Solve Somewhat Compicated 2-equation 2 unknown Problem

Hi guys, I’m running a simple predator vs. prey model and need to solve for two parameters. In the equations I’ve developed the variables of interest are R_1 and R_2; I’ve tried running them through Wolfram|Alpha to get a solution, but it times out before arriving at solution; selecting more time for calculation doesn’t seem to help. I don’t have any high end calculating software on my computer, so does anyone know of a simple method to solve the following system?

650 = (exp(R_1) - exp(R_2) )(419.125)/(R_2 - R_1) + 650exp(R_1),
1000 = (exp(11
R_1) - exp(11R_2) )(418.03)/(R_2 - R_1) + 650exp(11*R_1)

If it helps at all, I suspect that the R_1 and R_2 values should vary between -1 and +1. Any suggestions or calculations would be deeply appreciated.

Take the natural logs of both equations to get the variables about of the exponent, and then solve two simple linear equations.

ln(e^a) = a

Perhaps I’me mistaken, but the natural log of the right hand side of my equations do not reduce to linear form. Tell me if you can see other wise:

ln(650) = ln[(exp(R_1) - exp(R_2) )(419.125)/(R_2 - R_1) + 650*exp(R_1) ],

ln(1000) = ln[ (exp(11R_1) - exp(11R_2) )(418.03)/(R_2 - R_1) + 650exp(11R_1)]

Any other suggestions?

Oh, I didn’t see the variables in the denominator.

If you know how to do numerical programming and you’re just stuck because you don’t have the right software, you can grab a copy of R and do whatever you need to in there. Otherwise, your best bet is to find someone who knows how to do numerical programming and entice them to help you out.

Is there anything else you know about the relationship between R1 and R2? Is there anything about those variables that might let you make some approximations about some terms being close to zero?

I doubt you can solve those two equations analytically. Using Excel’s solver I get
r1 = -0.595968329 r2 = 0.063074332 which satisfies both equations to within .0002. Assuming there is a solution, and all you need is a numerical answer, you should be able to improve that if you need to do so by changing the options from Excel’s defaults. However, I don’t guarantee that you can do better considering the rounding error inherent in Excel.

Use solver in excel.

Aha, I forgot about solver. The estimates provided are certainly sufficient. Thanks so much!

Incidentally, the vast majority of equations do not have closed-form solutions, so a program or method that tries to find such solutions (such as Wolfram Alpha) will invariably fail on them, and might take a very long time to realize that it’s failing. In practice, many equations that one actually encounters in real problems will have solutions, especially when you take the step of inventing new functions just to serve as solutions to them, but you’ll never get all of them (or, in fact, any more than a minuscule portion of them).

If I were doing this I would make plots of the two equations on the same axes to get some insight (the place(s) where the curves intersect would of course be the solution(s)).