Is it always a coin toss whether a computer algebra system can solve a log
equation?
Should I not expect to make a career out of using Sage to solve nonlinear
equations?
cs
On Sunday, July 16, 2017 at 3:41:42 PM UTC-5, Emmanuel Charpentier wrote:
>
> Wups... My bad : I wasn't really awake, it s
Wups... My bad : I wasn't really awake, it seems...
Anyway, as suggested by Dominique, you can do :
sage: E=log(y) == C + log(x) + log(y-1);E
log(y) == C + log(x) + log(y - 1)
sage: S=E.solve(x)[0].solve(y);S
[y == x*e^C/(x*e^C - 1)]
sage: bool(E.subs(S).expand_log())
True
which checks.
Again,
Dominique
THANK YOU! Without or without declaring x your way works
This...
var("y C")
solve( log(y) == C + log(x) + log(y-1),x)
solve( x == y/(y*e^C - e^C), y)
Gives...
[y == x*e^C/(x*e^C - 1)]
What is amazing is that simply having y appear in 2 places makes it
unsolvable directly withou
Why not adding "x" ? (and of cause declaring x in the same way than y and C)
Because
var("x y C")
solve( log(y) == C + log(x) + log(y-1),x,y)
returns
([x == y/(y*e^C - e^C)], [1])
Dominique
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Emmanuel
Thank you for your reply but you solved a DIFFERENT equation. Notice mine
has an x variable in it.
I can get your's to work but not mine.
cs
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Works for me :
sage: reset()
sage: var("y,C")
(y, C)
sage: E=log(y)==C+log(y)+log(1-y);E
log(y) == C + log(y) + log(-y + 1)
sage: S=solve(E,y);S
[y == (e^C - 1)*e^(-C)]
Let's check this unique solution :
sage: y0=S[0].rhs()
sage: E.subs(y==y0)
log((e^C - 1)*e^(-C)) == C + log((e^C - 1)*e^(-C)) +
