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

Mark Clements wrote:
>
> 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
>
> -- 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)
>
> Can anyone come up with a better solution?
>
> Kindly, Mark Clements.
>
> ------------------------------------------------------------------------
>
> _______________________________________________
> Axiom-mail mailing list
> [email protected]
> http://lists.nongnu.org/mailman/listinfo/axiom-mail
>   
a:=(3/7)*(4*x-5)*(7/3)*(2*x-7)-1
b:=zerosOf(a)

    +--+         +--+
   \|89 + 19, - \|89 + 19
  [---------, -----------]
       8           8
                                 Type: List AlgebraicNumber


eval(a,x=b.1)
    0

eval(a,x=b.2)
    0