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I have the following equation:

r^(1 + Rho) - r^Rho (-1 + Mu) - z Rho == 0

I searched the list but I couldn't come up with a solution. Any help is appreciated.

How do I solve for r?

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  • $\begingroup$ See the documentation for Solve and Reduce. $\endgroup$
    – IPoiler
    Jul 28, 2015 at 15:03
  • $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory Tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$
    – bbgodfrey
    Jul 28, 2015 at 15:06
  • $\begingroup$ An analytical solution may not exist. Try solving the equation numerically. $\endgroup$
    – bbgodfrey
    Jul 28, 2015 at 15:10

1 Answer 1

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I know nothing about the range of the data for your problem. Assuming that the values for Rho, Mu and z are real numbers you can gain insight into your problem by combining bbgodfrey's comment with a plot using Manipulate.

For example, if Rho and Mu are known parameters you can see how the solution varies as you change the value of z.

Manipulate[

 Column[{
   NSolve[r^(1 + Rho) - r^Rho (-1 + Mu) - z Rho == 0, r],

   Plot[r^(1 + Rho) - r^Rho (-1 + Mu) - z Rho, {r, -5, 5}]
   }],

 {{Rho, 1}, -1, 3, Appearance -> "Open"},
 {{Mu, 2}, 1, 4, Appearance -> "Open"},
 {{z, 1}, 0, 10, Appearance -> "Open"}
 ]

Mathematica graphics

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  • $\begingroup$ 0 < rho <= 1, 0 < r < 20 $\endgroup$
    – Robert Reynolds
    Jul 28, 2015 at 16:36
  • $\begingroup$ Thanks this is really helpful :-) $\endgroup$
    – Robert Reynolds
    Jul 28, 2015 at 16:42

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