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?

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.

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

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?