Re: Easy Problem. Can this be done in Maxima/Axiom?

I'm responding also to fricas-devel, since a problem shows up there...

"Mark Clements" <[email protected]> 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