Am Samstag, den 02.07.2011, 22:28 +0900 schrieb Part Marty:
________________________________________________________________________
> From: shinabe.munehiro at hotmail.co.jp
> To: drdieterkaiser at web.de
> Subject: RE: [Maxima] (a^b)^c--->a^(b*c)
> Date: Sat, 2 Jul 2011 22:06:54 +0900
>
> Thank you,Dear Dieter Kaiser
> More,let me know!
>
> How about this?
>
> exp(%i*c)^n/exp(%i*c)-->exp(%i*c*(n-1))
Again setting the option variable radexpand to the value true will help.
The point is, that in general (a^b)^b is not equal to a^(b*c). You have
to know or to assume something about the domain of the variables a, b,
and c. Setting radexpand to the value true is equivalent to the
assumption that a,b, and c > 0.
Perhaps n is an integer. Then it is possible to declare n to be an
integer. For this case you get the desired simplification too, e.g.
(%i4) declare(n, integer)$
(%i5) exp(%i*c)^n/exp(%i*c);
(%o5) %e^(%i*c*n-%i*c)
At last, you can use a function like radcan to simplify expressions,
e.g.
(%i3) exp(%i*c)^n/exp(%i*c);
(%o3) %e^-(%i*c)*(%e^(%i*c))^n
(%i4) radcan(%);
(%o4) %e^(%i*c*n-%i*c)
Dieter Kaiser