1. ## Population models.

So working on a exp growth word problem. I see the function should be P(t) = P0e^kt

however they give me a function in the problem already

f(t) = 100,000/(1+5000e^-t)

so when solving should I just use the function they gave me instead of the P(t) = ?

they want initial start of population, population after 4 weeks, and how many weeks to reach 40,000

I get this for the first

1) f(t) = 100,000/(1+5000e^0) = 100,000/5001
2) f(t) = 100,000/(1+5000e^4) = correct form to put into calcualtor?
3) this is where we need to solve for t correct?

im completly lost on how to solve for t. I know 1^-x = 1/x..but will this work for e^-1 or should I just move it in front. blah :/

2. ## Re: Population models.

Originally Posted by RYdis
So working on a exp growth word problem. I see the function should be P(t) = P0e^kt

however they give me a function in the problem already

f(t) = 100,000/(1+5000e^-t)

so when solving should I just use the function they gave me instead of the P(t) = ?

they want initial start of population, population after 4 weeks, and how many weeks to reach 40,000

I get this for the first

1) f(t) = 100,000/(1+5000e^0) = 100,000/5001
2) f(t) = 100,000/(1+5000e^4) = correct form to put into calcualtor?
3) this is where we need to solve for t correct?

im completly lost on how to solve for t. I know 1^-x = 1/x..but will this work for e^-1 or should I just move it in front. blah :/
So you have $\frac{100000}{(1+5000e^{-t})}$

For 1), you are correct...t=0
For 2), it should be $e^{-4}$, not $e^4$

$e^{-t}$ = $\frac{1}{e^t}$, so for 3), set f(t) = 40,000, cross multiply and after you rearrange the data you should get $e^{-t} = \frac{1.5}{5000}$. Can you solve for t from there?

3. ## Re: Population models.

Originally Posted by Youkla
So you have $\frac{100000}{(1+5000e^{-t})}$

For 1), you are correct...t=0
For 2), it should be $e^{-4}$, not $e^4$

$e^{-t}$ = $\frac{1}{e^t}$, so for 3), set f(t) = 40,000, cross multiply and after you rearrange the data you should get $e^{-t} = \frac{1.5}{5000}$. Can you solve for t from there?
not with any success yet :/

I was able to come up with the same number after some one . but 1.5/5000 doesn't reallly give me a time frame for the answer. Answer 1 we were given t to solve for population, same with 2. 3 we are given population and told to find time. The first was intial time. So =0. Second was after 4 weeks. So t=4. However i can't see to use this information we have to get a week time frame. Maybe im missing something really obvious..sorry :/

so I did e^-t = 1.5/5000 and changed to 1/e^t = 1/1.5/5000 = 5000/1.5 gives me around 3,333. It just dont seem like the correct answer :/

4. ## Re: Population models.

So we have this:
$e^{-t} = \frac{1.5}{5000}$

This can then be re-written as
$\frac{1}{e^t} = \frac{1.5}{5000}$

From here we can then take the reciprocals of each and get
$e^t = \frac{5000}{1.5}$

Then to solve for t you take the natural log of both sides (have you learned this?). So you get
$\ln({e^t}) = \ln({\frac{5000}{1.5}})$

then

$t = \ln({\frac{5000}{1.5}})\approx{8.1}$ weeks.