I try to solve this equation symbolically:

Solve[A*s - b*Log[Sqrt[(d - e*s)^2 + 1] + d - e*s] - x + b*c == 0, s]

(yes: ^2 = cntrl-6)

But M. (trial version) is running forever.

I need an expression for s(x).

Can anyone set me on the right track? I'm new to M., obviously.

  • 5
    $\begingroup$ A transcendental equation like that doesn't seem likely to have a closed-form solution. $\endgroup$ Mar 16, 2019 at 15:50

1 Answer 1


As pointed out in the comments, this is a transcendental equation and is unlikely to have a symbolic/closed form solution for s. This is above the power of Solve[].

Assuming your function is solely dependent on s (like you said), and not something else one can beckon FindRoot[] to your cause for atleast a numerical solution.

If you know the rest of your variables, maybe you can get to a point where you have an idea of what s could possibly be depending on what you're doing/requiring.

I took your equation, and simplified it,

FullSimplify[A s - b Log[Sqrt[(d - e s)^2 + 1] + d - e s] - x + b c == 0]
(*b c + A s == x + b ArcSinh[d - e s]*)

Plot[(-b c - A s + x + b ArcSinh[d - e s] /. {e -> 1, d -> 1, b -> 1, x -> 1, A -> 1, c -> 1}, {s, -5, 5}, PlotRange -> All]



This just looks like somewhere around 1.

and then used FindRoot with s -> 1

FindRoot[b c + A s == x + b ArcSinh[d - e s] /. {e -> 1, d -> 1, b -> 1, x -> 1, A -> 1, c -> 1}, {s, 1}]

Giving me the solution of s -> 0.49....Of course, this will change drastically as your constants change!

As a tip, generally if mathematica takes longer than a few seconds to calculate a solution, it probably can't find a solution...

Hope this is helpful to get you further along.


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