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woodyp20 · Feb 16, 2011, 11:54 PM

The sales of a new model of notebook computer are approximated by: s(x)
= 5000-13000e^-x/9 , where x represents the number of months the computer has been on the market and S represents sales in thousands of dollars. In how many months will the sales reach $2,600,000?

ebaines · Feb 17, 2011, 07:16 AM

I assume you mean this formula:

$ S(x) = 5000-13000 e^{-x/9} $

You need to rearrange this to get x by itself. One concept to remember is that the inverse of the exponential function e^x is the natural logarithm, ln(x). Thus if A = e^b, then b = ln A. Other than that little trick the rest is straightforward algebra. Try it, and post back with what you get.

woodyp20 · Feb 17, 2011, 10:34 PM

I'm still confused with the inverse, could you set it up for me so I can see how its done?

ebaines · Feb 18, 2011, 07:17 AM

I'll give you an example of a similar problem - that way you can see how it's done and then apply the same technique to this problem.

Suppse you have:

$ y(x) = 10-3e^{-2x} $

Its inverse is found by getting the x variable all by itself:

1. Subtract 10 from both side:

$ y-10 = -3e^{-2x} $

2. Divide both sides by -3:

$ -\frac 1 3 (y-10) = e^{-2x} $

3. Now take the natural log of both sides:

$ \ln(-\frac 1 3 (y-10)) = \ln (e^{-2x}) $

4. Since $ln(e^a) = a$ , this means:

$ \ln(-\frac 1 3 (y-10)) = -2x $

5. Divide by -2:

$ - \frac 1 2 \ln(-\frac 1 3 (y-10)) = x $

So now if you are given a value for y, you can determine the corresponding value of x. For example - suppose y = 4; then:

$ x \ =\ - \frac 1 2 \ln(- \frac 1 3 (4-10))\\ = - \frac 1 2 \ln(2) \ =\ - \frac {0.693} 2\ = \ -0.347 $

Hope this helps - can you apply this same technique to your problem?