I have an equation of the form

eq := 4/\[Pi] y x^2 - 1/16 c^2 Log[(2 x)/(\[Pi] w)] == cons

with the following set of parameters

parameters = {w -> 100, c -> 1, cons -> 0.46, y -> 0.5};

Replacing the parameters and solving the equation, gives

Solve[eq /. parameters, x]

$x = 0.114106, x = 0.351317$

But if I first solve the equation analytically and then replace the parameters, I get

 xans = x /. Solve[eq, x];
 xans /. parameters // Chop

$x=0.114106,x= -0.114106$

My question is that why the second answer appears in the first way, i.e., $x= 0.351317$ is missed in the 2nd way? How I can I find the analytic solution with this x as the answer?

  • $\begingroup$ Think of xans = x /. Solve[eq, x, Reals] xans /. parameters and compare with xans = x /. Solve[eq, x] xans /. parameters. $\endgroup$
    – user64494
    Commented May 29, 2021 at 15:19
  • $\begingroup$ x==- 0.114106 is not a solution in view of 4/\[Pi] y x^2 - 1/16 c^2 Log[(2 x)/(\[Pi] w)] - cons /. {w -> 100, c -> 1, cons -> 0.46, y -> 0.5, x -> -0.114106} which outputs 1.74124*10^-7 - 0.19635 I. The Solve command uses nonequivalent transformations in some cases. $\endgroup$
    – user64494
    Commented May 29, 2021 at 15:51
  • $\begingroup$ Reduce[eq, x, Reals] performs a warning "Reduce::nsmet: This system cannot be solved with the methods available to Reduce." and returns the input. Reduce[eq, x] has no chance in view of four parameters. $\endgroup$
    – user64494
    Commented May 29, 2021 at 15:58
  • $\begingroup$ @ user64494 Thanks for your comments. In fact I need the analytic solution which leads to x=0.351317. $\endgroup$
    – Kheeyal
    Commented May 29, 2021 at 17:12

1 Answer 1


When you solved it analytically by Solve, I think you got the warning message:

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

So it is not guaranteed that all the analytic solutions are obtained in this way. (Note that the Lambert W function is multivalued.) As user64494 pointed out, the second analytic solution leads to a wrong numerical answer in your case.

To get the second analytical solution, you need to select another branch for ProductLog (see also this post):

x /. First[Solve[eq, x]] /. ProductLog[z_] -> ProductLog[-1, z] /. parameters  (* 0.351317 + 0. I *)
  • 1
    $\begingroup$ Thank you so much for your great help. $\endgroup$
    – Kheeyal
    Commented May 29, 2021 at 17:45

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