Works perfect now. Thanks David and Mike for your replys, they were very
helpful.

Bernard thanks for the LN(2) suggestion and you are right, the value is and
should be 197.464 (example I was given is wrong).

First time I visited here and really appreciate the great help from each of
you.

"Bernard Liengme" wrote:

Surely you mean =$B$1*EXP(-(0.693*((A5-B2)/(30*365))))
which gives 197.460
What makes you think the "correct" value is 199.70?
If I replace 0.693 by LN(2) to improve the precision and 365 by 365.25, I
get only 197.464
I think you have been given the wrong answer! Take it from a retired chem.
prof that textbook errors are not rare.

Note that my formula does give a result that decrease as time does on (ie as
A5 increases)
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"me28" wrote in message
...
Thanks so much for a quick response. The information you provided is a
good
starting point but if possible let me provide additional information:

Initial Activity = 206 (cell B1)
Initial Date = 1/31/05 (cell B2)
Current Date = 12/1/06 (cell A5)
Half-Life = 30 (hard-coded in the formula)

My current formula is =$B$1*EXP(-(0.693*((B2-A5)/30)))
I need this to give me the decay value for 12/1/06. The correct value is
197.46, however my current formula returns 199.70 and is actually
increasing
each day when it should be decreasing.

Thanks for you time.

"Mike Middleton" wrote:

me28 -

With the value of A in cell A1, value of t in cell A2, and value of T
Half-Life in cell A3, in some other cell use

=A1*EXP(-0.693*A2/A3)

- Mike
http://www.mikemiddleton.com


"me28" wrote in message
...
I need to create a worksheet that takes an active value and calculates
the
timed decay. The formula is A e-(0.693t/T½) (e to the power of
(0.693t,
though LN may work). Value of A = Initial Value, T = Timed decay.