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Has Mathematica 12 gotten worse at solving simple equations?

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13

$begingroup$

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?

equation-solving

share|improve this question

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

$endgroup$

• 
  
  
  

  
  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 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 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 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 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.
  $endgroup$
  – Thomas Ahle
  13 hours ago
  

  
  
  
  

add a comment | 

13

$begingroup$

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?

equation-solving

share|improve this question

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

$endgroup$

• 
  
  
  

  
  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 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 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 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 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.
  $endgroup$
  – Thomas Ahle
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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?

equation-solving

share|improve this question

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

$endgroup$

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

share|improve this question

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

share|improve this question

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

edited Jun 11 at 13:05

user64494

3,96821323

edited Jun 11 at 13:05

user64494

3,96821323

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

asked Jun 10 at 12:57

Thomas AhleThomas Ahle

18817

1 Answer
1

active

oldest

votes

18

$begingroup$

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 ($kinmathbbC$). 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 $p=fracln^2kk$ is actually valid for any $lvert krvert>1$, 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]

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.

share|improve this answer

edited Jun 10 at 16:30

answered Jun 10 at 13:11

RomanRoman

11.3k11944

$endgroup$

• 
  
  
  

  
  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 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 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!
  $endgroup$
  – Thomas Ahle
  Jun 10 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].
  $endgroup$
  – Chip Hurst
  Jun 10 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.
  $endgroup$
  – Chip Hurst
  Jun 10 at 14:43
  
  

  
  
  
  

add a comment | 

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18

$begingroup$

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 ($kinmathbbC$). 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 $p=fracln^2kk$ is actually valid for any $lvert krvert>1$, 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]

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.

share|improve this answer

edited Jun 10 at 16:30

answered Jun 10 at 13:11

RomanRoman

11.3k11944

$endgroup$

• 
  
  
  

  
  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 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 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!
  $endgroup$
  – Thomas Ahle
  Jun 10 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].
  $endgroup$
  – Chip Hurst
  Jun 10 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.
  $endgroup$
  – Chip Hurst
  Jun 10 at 14:43
  
  

  
  
  
  

add a comment | 

18

$begingroup$

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 ($kinmathbbC$). 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 $p=fracln^2kk$ is actually valid for any $lvert krvert>1$, 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]

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.

share|improve this answer

edited Jun 10 at 16:30

answered Jun 10 at 13:11

RomanRoman

11.3k11944

$endgroup$

• 
  
  
  

  
  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 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 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!
  $endgroup$
  – Thomas Ahle
  Jun 10 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].
  $endgroup$
  – Chip Hurst
  Jun 10 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.
  $endgroup$
  – Chip Hurst
  Jun 10 at 14:43
  
  

  
  
  
  

add a comment | 

$begingroup$

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 ($kinmathbbC$). 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 $p=fracln^2kk$ is actually valid for any $lvert krvert>1$, 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]

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.

share|improve this answer

edited Jun 10 at 16:30

answered Jun 10 at 13:11

RomanRoman

11.3k11944

$endgroup$

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

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

share|improve this answer

edited Jun 10 at 16:30

answered Jun 10 at 13:11

RomanRoman

11.3k11944

answered Jun 10 at 13:11

RomanRoman

11.3k11944

answered Jun 10 at 13:11

RomanRoman

11.3k11944