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I'm quite new to mathematica, so apologies if the question it soo trivial.

I'm trying to symbolically solve an equation for R, and would like at the same time to mix in a second equation. What I've done is the following:

Remove["Global`*"];
L == M^2/R^4;
Solve[L == M^5.5 R^-5.5, R]

The result is:

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.
{{R -> 1./(L/M^5.5)^0.1818181818181818}}

Two questions:

  1. What is mean with this warning / error message?
  2. Is there a way to get the result as a fraction in the exponent?
  3. Is there any more elegant way to get to the result, so I don't always have to use Remove Global?
  4. Am I correct to use double equal == in this context?

To clarify, I'd like to express L(M), by expressing R in terms of M. How can this be achieved? It's essentially by putting the first equation into the second and solving it for L. – Any suggestions are appreciated.

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  • $\begingroup$ L == M^2/R^4; Reduce[L == M^[email protected] R^[email protected], R, Reals] $\endgroup$
    – cvgmt
    Oct 28, 2020 at 12:12
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    $\begingroup$ L == M^2/R^4; Solve[L == M^(11/2) R^(-11/2), R] $\endgroup$
    – cvgmt
    Oct 28, 2020 at 13:01
  • $\begingroup$ To clarify, I'd like to express L(M), by expressing R in terms of M. How can this be achieved? It's essentially by putting the first equation into the second and solving it for L. $\endgroup$
    – Nickpick
    Oct 28, 2020 at 18:59
  • $\begingroup$ Ls[M_] = L /. First@ Solve[{L == M^2/R^4, Rationalize[L == M^5.5 R^-5.5]}, L, {R}, Reals] yields ConditionalExpression[Root[-1 + M^22 #1^3 &, 1], M > 0] .The L, {R} in Solve solves for L and at the same \ time eliminates R from the two equations. LogPlot[Ls[M], {M, 0, 10}, PlotRange -> All] and e.g. Ls[7] // ToRadicals $\endgroup$
    – Akku14
    Oct 28, 2020 at 20:22
  • $\begingroup$ ok I think that goes into the right direction, but why the Plot etc? I just want to solve the equation. How can I then pick up the L, {R}? thanks $\endgroup$
    – Nickpick
    Oct 28, 2020 at 23:36

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