ASKSAGE: Sage Q&A Forum - Individual question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 28 Apr 2011 12:00:41 -0500exponential equationhttps://ask.sagemath.org/question/7719/exponential-equation/sage: solve(exp(-x)+exp(x) == 2,x)
[x == 0]
sage: solve(exp(-2*x)+exp(2*x) == 2,x)
[]
Can tell me anyone why the second equation has no
solution?
Sat, 02 Oct 2010 06:35:11 -0500https://ask.sagemath.org/question/7719/exponential-equation/Comment by Kelvin Li for <p>sage: solve(exp(-x)+exp(x) == 2,x)
[x == 0]
sage: solve(exp(-2<em>x)+exp(2</em>x) == 2,x)
[]</p>
<p>Can tell me anyone why the second equation has no
solution?</p>
https://ask.sagemath.org/question/7719/exponential-equation/?comment=21776#post-id-21776For the record, this question is closely related to http://ask.sagemath.org/question/156/exponential-equation-real-solutionThu, 28 Apr 2011 12:00:41 -0500https://ask.sagemath.org/question/7719/exponential-equation/?comment=21776#post-id-21776Answer by kcrisman for <p>sage: solve(exp(-x)+exp(x) == 2,x)
[x == 0]
sage: solve(exp(-2<em>x)+exp(2</em>x) == 2,x)
[]</p>
<p>Can tell me anyone why the second equation has no
solution?</p>
https://ask.sagemath.org/question/7719/exponential-equation/?answer=11709#post-id-11709I unfortunately don't have time to check why this doesn't work, but the following keyword argument (useful to know in any case) does work:
sage: (exp(-2*x)+exp(2*x) == 2).solve(x,to_poly_solve=True)
[x == I*pi*z15]
which should be interpreted as saying `z15` is a free integer variable (hence `z`). One can also do
sage: (exp(-x)+exp(x) == 2).solve(x,to_poly_solve='force')
[x == 2*I*pi*z24]
for all possible solutions to the original one. Unfortunately, finding the documentation for this (currently) requires using the method notation instead of functional
sage: solve?
notation.Sat, 02 Oct 2010 15:47:01 -0500https://ask.sagemath.org/question/7719/exponential-equation/?answer=11709#post-id-11709