# Growth of a population of wolves

## Homework Statement

A population of wolves grows according to the function N(t)=(10N0)/(1+9e^(-t/4)), where t is in years and N0 is the initial amount. How long from t=0 will the population take to double?

## The Attempt at a Solution

Not sure if this is a completely calculus based question or not, but I threw it in here because it's listed in my calculus problems. Basically this one had stumped me, as I have no idea how to go about solving it. My first attempt was to try and find the initial population by using t=0, and then use that value to calculate the time it would take for it to double. This got me nowhere, for all I ended up with was N(t)=N0. I tried to use this by making N(t)=2N(t) and solving for t, but this only got close to the correct answer of 4ln10. Any help you guys can provide would be greatly appreciated on this question, thanks in advance

HallsofIvy
Homework Helper

## Homework Statement

A population of wolves grows according to the function N(t)=(10N0)/(1+9e^(-t/4)), where t is in years and N0 is the initial amount. How long from t=0 will the population take to double?

## The Attempt at a Solution

Not sure if this is a completely calculus based question or not, but I threw it in here because it's listed in my calculus problems. Basically this one had stumped me, as I have no idea how to go about solving it. My first attempt was to try and find the initial population by using t=0, and then use that value to calculate the time it would take for it to double. This got me nowhere, for all I ended up with was N(t)=N0.
If you set t= 0 then, of course, you got N(0)= N0. That is the initial population. So you need to solve $10N_0/(1+ 9e^{-t/4}= 2N_0$ for t.

I tried to use this by making N(t)=2N(t) and solving for t, but this only got close to the correct answer of 4ln10. Any help you guys can provide would be greatly appreciated on this question, thanks in advance
Obviously, N(t)= 2N(t) has solution N(t)= 0! So that isn't what you mean. You really mean to solve N(a+ t)= 2N(a) for t, the time it take for the population at t= a to double.
But your first idea works: divide both sides of $10N_0/(1+ 9e^{-t/4}= 2N_0$ by N0 to get $10/(1+ 9e^{-t/4}= 2$ and then $10= 2+ 18e^{-t/4}$. But we can't say what you did wrong if you don't tell us what you did!

Could you fix the latex images HallsofIvy? It says there is an error in them :tongue:

Mark44
Mentor
You need to solve N(t) = 2N0 for t.

I solved N(t)=2N0 and came up with the answer of -4ln(4/9), which is the answer I came up with before (I just had my variables mixed up), so I am still guessing I am still missing something here.

Mark44
Mentor
This is basically an exercise in algebra, not calculus, so if you're not getting the answer that's posted, then you've made a mistake or the answer in the book is wrong.

You can check both answers by plugging them in your function. A correct value of t should produce 2N0.

My answer is correct, I just plugged it in and got 2N0, so the answer sheet must be wrong for that question. Thanks for the help Mark and HallsofIvy, even if it was a more algebra based question :tongue: