The antilog of x is 10^x, so all you need to do is used the ^ operator.
If you are doing the antilog for some other base, there is formula to do
that as well, but I'm forgetting it.
Regards,
Yasir
On Fri, 26 Dec 2003, Martin Marques wrote:
> Date: Fri, 26 Dec 2003 19:34:35 -0300
> From: Martin Marques <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED],
> Sai Hertz And Control Systems <[EMAIL PROTECTED]>,
> [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: Re: [SQL] Anti log in PostgreSQL
>
i> El Vie 26 Dic 2003 19:12, Sai Hertz And Control Systems escribió:
> > Dear all ,
> >
> > In one of our project I require to calculate antilog of (3.3234)
> > But I could not find any functions in Documentation for the same.
> >
> > In mathematics I would have written it something like
> >
> > A = antilog (3·3234) = 2144
>
> As I can understand, this is a 10 base log, so that what you want is
> 10^(3.3234)?
>
> For that you have the exponential operator ^.
>
> --
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> Martín Marqués | [EMAIL PROTECTED]
> Programador, Administrador, DBA | Centro de Telemática
> Universidad Nacional
> del Litoral
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