1. ## Exponential function

I'm trying to solve the following problem...

Three hundred students attended the dedication ceremony of a new building on a college campus. The president of the traditionally female college announced a new expansion program, which included plans to make the college co-educational. The number of students who learned of the new program 't' hours later is given by the function...

$f(t) = 3000 / (1 + Be^{-kt})$

If 600 students on campus had heard about the new program 2 hours after the ceremony, how many students had heard about the policy after 4 hours?

I've tried to tackle this two ways - by solving for b, and solving for k.

I'm assuming here that $f(0) = 3000 / (1 + Be^{-k[0]}) = 300$ and that $f(2) = 3000 / (1 + Be^{-2k}) = 600$

If f(0) = 300, then 3000 / 10 = 300. Therefore, '1 + B e^-k(0)' has to equal 10. This would mean that B = 9 (as e^0 = 1).

If I use this value of b in the equation $f(2) = 3000 / (1 + 9e^{-2k}) = 600$ then try to solve for k.

$3000 / 600 = 1 + 9e^{-2k}
= 4 = 9e^{-2k}
= ln 4/9 = -2k
= ln 4/9(-2) = k
k = 1.622$

If I use this value of k in f(2) = 3000 / 1 + 9e^{-2*1.622} I get 2,220, nowhere near the 600 I should be getting.

Instead, I tried to solve for B and k using f(2) = 600.

$f(2) = 3000 / 1 + Be^{-kt} = 600$

So now I'm assuming that '1 + Be^(-kt) is equal to 5. One of the problems I'm having is to figure out how I can find out the value of B and k in that value of 5 (or 4, if you take into account the + 1 at the start)

2. ## Re: Exponential function

Originally Posted by astuart
I'm trying to solve the following problem...

Three hundred students attended the dedication ceremony of a new building on a college campus. The president of the traditionally female college announced a new expansion program, which included plans to make the college co-educational. The number of students who learned of the new program 't' hours later is given by the function...

$f(t) = 3000 / (1 + Be^{-kt})$

If 600 students on campus had heard about the new program 2 hours after the ceremony, how many students had heard about the policy after 4 hours?

I've tried to tackle this two ways - by solving for b, and solving for k.

I'm assuming here that $f(0) = 3000 / (1 + Be^{-k[0]}) = 300$ and that $f(2) = 3000 / (1 + Be^{-2k}) = 600$

If f(0) = 300, then 3000 / 10 = 300. Therefore, '1 + B e^-k(0)' has to equal 10. This would mean that B = 9 (as e^0 = 1).

However, if I use 9 as B, when i plug that into the original equation and try and solve f(4), I get the wrong answer - about 2848, when I should be getting 1080.

I then tried to solve f(2) = 600, then solve for B and k.

$f(2) = 3000 / (1 + Be^{-k[2]}) = 600$

I reach an issue where using 9 for B (As in the original equation) doesn't work, and I can't figure out how to isolate one variable without knowing the other..
How can you get an answer for f(4) when you haven't evaluated k yet?

3. ## Re: Exponential function

Originally Posted by Prove It
How can you get an answer for f(4) when you haven't evaluated k yet?
Whoops, that's an error. I'll fix up the original post..

4. ## Re: Exponential function

Fixed that up, sorry about that..

Anybody?

6. ## Re: Exponential function

Surely \displaystyle \begin{align*} -2k = \ln{\left(\frac{4}{9}\right)} \end{align*} means that \displaystyle \begin{align*} k = -\frac{1}{2}\ln{\left(\frac{4}{9}\right)} \end{align*}...

7. ## Re: Exponential function

Originally Posted by Prove It
Surely \displaystyle \begin{align*} -2k = \ln{\left(\frac{4}{9}\right)} \end{align*} means that \displaystyle \begin{align*} k = -\frac{1}{2}\ln{\left(\frac{4}{9}\right)} \end{align*}...
Whoops. That was meant to be ln (4/9) / -2 (or * 1/2). Don't know how I didn't pick that up this morning - I can understand how I missed it last night.. Really need to double check calculations better.

Thanks!