       Re: making a e-funtion of two coördinates

• To: mathgroup at smc.vnet.net
• Subject: [mg31639] Re: making a e-funtion of two coördinates
• From: "Robert Nowak" <robert.nowak at ims.co.at>
• Date: Fri, 23 Nov 2001 05:46:02 -0500 (EST)
• References: <9t86cv\$r6l\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```taking the log on both sides may help
note e is not eulers number
you have to take E for eulers number

y == U * e^(a * t) // Log /@ # &
% // PowerExpand
% // Solve[{# /. {t -> t1, y -> y1}, # /. {t -> t2, y -> y2}}, {U, a}] &

Out=
Log[y] == Log[e^(a*t)*U]

Out=
Log[y] == a*t*Log[e] + Log[U]

Out=
{{a -> (-Log[y1] + Log[y2])/((-t1 + t2)*Log[e]),
U -> E^(((-t2)*Log[y1] + t1*Log[y2])/(t1 - t2))}}
--
---
Robert Nowak (robert.nowak at ims.co.at)
IMS Nanofabrication GmbH
A-1020 Wien, Schreygasse 3, Austria
Phone: (+43 1)2144894-32, Fax: (+43 1)2144894-99

TheSquaredBun <veezdREMOVETHIS at hotmail.com> wrote in message
news:9t86cv\$r6l\$1 at smc.vnet.net...
>
> I have got two coordinates out of graph. The line through the
> coordinates could be described as:
>
> y = U * e^(a * t)
>
> (e is Exponetial e (approx. 2.37))
>
> t is the x-value, I have give two points (t1,y1) and (t2,y2)
>
> I have tried several things (Solve and Fit for example), but I can't
> get the function.
> Does anyone know what should be the input to let Mathematica solve the
> a and U in the formula.
>
> Much thanks in advance,
>
> Roderik
> r.f.emmerink at st.hanze.nl
>

```

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