Re: Ada83 Exponentiation HELP !!!

[Robert Dewar <robert_dewar@my-deja.com>]

In article <m3puw9uacy.fsf@kiuk0156.chembio.ntnu.no>,
  Preben Randhol <randhol@pvv.org> wrote:
> "DuckE" <nospam_steved@pacifier.com> writes:
>
> | It's been years since I've dealt with evaluating things like
this, but as I
> | recall 6^(1.3) is equivalent to:
> |
> |    Exp( 1.3 * Log( 6 ) )
>                 ^^^
> Just to make it clear, Log here must be the natural logarithm
Ln
>
> Ln(a^b) = b*Ln(a)
>
> a^b = exp(b*Ln(a))       ; a > 0
>
> Isn't this correct?


No it is completely wrong. See chapter 5 on IEEE arithmetic
in my microprocessors book if you can find a copy around :-)
In that chapter I use this example to motivate the need for
IEEE extended format, but of course only the ia32 and ia64
architectures provide this in hardware among commonly used
processors, so the above algorithm should almost never be
used. Even if the log and exp routines are last bit accurate
the result of the exponentiation can be very far off being
last bit accurate, and there are perfectly good accurate
algorithms available.h


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