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)) + log(-(e^C - 1)*e^(-C)
+ 1)
Not nice...
sage: E.subs(y==y0).expand().simplify()
log(-e^(-C) + 1) == log(-e^(-C) + 1)
Nicer. And indeed :
sage: bool(E.subs(y==y0).expand().simplify())
True
Whereas :
sage: bool(E.subs(y==y0))
False
Sometime, sage needs a little help : systematically simplifying
intermediate results leads to dead ends, and that's why sage doesn't do
that.
HTH,
--
Emmanuel Charpentier
Le dimanche 16 juillet 2017 00:44:03 UTC+2, Chris Seberino a écrit :
>
> This does not solve...
>
> var("y C")
> solve( log(y) == C + log(x) + log(y-1),y)
>
> It returns....
>
> [log(y) == C + log(x) + log(y - 1)]
>
> Any ideas?
>
> Thanks!
>
> Chris
>
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