# exponential equation

 0 sage: solve(exp(-x)+exp(x) == 2,x) [x == 0] sage: solve(exp(-2x)+exp(2x) == 2,x) [] Can tell me anyone why the second equation has no solution? asked Oct 02 '10 amalea 71 ● 1 ● 7 ● 10 Kelvin Li 443 ● 10 ● 17 For the record, this question is closely related to http://ask.sagemath.org/question/156/exponential-equation-real-solutionKelvin Li (Apr 28 '11)

 1 I 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. posted Oct 02 '10 kcrisman 7427 ● 17 ● 76 ● 166

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