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I have to see the x for this case how should be?!

    Reduce[{( Sqrt[ f + a^2/(4 b^2)]/ f Log[12 x b^2 f + 3 x a^2 + Sqrt[3]
      x Sqrt[(4 b^2 f + a^2) (4 b (((2 - 3 b) k)/x^2 + 3 b f) + 
        3 a^2)]] - a/(2 b f)ArcTanh[(x a)/Sqrt[4/3 (2 - 3 b) b k + x^2 (4 b^2 f + a^2)]] - a/(4 b f) Log[(-2 + 3 b) k - 3 x^2 b f]) > 0, f > 0, b > 0, a > 0}, x, Reals]

thank you

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  • $\begingroup$ After all this time and +7 upvotes on the answer, you probably shouldn't change the question. Just chalk it up to a mistake and ask a new one. See mathematica.meta.stackexchange.com/questions/1614/… $\endgroup$
    – Michael E2
    Jun 29 at 21:21
  • $\begingroup$ Sorry, I just realized my bad. $\endgroup$ Jun 29 at 21:25
  • $\begingroup$ That's ok. You can roll back the edit, or I can do it. $\endgroup$
    – Michael E2
    Jun 29 at 21:26

2 Answers 2

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Let's first check the conditions on the parameters by including the function domain restrictions (implied by specifying Reals) but omitting at first the main inequality. We'll simplify this bit and come back to the main question. There are three complicated, transcendental functions in the main term, two involving Log and one involving ArcTanh. We'll add their function domains to the restrictions. These are already implied, but we're helping Reduce out by singling them out:

Reduce[{
  (*(Sqrt[f+a^2/(4 b^2)]/f Log[12 x b^2 f+3 x a^2+Sqrt[
  3] x Sqrt[(4 b^2 f+a^2) (4 b (((2-3 b) k)/x^2+3 b f)+3 a^2)]]-
  a/(2 b f)ArcTanh[(x a)/Sqrt[4/3 (2-3 b) b k+x^2 (4 b^2 f+a^2)]]-
  a/(4 b f) Log[(-2+3 b) k-3 x^2 b f])>0,*)
  f > 0, b > 0, a > 0, 
  FunctionDomain[
   Log[12 x b^2 f + 3 x a^2 + 
     Sqrt[
       3] x Sqrt[(4 b^2 f + a^2) (4 b (((2 - 3 b) k)/x^2 + 3 b f) + 
          3 a^2)]], x],
  FunctionDomain[
   ArcTanh[(x a)/Sqrt[4/3 (2 - 3 b) b k + x^2 (4 b^2 f + a^2)]], x],
  FunctionDomain[Log[(-2 + 3 b) k - 3 x^2 b f], x]},
 x, Reals]

(*  False  *)

The result False means there are no real x satisfying the constraints. Adding the main inequality could only restrict the answer more, but of course, starting from "no solutions," the only possible result is still "no solutions."

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  • $\begingroup$ Therefore, under which condition this equation will be positive? $\endgroup$ Jun 26 at 19:26
  • $\begingroup$ @ArminSharafi I doubt it can be solved, the system being transcendental, but to find out, you should solve for the parameters, not for x. $\endgroup$
    – Michael E2
    Jun 26 at 21:40
  • 1
    $\begingroup$ @ArminSharafi When I try solving for {a, b, k, f}, it still says no real solutions (False). Unless you want to relax the constraints.... $\endgroup$
    – Michael E2
    Jun 26 at 21:43
  • $\begingroup$ I was thinking maybe ContourPlot will be useful?!! $\endgroup$ Jun 29 at 19:30
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    $\begingroup$ I just realized that the equation it was not correct, I changed it $\endgroup$ Jun 29 at 21:06
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   g[x_] := (Sqrt[4 a b^2 + f^2]
 ArcCoth[(Sqrt[3] Sqrt[4 a b^2 + f^2] x)/Sqrt[4 b (-2 + 3 b) k + 3 (4 a b^2 + f^2) x^2]] - f Log[(3 f + b Sqrt[36 a + (3 (3 f^2 + (4 b (-2 + 3 b) k)/x^2))/b^2]) x])/(2 a b);

a = 0.7; f = 0.2; b = 1; Plot[g[x] /. {k -> -0.04}, {x, 0.01, 10}, AspectRatio -> 1]

enter image description here

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