> CAD <[log in to unmask]> writes:
>> 1) 1.32 10^-3 = d (1-EXP(-15/a))
>> 2) 2.00 10^-3 = d (1-EXP(-5865/a))
RG> ... put q = EXP(-15/a) and form a new equation
RG> 1.32*q^391 - 2*q + 0.68= 0, which you can see has
RG> two solutions. but only one gives real a.
Oops. It's just been pointed out to me - thanks, Paul - that there's
an obvious third solution: q=1, corresponding to a->infinity. (The
steep gradient hid that section of the plot, but I should have seen
it anyway just by looking at the equation).
Ray
--
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http://www.users.zetnet.co.uk/rgirvan/ +++ The Apothecary's Drawer
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