another modeling with exponential and log functions

[pgotm]

After discontinuing all advertising for a certain product (in 1995),
the manufacturer of the product found that the sales began to drop according to the model
S=500,000
_______
1+ o.6e^kt

0 less than equal to t less than equal to 6

where S represents the number of units sold and t represents the calendar year, with t = 0 corresponding to 1995.

Find k if the company sold 300,000 units in 1997. (round to 3 decimal places)
I got 1.131 but that was marked wrong, so then i got the next part wrong also.

According to this model, what were the sales in 2000?
Sorry for asking so much help, but I'm terrible at these problems.

 

[pka]

Pgot, here is a suggestion. At the top of this web-page is a ‘Forum Help’ tab. On it are several links for TeX/LaTeX. I think proper formatting of your mathematical expressions will make things easier. Here is an example of what I mean.
\(\displaystyle \L
S(t) = \frac{{500000}}{{1 + e^{kt} }}\) now we have S(2)=300000 (t=2 in 1997); solve: \(\displaystyle \L
\frac{{500000}}{{1 + e^{2k} }} = 30000\) for k.

 

[steve_b]

After discontinuing all advertising for a certain product (in 1995),
the manufacturer of the product found that the sales began to drop according to the model
S=500,000
_______
1+ o.6e^kt

0 less than equal to t less than equal to 6

where S represents the number of units sold and t represents the calendar year, with t = 0 corresponding to 1995.

Find k if the company sold 300,000 units in 1997. (round to 3 decimal places)
I got 1.131 but that was marked wrong, so then i got the next part wrong also.

According to this model, what were the sales in 2000?

--------------------

I think "pka" shuffled your problem a bit. Let me try. Your modeling equation is:

S = 500000 / [1+.6e^(kt)]

You're given that, when t=2, s = 300000:

300000 = 500000 / [1 + .6e^(2k)]

To solve for k, first "cross multiply" to get:

1 + .6e^(2k) = 500000 / 300000

1 + .6e^(2k) = 5/3

Subtract 1 from both sides:

.6e^(2k) = 5/3 - 1

Divide both sides by .6:

e^(2k) = (5/3 - 1)/.6 ~= 1.11111

Take natural log of both sides:

2k ~= .1053605

k ~= .05268

In the year 2000, t = 5 so:

S = 500000 / [1 + .6e^(.05268 * 5)] ~= 280771

Hope that helps...

Steve