[Leica] Need help with formulae

[afirkin at afirkin.com (Alastair Firkin)]

Thanks Bob, I was struggling with this problem and then thought "someone 
will know". I'll plug in the formula and thanks again

Alastair

--- leica@web-options.com wrote:

From: "Bob W" <leica@web-options.com>
To: "'Leica Users Group'" <lug@leica-users.org>
Subject: RE: [Leica] Need help with formulae
Date: Sat, 31 Jan 2009 09:41:26 -0000


x := (x*(2^y))+(x*(2^y)*z) 

your example:

(10*(2^2))+((10*(2^2)*0.8))

Bob

> -----Original Message-----
> From: lug-bounces+leica=web-options.com@leica-users.org 
> [mailto:lug-bounces+leica=web-options.com@leica-users.org] On 
> Behalf Of Bob W
> Sent: 31 January 2009 08:57
> To: 'Leica Users Group'
> Subject: RE: [Leica] Need help with formulae
> 
> how can a number double .n of a time?
> 
> Bob 
> 
> > 
> > G'day all,
> > 
> > does anyone know a formula which allows you to calculate a 
> > value in a geometric progression. For example if a number 
> > doubles 2.8 times in 10 years, what will that number be. At 
> > the moment the only way I can do it is "long hand" ie if 10 
> > is going to double 2.8 times then it would be 10 doubled to 
> > 20, twenty doubled to 40 then add 80% of the next double ie 
> > 32 plus the 40 equals 72, BUT is there a formula, so I can 
> > put in other times of doubling. 
> > 
> > Troubled
> > 
> > Alastair
> 
> 
> 
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> 
> 


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