On Tue, 22 Dec 1998, Ricardo Jurczyk Pinheiro wrote:
> >> Well, well, well...
> >>
> >> X X
> >> A => ln( exp( A )) => ln(X*exp(A))
> >>
> >> Hurrah to the good old math!
> >
> >This is AWFUL math!!! This expression is completely wrong, since exp(A^X)
> >is totally different from X*exp(A).
>
> Awful? Are u really up to your mind? It's a matter of opinion, only.
No! I'm not talking about opinions, I'm just saying that your expression
ir wrong! Think well, the first pass is right, but exp(A^X) is completely
different (except for special cases) of X*exp(A). Remember that the
logarithm function is monotonically crescent. So, if ln(x)=ln(y) then x=y.
> >I think that you should say:
> > X X
> > A => exp( ln( A )) => exp(X*ln(A))
> >
> >This should give a good result (with some small imprecision).
>
> Well, as u may know, there isn't any difference in the mathematical
> way. Maybe your version of the equation works better with computers. And
> I'm a mathematician, u know, so I know at least a bit of arithmetics.
I know exactly that ln(A^X) is equal to X*ln(A), but exp(A^X) is not equal
to X*exp(A). I don't understant why you insist in saying that! Let's do a
numerical example, using base 10 to simplify the calculations:
5 5
You say that 2 = log(10^(2 )) = log(5*10^(2))
32 = log(10^32) = log(5*100)
32 = log(100000000000000000000000000000000) = log(500)
32 = 32 = 2.4771
Well, this shows that your expression is completely wrong!
Let's try my expression:
5 5
I say that 2 = 10^log(2 ) = 10^(5*log(2))
32 = 10^log(32) = 10^(5*0.3010)
32 = 10^1.505 = 10^(1.505)
32 = 32 = 32
Is that clear now?
[]'s
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Marco Antonio Simon Dal Poz http://www.lsi.usp.br/~mdalpoz
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