# Solving a system of equations

Mathematica doesn't want to solve my exact system with 3 equations and 3 variables.

AdjustPar20[x_] := a x^b Exp[c x]
AdjustPar20[0] == 0, {a, b, c}]


As an explanations why this cannot be solved the program replies:

{}
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found;
use Reduce for complete solution information. >>


Any ideas?

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Solve[] works much better with exact arithmetic:

AdjustPar20[x_] := a x^b Exp[c x]
p = 937126/100000;
sol = Solve[
((D[AdjustPar20[t], t] == 0) /. t -> 5/100) &&
AdjustPar20[0] == 0, {a, b, c}, Reals]

ParametricPlot3D[{a, b, c} /. sol, {a, p, 10 p},
PlotRange -> {Automatic, {-1, 1}, Automatic},
AxesLabel -> {a, b, c}, BoxRatios -> 1]


Please note that AdjustPar20[0] == 0 isn't adding any information since your selected function form already ensures it

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Note that AdjustPar20[0] == 0 isn't adding information since your selected function form already ensures it. –  belisarius Apr 29 '13 at 12:46
How does one know to solve with these circumstances (rational numbers, real domain)? –  BoLe Apr 29 '13 at 12:51
@BoLe, in general, inexact numbers and Solve[] don't mix too well... –  Ｊ. Ｍ. Apr 29 '13 at 12:57
@ belisarius thanks for you suggestion, but my version of Mathematica, 8, cannot solve this. Solve::ivar: 298.48501198004965` is not a valid variable. >> Solve[2.687648609756841*10^-437 a == 9.37126 && 1.007466030596621*10^-433 a == 0 && 3.211110349737091*10^-722 a == 0.01, {a, 298.485, -2221.2}, Reals] Do you think I should define one more equation(instead of the 3rd one), in order to get one exact value of a? –  Gasper Apr 29 '13 at 13:08
@Gasper Yes, you need one more equation. But we can't help with that since the equation should represent your problem conditions ... and we ignore them- –  belisarius Apr 29 '13 at 13:19