I have an equation

y[a, b, x] = a^2 b x^-2 (Exp[b/x] - 1)^-1

and when I plot, it gives the expected curve as shown in the image attached.

Equation Plot

I need to find the value of x at the intersecting point, marked with a circle. For a large number of the parameters a and b, I could find the solution and select the value of x at x=1, but it will give two different points.

For varying a and b, the line will also shift accordingly. How should I do it?

  • $\begingroup$ Please post the Mathematica code about your picture. $\endgroup$
    – cvgmt
    2 days ago
  • $\begingroup$ The equation is given in the plot. It is y[a,b,x]=a^2*bx^{-2} Exp[(b/x)-1]^{-1}, then just did the LogLogPlot $\endgroup$
    – Pritam
    2 days ago
  • 1
    $\begingroup$ @Pritam next time, please do make some effort to craft a reasonably good question. $\endgroup$
    – rhermans
    2 days ago
  • $\begingroup$ Thank you for your efforts. $\endgroup$
    – Pritam
    2 days ago
  • $\begingroup$ You write you wanted to find the value of x but selected an answer that does not do that. That's very confusing. $\endgroup$
    – Michael E2
    21 hours ago

3 Answers 3


Notice you have a syntax problem in your code, the correct way to define your function is like this

y[a_, b_, x_] = a^2 b x^-2 (Exp[b/x] - 1)^-1

Now you can Solve for the parameter a, for any b and x

Solve[y[a, b, x] ==1,{a}]

enter image description here

Now you have an expression for a, that you can replace with any x and b that makes sense.

If you want only the solution to the left, then you can use Min to get

Min[a/.Solve[y[a, b, x] ==1,{a}]]

enter image description here

  • 1
    $\begingroup$ Depending on what the OP is doing, it might be numerically advantageous to reformulate the solution as x Exp[b/(4 x)] Sqrt[2/b Sinh[b/(2 x)]] (and its negative). $\endgroup$ 2 days ago

Assuming (1) you really did mean, "I need to find the value of x....", (2) that "marked with a circle" is code for the minimum positive value of x, then Solve[] can do it. (I'll use the variant SolveValues[] for convenience.)

solveForX[a_, b_] := Min@SolveValues[
    a^2 b x^-2 (Exp[b/x] - 1)^-1 == 1 && x > 0,

solveForX[1, 1/10]
Root[{-1 - 10 #1^2 + 10 E^(1/(10 #1)) #1^2 &, 

If you don't like Root objects or have to use the result outside of Mathematica, then N[] may be used to convert the solution to any desired precision.

N[solveForX[1, 1/10]]
(*  0.0171457  *)

Or if you use floating-point inputs, you'll get a floating-point output:

solveForX[1., 0.1]
(*  0.0171457  *)

We set a,b and the leval set c varying and find the root x.

f[{a_, b_}][x_] = a^2 b x^-2 (Exp[b/x] - 1)^-1;
root[p_?VectorQ][c_] := NSolveValues[f[p][x] == c, x, Reals];
pt[p_?VectorQ][c_] := 
  If[Length[ root[p][c]] >= 2, {Log@First@root[p][c], Log@c} // Point,
 Show[Plot[{f[p][x], c} // Evaluate, {x, 0.02, 1000}, 
   ScalingFunctions -> {"Log", "Log"}, PlotRange -> {0, 4}, 
   AspectRatio -> 1, PerformanceGoal -> "Quality"], 
  Graphics[{Red, AbsolutePointSize[5], pt[p][c]}]], {{p, {3, 2}}, {2, 
   2}, {10, 10}}, {c, .5, 1.5}, ControlPlacement -> Left]

enter image description here


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