Well, though a=b of course implies exp(a)=exp(b), the converse is not true
(e.g. exp(0)=exp(2*%pi*%i)).
But we already do sin(a=b) => sin(a)=sin(b), so apparently that is not the
general reasoning.
So I guess it would be OK to have x^(a=b) => x^a = x^b.
-s
On Thu, Dec 20, 2012 at 9:38 AM, Jorge Calvo <Jorge.Calvo at avemaria.edu>wrote:
> Hello Volker:
>
> Thanks for your response. Unfortunately, I don't usually know a priori
> when solve will fail so I was looking for a solution that I use directly
> with the result of solve and does not involve going back to the original
> equation. In this case, the brackets get in the way of map:
>
> (%i2) solve(log(x) + log(x+9) = 2*log(6));
> (%o2) [log(x) = 2*log(6)-log(x+9)]
> (%i3) map(exp, %);
> (%o3) [%e^(log(x) = 2*log(6)-log(x+9))]
> (%i4) solve(%);
> (%o4) []
>
> I suppose that I can use indexes to get rid of the brackets:
>
> (%i5) solve(log(x) + log(x+9) = 2*log(6));
> (%o5) [log(x) = 2*log(6)-log(x+9)]
> (%i6) map(exp, %[1]);
> (%o6) x = 36/(x+9)
> (%i7) solve(%);
> (%o7) [x = -12,x = 3]
>
> but that still more cumbersome than the comparable use of log. So I am
> still wondering whether there are any good reasons why the behavior for exp
> should not be changed to:
>
> exp(A=B) --> e^A = e^B
>
> Many thanks!
>
> Jorge
>
> On Dec 19, 2012, at 11:47 AM, Volker van Nek wrote:
>
> > Alternatively you can map exp and log over the equations:
> >
> > (%i2) eq : log(x) + log(x+9) = 2*log(6)$
> > (%i3) map(exp, eq);
> > (%o3) x*(x+9) = 36
> > (%i4) solve(%);
> > (%o4) [x = -12,x = 3]
> >
> > (%i5) eq : 3^(2*x) = 3^(x^2)$
> > (%i6) map(log, eq);
> > (%o6) 2*log(3)*x = log(3)*x^2
> > (%i7) solve(%);
> > (%o7) [x = 0,x = 2]
> >
> > Volker van Nek
> >
>
>
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