[Top][All Lists]
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [Axiom-mail] Easy Problem. Can this be done in Maxima/Axiom?
From: |
Martin Rubey |
Subject: |
Re: [Axiom-mail] Easy Problem. Can this be done in Maxima/Axiom? |
Date: |
Tue, 17 Nov 2009 08:12:26 +0100 |
User-agent: |
Gnus/5.11 (Gnus v5.11) Emacs/22.3 (gnu/linux) |
I'm responding also to fricas-devel, since a problem shows up there...
"Mark Clements" <address@hidden> writes:
> The following problem came up on the Maxima mail-list (translated here into
> Axiom):
>
> -- solve for x (over the Reals?)
>
> ex := (3/7)^(4*x-5)*(7/3)^(2*x-7)=1
> I came up with several overly long solutions (using Fricas 1.0.3):
>
> rule1 := (rule log(7/3)==-log(3/7))
>
> rule1 rhs solve(map(expandLog,map(log,ex)), x).1
curiously, this doesn't work here, the solve won't yield a solution :-(
If this really used to work, we should add it as a regression to
bugs2009.input.pamphlet.
> -- or, similarly
>
> rule1 rhs solve(expandLog log lhs ex, x).1
>
> rule2 := rule((a/b)^c*(b/a)^d==(a/b)^(c-d))
>
> solve((rule2 lhs ex)=(rhs ex),x)
this works here, too.
I'm not sure whether the following yields all the solutions, but at
least you get one:
(1) -> ex := (3/7)^(4*x-5)*(7/3)^(2*x-7)=1
3 4x - 5 7 2x - 7
(1) (-) (-) = 1
7 3
(2) -> res := solve(ex, x)
3 19683
7log(-) - log(--------)
7 40353607
(2) [x= -----------------------]
3
2log(-)
7
(3) -> X := normalize rhs(res.1)
(3) - 1
(4) -> eval(ex, x=X)
(4) 1= 1
normalize is quite powerful! (and that huge fraction is just (3/7)^9,
no idea why it doesn't get simplified immediately.)
Martin