# Homework Help: Double ln

1. Jan 25, 2009

### Gogsey

Can you help me solve this equation? Its the Gompert equation, dp/dt=cln(K/p)p.

I used substution to get -ln(u) = ct+b, where c is a constant, and bi s a constant of integration.

Next we have -ln(ln(k/p)) evaluated from pt to po = ct+b

Then ln(k/p) from pt to po = Bexp(-ct).

This is where I am stuck and don't know how to evaluate the rest of this to get an expression for pt.

2. Jan 25, 2009

### Dick

Just exponentiate again and solve for p, right?

3. Jan 25, 2009

### Gogsey

Then I get k/pt - k/po = exp(bexp(-ct), right? and then solving for pt i get

pt = k/exp(bexp(-ct) + po. But this is different from what my buddy got, and his is correct. We have to enter it online in the space provided, and it tells us if its correct once we submit it.

He got Kexp((ln(b))exp(-ct)) before he solved for b. I want to understand how he got this, not just plug the answer in without knowing.

4. Jan 25, 2009

### Dick

I would say you get k/p(t)=exp(B*exp(-ct)), or p(t)=k/exp(B*exp(-ct)), which you can also write as p(t)=k*exp(-B*exp(-ct)). That's p as a function of t. I'm not very sure what p0 and pt are supposed to be.

5. Jan 25, 2009

### Gogsey

Don't worry about it. This was initially what I had, but chaged it for some reason, thinking I had to evaluate the integral between pt and po, instead of pt and 0, lol. po is the IVP.

So I actually had the same answer as you got except I left the exp function in the denominator, and left is a a positive exp.

Thanks