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I am trying to solve this equation for P:

(P - 0.3)*(1 + (((1013.25^0.190284*0.0065)/
   288)*(A/(P - 0.3)^0.190284)))^(1/0.190284) == Q

When I run:

Solve[(P - 
 0.3)*(1 + (((1013.25^0.190284*0.0065)/
    288)*(A/(P - 0.3)^0.190284)))^(1/0.190284) == Q , P]

Solve does not return a solution for P, it merely returns the original equation.

Update

In response to Michael E2's question, a screenshot of the result I receive follows below:

enter image description here

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  • $\begingroup$ I get an error containing a hint on how to proceed, when I execute your code. Do you? $\endgroup$
    – Michael E2
    Commented Apr 10, 2018 at 4:06
  • $\begingroup$ I do not. I will update my question with a screenshot of what I get. $\endgroup$
    – Joe Crivello
    Commented Apr 10, 2018 at 4:15
  • $\begingroup$ Well, I tried the advice, but it didn't work. (I'm on V11.3. I suppose you have a different version.) $\endgroup$
    – Michael E2
    Commented Apr 10, 2018 at 4:20
  • $\begingroup$ Interestingly enough, I am on 11.3 as well. Perhaps Mathematica needs a restart... unfortunate that the advice did not work. I don't see why this equation shouldn't be solvable. I also tried bounding the problem with some inequalities but that did not help. Edit: Ok, now I am getting the hint you spoke of after the restart of Mathematica. However, using exact fractions doesn't seem to fix the problem either... $\endgroup$
    – Joe Crivello
    Commented Apr 10, 2018 at 4:25

1 Answer 1

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Some massaging of the equation:

(P - 0.3)*
 ((1 + (((1013.25^0.190284*0.0065)/288)*(A/(P - 0.3)^0.190284)) //
    Together)^Hold@(1/0.190284) // PowerExpand) == Q // ReleaseHold
(*  1. (0.0000842288 A + (-0.3 + P)^0.190284)^5.2553 == Q  *)

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

Solve[%, P]
(*  {{P -> 0.3 + 1. (-0.0000842288 A + 1. Q^0.190284)^5.255302600323726}}  *)

Basically, it's difficult to solve symbolic equations that have approximate (floating-point) real powers. Mathematica treats them as functions of a complex variable. PowerExpand treats everything as a positive real number, which I used to manipulate the exponents. It's sort of a BFH hammer, so you might want to verify the solution in some way.

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1
  • $\begingroup$ Thank you!! This works perfectly. $\endgroup$
    – Joe Crivello
    Commented Apr 10, 2018 at 4:35

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