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exponential equation

asked 14 years ago

amalea gravatar image

updated 13 years ago

Kelvin Li gravatar image

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?

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For the record, this question is closely related to http://ask.sagemath.org/question/156/exponential-equation-real-solution

Kelvin Li gravatar imageKelvin Li ( 13 years ago )

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answered 14 years ago

kcrisman gravatar image

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.

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Asked: 14 years ago

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Last updated: Oct 02 '10