P
1=2772/700P-396/700+3/P
4872/700P=1096/700
P=4.445

That's not right, what am I missing?
Thanks.

You have this equation: $$\frac{dP}{dt} = \frac {6}{700}P(7-P)$$. You cannot simply multiply by dt and integrate, since you have not integrated the terms including P wrt P! You must rearrange the equation to give: $$\int \frac{dP}{P(7-P)}=\int\frac{6}{700}dt +C$$

Do you know how to solve this?

Last edited by a moderator:
You did put +C accidentally right?

Last edited:
Yeah, I can solve that. Didn't think to separate variables for some retarded reason. Thanks for the help.

ranger said:
You did put +C accidentally right?

Yea, I guess I haven't really integrated anything yet, so strictly the constant doesn't appear until the next line!

glid02 said:
Yeah, I can solve that. Didn't think to separate variables for some retarded reason. Thanks for the help.

You're welcome!

OK I lied, I'm still not getting the right answer.

dP/P(7-P)=6/700dt
1/7log(P)-1/7log(7-P)=6t/700+c
log(P)-log(7-P)=6t/100+c
Using e
P-7+P=e^(6t/100)+c
P=(e^(6t/100)+7)/2+c
P(0)=3 so c=-1
Subbing 66 for t, I get 28.729

Still not right, what am I doing wrong now?
Thanks again.

glid02 said:
OK I lied, I'm still not getting the right answer.

dP/P(7-P)=6/700dt
1/7log(P)-1/7log(7-P)=6t/700+c
log(P)- log(7-P)=6t/100+c
Using e
P-7+P=e^(6t/100)+c
What you've done here is wrong. You must collect the logarithmic terms before you can take the exponential of both sides.

You lost me with the collecting. Can you give me another example?

Well, have you come across the general rule: log(a)+log(b)=log(ab) ?

If I had I'd forgotten it. I should be able to solve from here (again). Thanks again for helping.