get value at a point on an exponential curve I'm not super with math but I need to make a function in my web app to get the value of a point on a curve when I know the curve points that are set.  Here is what I did, I put a set of point with the x and y set at the know points then did an exponential trend line with limits of 1.15 to 3 for the x value.  The y value ranges from 0-1000.  I get for the formula output from excel as
y = 5.3785e0.7204x
R² = 0.9898 

Here are the points I have 
x        y
10      3
25      2.95
50      2.75
100     2.5
200     2
300     1.5
1000    1.15

I short i need to have my web app say, what is the y value when x is 176.  I know this is probably simple for you math guys but I'd be thankful the help. Cheers
 A: Note that your text says y ranges from 0 to 1000 and x from 1.15 to 3, but your data is the reverse.  When I plot it, the data doesn't fit an exponential at all.  The last point is way off, with the rest fitting a straight line very well.  If I had to use this data I would either throw away the point at 1000 or use a pair of linear fits, one from 10 to 300, and another from 300 to 1000.
The way to read your output from excel is $y=5.3785e^{0.7204x}$ and in the computer languages I have used you would write y=5.3785*exp(0.7204*x), but that can't be right because y increases with x while the data goes the other way.  When I fit the data as presented to an exponential in excel, I get $y=2.7032e^{-.001x}$ and if I transpose x and y I get $y=10046e^{-2.054x}$
A: Using Octave, I used least squares to fit a line of the form $x \mapsto ax+b$ to the data points $(x_i, \ln y_i)$ above, and ended up with $a \approx -0.95815 \times 10^{-3}$, $b \approx 0.99445$. This corresponds to a model $y = K e^{\alpha x}$ where $\alpha = a \approx -0.95815 \times 10^{-3}$ and $K = e^b \approx 2.7032$.
So the model should be $y = 2.7032 e^{-0.95815 \times 10^{-3} x}$.
