Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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 $y=10046e^{-2.054x}$

share|improve this answer
Yes your right this really doesn't fit an exponential function does it. Well those are the points.. um.. well i tried to adding 5.3785*exp(0.7204*A2) were A2 is a cell that is value of 10 so it should solve to 3 but it does not. I get 7233.00259.. for 2.7032*-EXP(0.001*A2) i get -2.730367612 .. all 3 you put don't seem to get the right number. The way i got the 5.3785e0.7204x was from the trend line. .. not she here. –  jeremy.bass Aug 28 '12 at 0:04

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 $y = 2.7032 e^{-0.95815 \times 10^{-3} x}$.

share|improve this answer
not 100% here but i used =2.7032*-EXP(0.95815*A2*10^-10*A2) in excell to test it were X is 10 expecting 3 but i got -2.7032000026 (A2 is the x) –  jeremy.bass Aug 28 '12 at 0:11
I have really no idea what you are doing. Why would you not use 2.7032*exp(-0.95815e-3*A2) (assuming $x$ is in A2)? I mean, where did the 10^10 and extra A2 come from? Also, you put the minus sign outside the exponential. –  copper.hat Aug 28 '12 at 0:14
well lol. i just put in what i thought you had but left a typo from the other anwser. I put 2.7032*exp(-0.95815e-3*A2) and got 2.677422978 where A2 = 10 .. so doesn't seem right. –  jeremy.bass Aug 28 '12 at 0:29
Well, I doubt you will get an exact match, and a value of 2.677 is not too bad compared to the expected 3? –  copper.hat Aug 28 '12 at 0:38
Well i don't really have a choice here to be close 10 has to come out to 3, 25 has to be 2.95 and 50 need to be 2.75 etc on the rest already noted. –  jeremy.bass Aug 28 '12 at 0:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.