Hello there:
I just noticed that log and exp seem to treat equations differently, and this creates (apparently) unnecessary difficulties for solve. For example, I can use log to help out solve deal with some exponential equations:
(%i2) solve(3^(2*x) = 3^(x^2));
(%o2) [3^x^2 = 3^(2*x)]
(%i3) log(%);
(%o3) [log(3)*x^2 = 2*log(3)*x]
(%i4) solve(%);
(%o4) [x = 0,x = 2]
I would like to use exp in a similar way for logarithmic equations, but instead I get:
(%i5) solve(log(x) + log(x+9) = 2*log(6));
(%o5) [log(x) = 2*log(6)-log(x+9)]
(%i6) exp(%);
(%o6) [%e^(log(x) = 2*log(6)-log(x+9))]
(%i7) solve(%);
(%o7) []
The problem seems to be that
log(A=B) --> log(A) = log(B)
while
exp(A=B) --> exp(A=B) instead of exp(A) = exp(B).
Is there a reason for this? Could it (or should it) be fixed?
Thanks!
Jorge
PS. Of course, I can force the desired behavior by typing exp(lhs(%[1] = rhs(%[1]) instead of exp(%), but this seems silly. I can also do
(%i8) solve(log(x) + log(x+9) = 2*log(6));
(%o8) [log(x) = 2*log(6)-log(x+9)]
(%i9) %-log(x);
(%o9) [0 = -log(x+9)-log(x)+2*log(6)]
(%i10) logcontract(%);
(%o10) [0 = log(36/(x*(x+9)))]
(%i11) solve(%);
(%o11) [x = -12,x = 3]
but this is working too hard.
--
Dr. Jorge Alberto Calvo
Associate Professor of Mathematics
Department of Mathematics and Physics
Ave Maria University
Phone: (239) 280-1608
Email: jorge.calvo at avemaria.edu
Web: http://sites.google.com/site/jorgealbertocalvo