Daniel Lemire write:
>Why won't it do the same thing for exp(a * x) = C * exp(b *x) ?
>What's the philosophy here?
(C1) solve(x^a -c*x^b,x);
a
b x
(D1) [x = --]
c
I think it's right,at least reasonable,now.
But for the first time I happend to encounter that,
I was surprised.
Maybe the philosophy is that if terams is increased,example
solve(x^a-c*x^b-d*x^e,x),even if a,b,e are positive interger,
x^a-c*x^b-d*x^e=0 has too many patterns for equation.
Not like as x^3-a*x+b=0,x^a-c*x^b-d*x^e has not a single standard
pattern,so If we intend to solve with symbolic manipulation,we must
classify in some or other case.
If we know well the problem which we want to solve,that efforts are
no use,I think.
gosei furuya (go_furuya@infoseek.jp)