Easy Problem. Can this be done in Maxima?

It's unfortunate that core Maxima can't do this directly, but it is easy
enough to use Maxima to solve it:

(%i1) ex: (3/7)^(4*x-5)*(7/3)^(2*x-7)=1$
(%i2) map(log,ex),logexpand:all;
(%o2) log(3)*(2*x+2)+log(7)*(-2*x-2) = 0
(%i3) solve(%,x);
(%o3) [x = -1]

to_poly_solver solves it:

(%i1) load(to_poly_solver);
Warning - you are redefining the Maxima function prog1
Warning - you are redefining the Maxima function symbolcheck
Warning - you are redefining the Maxima function push
Warning - you are redefining the Maxima function pop
Warning - you are redefining the Maxima function tr_ev
(%o1)
"C:/PROGRA~2/MAXIMA~1.2/share/maxima/5.19.2/share/contrib/to_poly_solver.mac"
(%i2) eq: (3/7)^(4*x-5)*(7/3)^(2*x-7)=1;
(%o2) 7^(-2*x-2)*3^(2*x+2) = 1
(%i3) sol: to_poly_solve(eq,x);
(%o3) %union([x = (2*%i*%pi*%z51+log(9/49))/(log(49)-log(9))])

But unfortunately there doesn't seem to be any straightforward way to
simplify this last expression to

         %union([x = (2*%i*%pi*%z51/(log(49)-log(9)) - 1 ])

The following does it:
        substpart(map(radcan,multthru(piece)),sol,1,1,2)

but I wouldn't call that straightforward....

              -s

On Sun, Nov 15, 2009 at 4:41 PM, Richard Hennessy <rich.hennessy at verizon.net
> wrote:

>  Hi List,
>
> I recently took a placement test with this problem in it.  It was very easy
> to solve but after getting home I suspected Maxima could not do it.  At
> least not directly.
>
> (%i1) solve((3/7)^(4*x-5)*(7/3)^(2*x-7)=1,x);
>
> (%o1) [3^(2*x+2) = 7^(2*x+2)]
>
> Is there an easy way Maxima can do it?  The answer is -1.
>
> Rich
>
>
>
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>