# Find exponential equation given points?

• Oct 13th 2008, 11:29 PM
gyan1010
Find exponential equation given points?
First time I'm asking a math question on any forum that's not related to homework, but I got interested in trying to solve a pattern. Not entirely important but it peaked my curiosity.

It is an exponential pattern where I have values for x = 2 to 100. It is possible the y values are truncated/rounded decimals as this is from a computer game.

I can plot the values in excel fairly easily, and get a nice looking exponential graph. What method would i use to find the equation used to generate the numbers though?

Here they are in 2 columns to save space:
2 : 42 ~~~~~~~~~~~~~~~~~~~ 52 : 2165947
3 : 104 ~~~~~~~~~~~~~~~~~~ 53 : 2368499
4 : 280 ~~~~~~~~~~~~~~~~~~ 54 : 2586879
5 : 364 ~~~~~~~~~~~~~~~~~~ 55 : 2822194
6 : 612 ~~~~~~~~~~~~~~~~~~ 56 : 3075629
7 : 976 ~~~~~~~~~~~~~~~~~~ 57 : 3348452
8 : 1490 ~~~~~~~~~~~~~~~~~ 58 : 3642020
9 : 2193 ~~~~~~~~~~~~~~~~~ 59 : 3957785
10 : 3128 ~~~~~~~~~~~~~~~~ 60 : 4297300
11 : 4341 ~~~~~~~~~~~~~~~~ 61 : 4662227
12 : 5885 ~~~~~~~~~~~~~~~~ 62 : 5054090
13 : 7818 ~~~~~~~~~~~~~~~~ 63 : 5474505
14 : 10202 ~~~~~~~~~~~~~~~ 64 : 5925185
15 : 13106 ~~~~~~~~~~~~~~~ 65 : 6407945
16 : 16606 ~~~~~~~~~~~~~~~ 66 : 6924709
17 : 20780 ~~~~~~~~~~~~~~~ 67 : 7477218
18 : 25724 ~~~~~~~~~~~~~~~ 68 : 8067299
19 : 31540 ~~~~~~~~~~~~~~~ 69 : 8696824
20 : 38331 ~~~~~~~~~~~~~~~ 70 : 9367757
21 : 46224 ~~~~~~~~~~~~~~~ 71 : 10082146
22 : 55350 ~~~~~~~~~~~~~~~ 72 : 10842126
23 : 65845 ~~~~~~~~~~~~~~~ 73 : 11650062
24 : 77869 ~~~~~~~~~~~~~~~ 74 : 12508554
25 : 91588 ~~~~~~~~~~~~~~~ 75 : 13420451
26 : 107175 ~~~~~~~~~~~~~~ 76 : 14388857
27 : 124828 ~~~~~~~~~~~~~~ 77 : 15417146
28 : 144750 ~~~~~~~~~~~~~~ 78 : 16508977
29 : 167160 ~~~~~~~~~~~~~~ 79 : 17668295
30 : 192295 ~~~~~~~~~~~~~~ 80 : 18899355
31 : 220406 ~~~~~~~~~~~~~~ 81 : 20206727
32 : 251753 ~~~~~~~~~~~~~~ 82 : 21595301
33 : 286627 ~~~~~~~~~~~~~~ 83 : 23070505
34 : 325326 ~~~~~~~~~~~~~~ 84 : 24638311
35 : 368180 ~~~~~~~~~~~~~~ 85 : 26305266
36 : 415530 ~~~~~~~~~~~~~~ 86 : 28078511
37 : 467752 ~~~~~~~~~~~~~~ 87 : 29965813
38 : 525232 ~~~~~~~~~~~~~~ 88 : 31975578
39 : 588394 ~~~~~~~~~~~~~~ 89 : 34116885
40 : 657675 ~~~~~~~~~~~~~~ 90 : 36399509
41 : 733546 ~~~~~~~~~~~~~~ 91 : 38833942
42 : 816499 ~~~~~~~~~~~~~~ 92 : 41431421
43 : 907062 ~~~~~~~~~~~~~~ 93 : 44203958
44 : 1005781 ~~~~~~~~~~~~~ 94 : 47164362
45 : 1113245 ~~~~~~~~~~~~~ 95 : 50326392
46 : 1230060 ~~~~~~~~~~~~~ 96 : 53704797
47 : 1356876 ~~~~~~~~~~~~~ 97 : 57315344
48 : 1494365 ~~~~~~~~~~~~~ 98 : 61174862
49 : 1643244 ~~~~~~~~~~~~~ 99 : 65301264
50 : 1804255 ~~~~~~~~~~~~~ 100 : 69713595
51 : 1978188
• Oct 25th 2008, 10:22 PM
treetheta
I don't know if this will work, but you can try it out

since the fomula for an exponential formula is

$\displaystyle y = R^x$

where r is any real number and you have an x and a y value, you could porbably sub them in x = 2 y = 42 take log of both side, isolate for x and solve, .. just a hunch seems to easy to be real
• Oct 26th 2008, 12:16 AM
CaptainBlack
Quote:

Originally Posted by gyan1010
First time I'm asking a math question on any forum that's not related to homework, but I got interested in trying to solve a pattern. Not entirely important but it peaked my curiosity.

It is an exponential pattern where I have values for x = 2 to 100. It is possible the y values are truncated/rounded decimals as this is from a computer game.

I can plot the values in excel fairly easily, and get a nice looking exponential graph. What method would i use to find the equation used to generate the numbers though?

Here they are in 2 columns to save space:
2 : 42 ~~~~~~~~~~~~~~~~~~~ 52 : 2165947
3 : 104 ~~~~~~~~~~~~~~~~~~ 53 : 2368499
4 : 280 ~~~~~~~~~~~~~~~~~~ 54 : 2586879
5 : 364 ~~~~~~~~~~~~~~~~~~ 55 : 2822194
6 : 612 ~~~~~~~~~~~~~~~~~~ 56 : 3075629
7 : 976 ~~~~~~~~~~~~~~~~~~ 57 : 3348452
8 : 1490 ~~~~~~~~~~~~~~~~~ 58 : 3642020
9 : 2193 ~~~~~~~~~~~~~~~~~ 59 : 3957785
10 : 3128 ~~~~~~~~~~~~~~~~ 60 : 4297300
11 : 4341 ~~~~~~~~~~~~~~~~ 61 : 4662227
12 : 5885 ~~~~~~~~~~~~~~~~ 62 : 5054090
13 : 7818 ~~~~~~~~~~~~~~~~ 63 : 5474505
14 : 10202 ~~~~~~~~~~~~~~~ 64 : 5925185
15 : 13106 ~~~~~~~~~~~~~~~ 65 : 6407945
16 : 16606 ~~~~~~~~~~~~~~~ 66 : 6924709
17 : 20780 ~~~~~~~~~~~~~~~ 67 : 7477218
18 : 25724 ~~~~~~~~~~~~~~~ 68 : 8067299
19 : 31540 ~~~~~~~~~~~~~~~ 69 : 8696824
20 : 38331 ~~~~~~~~~~~~~~~ 70 : 9367757
21 : 46224 ~~~~~~~~~~~~~~~ 71 : 10082146
22 : 55350 ~~~~~~~~~~~~~~~ 72 : 10842126
23 : 65845 ~~~~~~~~~~~~~~~ 73 : 11650062
24 : 77869 ~~~~~~~~~~~~~~~ 74 : 12508554
25 : 91588 ~~~~~~~~~~~~~~~ 75 : 13420451
26 : 107175 ~~~~~~~~~~~~~~ 76 : 14388857
27 : 124828 ~~~~~~~~~~~~~~ 77 : 15417146
28 : 144750 ~~~~~~~~~~~~~~ 78 : 16508977
29 : 167160 ~~~~~~~~~~~~~~ 79 : 17668295
30 : 192295 ~~~~~~~~~~~~~~ 80 : 18899355
31 : 220406 ~~~~~~~~~~~~~~ 81 : 20206727
32 : 251753 ~~~~~~~~~~~~~~ 82 : 21595301
33 : 286627 ~~~~~~~~~~~~~~ 83 : 23070505
34 : 325326 ~~~~~~~~~~~~~~ 84 : 24638311
35 : 368180 ~~~~~~~~~~~~~~ 85 : 26305266
36 : 415530 ~~~~~~~~~~~~~~ 86 : 28078511
37 : 467752 ~~~~~~~~~~~~~~ 87 : 29965813
38 : 525232 ~~~~~~~~~~~~~~ 88 : 31975578
39 : 588394 ~~~~~~~~~~~~~~ 89 : 34116885
40 : 657675 ~~~~~~~~~~~~~~ 90 : 36399509
41 : 733546 ~~~~~~~~~~~~~~ 91 : 38833942
42 : 816499 ~~~~~~~~~~~~~~ 92 : 41431421
43 : 907062 ~~~~~~~~~~~~~~ 93 : 44203958
44 : 1005781 ~~~~~~~~~~~~~ 94 : 47164362
45 : 1113245 ~~~~~~~~~~~~~ 95 : 50326392
46 : 1230060 ~~~~~~~~~~~~~ 96 : 53704797
47 : 1356876 ~~~~~~~~~~~~~ 97 : 57315344
48 : 1494365 ~~~~~~~~~~~~~ 98 : 61174862
49 : 1643244 ~~~~~~~~~~~~~ 99 : 65301264
50 : 1804255 ~~~~~~~~~~~~~ 100 : 69713595
51 : 1978188

Assume that the data approximately satisfies an equation of the form:

$\displaystyle y=e^{\lambda (t+k)}$

Take logs:

$\displaystyle \ln(y)=\lambda t + \lambda k$

The write $\displaystyle Y$ for $\displaystyle \ln(y)$ and $\displaystyle K$ for $\displaystyle \lambda k$

Now plot $\displaystyle Y$ against $\displaystyle t$, if you have a reasonable approximation to a straight line then the slope is $\displaystyle \lambda$ and the intercept is $\displaystyle K$.

CB
• Oct 26th 2008, 07:46 AM
gyan1010
ok i took t as the 1-100 and Y as natural log of the larger and larger numbers.

I didn't get a strait line. Does that mean there are other unknown variables?

Here are my excel graphs, of both (t , y) and (t , ln(y)). I was unable to plot a trendline in excel to exactly match either of them.

http://www.mathhelpforum.com/math-he...1&d=1225035815http://www.mathhelpforum.com/math-he...1&d=1225035868
• Oct 26th 2008, 02:15 PM
tivaelydoc
Could you please explain the latter half of that?

I got to the natural log of y, but I couldn't follow you after that.
• Oct 26th 2008, 02:42 PM
CaptainBlack
Quote:

Originally Posted by gyan1010
ok i took t as the 1-100 and Y as natural log of the larger and larger numbers.

I didn't get a strait line. Does that mean there are other unknown variables?

Here are my excel graphs, of both (t , y) and (t , ln(y)). I was unable to plot a trendline in excel to exactly match either of them.

Well the plots of the data looks like $\displaystyle y=e^{kt^{1/2}}$ may be an acceptable representation. Now take logs to get:

$\displaystyle Y=\ln(y)=\lambda t^{1/2}$

or:

$\displaystyle Y^2=\lambda^2 t$

So now try plotting $\displaystyle Y^2$ against $\displaystyle t$, and if this is a line through the origin the slope will be $\displaystyle \lambda^2$

If this does not give you a straight line then you do not have a simple exponential model and rather more work would be needed to find a good representation.

CB
• Oct 26th 2008, 09:43 PM
gyan1010
Actually after posting that last one I did try that ln(t) squared which straitened it out a little. Then cubed even more so but it was varying above and below the strait line at that point. More of a wavy strait line.

My original theory was that it was some exponential that had a component of the previous result. A function with memory is the term i use in my signals & system course. I just don't really know how to test that.

This is all for fun anyway though. The values are actually levels in a video game and large number is the exact experience required to reach that level. I left out 1 because you start there. Figured i would use math to try and fulfill my curiosity about the equation the developers used.
• Oct 26th 2008, 09:51 PM
treetheta
since u did plot the points on excel,

there is an option on excel if u click the points then right click, and pick add a tread line then set it to exponential and say display equation on chart, you should get the equation (:
• Oct 26th 2008, 09:55 PM
gyan1010
I mentioned that i think. I could not find any trend line to match the points for either graph.
• Oct 27th 2008, 01:02 AM
CaptainBlack
Quote:

Originally Posted by gyan1010
Actually after posting that last one I did try that ln(t) squared which straitened it out a little. Then cubed even more so but it was varying above and below the strait line at that point. More of a wavy strait line.

My original theory was that it was some exponential that had a component of the previous result. A function with memory is the term i use in my signals & system course. I just don't really know how to test that.

This is all for fun anyway though. The values are actually levels in a video game and large number is the exact experience required to reach that level. I left out 1 because you start there. Figured i would use math to try and fulfill my curiosity about the equation the developers used.

It might be usefull if you could post the data in a form that can be dumped into Excel without retyping so that others can play with this.

Given its from a game, and the apparent smoothness of the curves there is probably some relativly simple relation being used to generate the data.

CB
• Oct 27th 2008, 02:00 AM
Laurent
Plotting "log log f(n)" against "log n", you get approximately a line.
I found the plot coincides pretty much with $\displaystyle \exp(4 n^{0.327})$, but It could be possible that for instance the original data is in fact of the form $\displaystyle C \exp( a n^b)$.