I'm not sure what logexpand does, but clearly, we are nowhere near
finding a solution to log(a*x)=C*log(b*x) for x!!!
See what I get...
(C1) logexpand:true;
(D1) TRUE
(C2) solve(log(a*x)=C*log(b*x),x);
LOG(a x)
(D2) [LOG(b x) = --------]
C
David Holmgren wrote:
> Hi - Try this for a start (on the expression with the logs):
> logexpand:true;
> solve(log(a*x)-C*log(b*x),x);
>
> I have not yet found anything similar for exponentials...
>
> Dave Holmgren
>
> Dr. David E. Holmgren,
> Imaging Scientist,
> SMART Technologies, Inc.
> Calgary, AB, Canada
> http://www.smarttech.com
> [403]-235-1452, ext. 251
> DavidHolmgren@smarttech.com
>
>
>
>>-----Original Message-----
>>From: Daniel Lemire [SMTP:lemire at ondelette]
>>Sent: Monday, October 01, 2001 5:12 PM
>>To: maxima@www.ma.utexas.edu
>>Subject: [Maxima] Newbie: problem solving basic equations
>>
>>Good day,
>>
>>I think the "solve" function is broken. I just installed Maxima and
>>tried solving a few elementary equations. In all cases but the very,
>>very elementary ones, Maxima is no good, it seems. (See below.)
>>
>>
>>What am I missing? Surely, Maxima can do this?
>>
>>[lemire@romeo lemire]$ maxima
>>GCL (GNU Common Lisp) Version(2.4.0) Wed May 9 12:02:00 CDT 2001
>>Licensed under GNU Library General Public License
>>Contains Enhancements by W. Schelter
>>Maxima 5.6 Wed May 9 12:01:49 CDT 2001 (with enhancements by W. Schelter).
>>Licensed under the GNU Public License (see file COPYING)
>>(C1) solve(exp(a*x)= C*exp(b*x),x);
>>
>> a x
>> b x %E
>>(D1) [%E = -----]
>> C
>>(C2) solve(log(a*x)=C*log(b*x),x);
>>
>> LOG(a x)
>>(D2) [LOG(b x) = --------]
>> C
>>(C3) solve(a^x=b^x,x);
>>
>> x x
>>(D3) [b = a ]
>>(C4) solve(m*x+b=y,x);
>>
>> y - b
>>(D4) [x = -----]
>> m
>>(C5)
>>--
>>Daniel Lemire, Ph.D.
>>
>>http://www.ondelette.com/
>>
>>
>>_______________________________________________
>>Maxima mailing list
>>Maxima@www.math.utexas.edu
>>http://www.math.utexas.edu/mailman/listinfo/maxima
>>
>
--
Daniel Lemire, Ph.D.
http://www.ondelette.com/