Ttdfjt

Is the specular reflection on a polished gold sphere white or gold in colour?

Has a life raft ever been successfully deployed on a modern commercial flight?

Why is oilcloth made with linseed oil?

Why does std::string_view create a dangling view in a ternary expression?

Definition of 'vrit'

How can I prevent a user from copying files on another hard drive?

Cut the gold chain

What triggered jesuits' ban on infinitesimals in 1632?

Syntax and semantics of XDV commands (XeTeX)

What does it cost to buy a tavern?

Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid?

What is "industrial ethernet"?

What is the "ls" directory in my home directory?

Dmesg full of I/O errors, smart ok, four disks affected

Am I legally required to provide a (GPL licensed) source code even after a project is abandoned?

What was the first third-party commercial application for MS-DOS?

Improve appearance of the table in Latex

Where should a runway for a spaceplane be located?

Methodology: Writing unit tests for another developer

What does this Swiss black on yellow rectangular traffic sign with a symbol looking like a dart mean?

How can a warlock learn from a spellbook?

How does DC work with natural 20?

Can the pre-order traversal of two different trees be the same even though they are different?

Why does Linux list NVMe drives as /dev/nvme0 instead of /dev/sda?

Has Mathematica 12 gotten worse at solving simple equations?

Why doesn't Mathematica solve $x=cos,x$ properly?How do I solve this equation?Failure message from ReduceDeriving least-squares equations in MathematicaHow do I solve for the roots of $sin(x) - x^2=0$?Exclusions found but Solve and Reduce don't find exclusionsReduce doesn't solve this equations with conditionGetting Solve::nsmet error messageHow to solve a system with GCD?How to solve this system with logarithms?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

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
  yesterday
  

  
  
  
  

 | 
show 1 more 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
  yesterday
  

  
  
  
  

 | 
show 1 more 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

11.4k11944

$endgroup$

• 
  
  
  

  
  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 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f200059%2fhas-mathematica-12-gotten-worse-at-solving-simple-equations%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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

11.4k11944

$endgroup$

• 
  
  
  

  
  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

11.4k11944

$endgroup$

• 
  
  
  

  
  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

11.4k11944

$endgroup$

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

11.4k11944

answered Jun 10 at 13:11

RomanRoman

11.4k11944

answered Jun 10 at 13:11

RomanRoman

11.4k11944