Failure on symbolic solve

edit

Here is a longish answer that will show you a few ways to get what you want... but won't exactly answer your question.

Apparently this is what Maxima is looking for (Sage uses Maxima for its assumptions and solving).

sage: x,y,z = var('x y z')
sage: assume(x>0)
sage: assume(y>0)
sage: assume(z>0)
sage: eqn = (x == y**z)
sage: eqn
x == y^z
sage: assume(z,'integer')
sage: solve(eqn,y)
[y == x^(1/z)]

I should point out that Maxima doesn't use its assumptions that much in its solving.

But Maxima then apparently doesn't care what the answer is, as long as x>0.

Maxima 5.22.1 http://maxima.sourceforge.net
using Lisp ECL 10.4.1
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) assume(y>0,z>0,x>0);
(%o1)                        [y > 0, z > 0, x > 0]
(%i2) eqn:x=y^z;
                                         z
(%o2)                               x = y
(%i3) solve(eqn,y);
Is z an integer?

n;
                                        1/z
(%o3)                             [y = x   ]

So this could be a bug - or feature - in Maxima. Of course, such functions are in general multivalued, so depending on your context this might not really be a good answer anyway. You might instead like

(%i5) load(to_poly_solver);

Loading maxima-grobner $Revision: 1.6 $ $Date: 2009/06/02 07:49:49 $
define: warning: redefining the built-in function prog1
define: warning: redefining the built-in function symbolcheck
define: warning: redefining the built-in function push
define: warning: redefining the built-in function pop
define: warning: redefining the built-in function tr_ev
(%o5) /Users/.../sage-4.6.2.alpha0/local/share/maxima\
/5.22.1/share/contrib/to_poly_solver.mac
(%i6) to_poly_solve(eqn,y);
                                          2 %i %pi %z5
                                          ------------
                                    1/z        z
(%o6)                  %union([y = x    %e            ])

where %z5 is an arbitrary integer. Normally this would come from solve(eqn,y,to_poly_solve=True) but the exception takes precedence in our current implementation.