Mathematica used to be easily able to solve an equation like this:

Reduce[Log[Sqrt[k p]/Log[k]] == 0, p]

(I can easily do it myself, at least I can find the solution p = log(k)^2/k.)

Now in Mathematica 12, all I get is Reduce::nsmet: This system cannot be solved with the methods available to Reduce.

I thought it might be an issue with assumptions, so I tried assuming both k and p were sufficiently large, but it didn't help.

Is there a way to get Mathematica 12 to produce useful output on the problem above?

  • 1
    $\begingroup$ Works fine with Solve instead of Reduce. A simpler version of this problem is Reduce[Sqrt[a x] == b, x], which shows how hard this problem is in all generality. $\endgroup$
    – Roman
    Jun 10, 2019 at 13:01
  • $\begingroup$ Fwiw I don't think this is particular to V12. I tried on 11.3 and it gave up there too. $\endgroup$
    – 1110101001
    Jun 11, 2019 at 3:03
  • $\begingroup$ I still keep my version 9.0.1 although I also have the newest version, but in a way 9.0.1 is the last really stable version. $\endgroup$
    – Yukterez
    Jun 11, 2019 at 3:54
  • 1
    $\begingroup$ Please do not use the bugs tag when posting new questions. See the tag description for why. $\endgroup$
    – Szabolcs
    Jun 11, 2019 at 6:40
  • $\begingroup$ @Roman my version of Mathematica is able to solve Sqrt[a x] == b using Reduce, though it does complain about the solution containing the "unsolved equation" 0 == b - Sqrt[b^2] as an assumption. $\endgroup$ Jun 16, 2019 at 11:09

1 Answer 1


Working only in the real numbers could be a solution:

Reduce[Log[Sqrt[k p]/Log[k]] == 0, p, Reals]
(*    k > 1 && p == Log[k]^2/k    *)

I think the branch cuts of the square root make this problem difficult to solve in all complex generality ($k\in\mathbb{C}$). Maybe previous versions of Mathematica were a bit less careful with these branch cuts? From what I hear there has been a lot of work done in Mathematica recently on how to deal with branch cuts.

The given solution $p=\frac{\ln^2k}{k}$ is actually valid for any $\lvert k\rvert>1$, not just for the positive real $k>1$:

ComplexPlot3D[Log[Sqrt[k p]/Log[k]] /. p -> Log[k]^2/k,
  {k, -2 - 2 I, 2 + 2 I}, PlotRange -> All, Exclusions -> None]

enter image description here

I guess it would still be nice to find a way to solve/reduce this equation that returns a solution like Abs[k] > 1 && p == Log[k]^2/k to show the most general solution.

  • 1
    $\begingroup$ Interestingly, while your answer works; Reduce[Log[Sqrt[k p]/Log[k]] == 0 && Element[{p, k}, Reals], p] does not. $\endgroup$
    – Bob Hanlon
    Jun 10, 2019 at 13:22
  • $\begingroup$ @BobHanlon Try p>0&&k>0 . The domain Reals means the variables and functions are assumed to be real. $\endgroup$
    – Michael E2
    Jun 10, 2019 at 14:03
  • $\begingroup$ @Roman Doesn't the assumptions k > 0 && p > 0 imply that the variables aren't complex? Adding Reals to reduce does the trick though. Thanks! $\endgroup$ Jun 10, 2019 at 14:25
  • 3
    $\begingroup$ @ThomasAhle yes, but the term inside the logarithm can still be negative, thereby allowing for cuts. Restricting k > 1 avoids this scenario and Reduce returns a result. Try Reduce[Log[Sqrt[k p]/Log[k]] == 0 && k > 1 && p > 0, p]. $\endgroup$
    – Greg Hurst
    Jun 10, 2019 at 14:41
  • 1
    $\begingroup$ Also read the few lines in the docs starting here: reference.wolfram.com/language/ref/Reduce.html#26975. Using the domain spec of Reduce (its third argument) restricts all values of all function evaluations to be real. Implicitly this means Reduce[Log[Sqrt[k p]/Log[k]] == 0, p, Reals] is restricting to k > 1. $\endgroup$
    – Greg Hurst
    Jun 10, 2019 at 14:43

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