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Chris......I'm pretty sure my ans for D is wrong.....soz


"Gags" <drgagnon@ozemail.com.au> wrote in message
news:hJaF7.4349$w61.207344@ozemail.com.au...
> The calculations are too complex to show.....they include exponentials,
> logs, and solving quadratics....I think what u want is:
>
> B = 100D[(Y/A)^(1/[CD])-1]
>
> C = log(Y)/[D(log(A) + log(BD/100 +1)]
>
> D = [(-100) (+-) sqroot(100^2 - 400BY/CA)] / 2B
>
> note: D is solution of a quadratic so there are two answers
> ie.  (+-) means do once with + and once with - to get 2 solutions for D.
>
> hope this helps
>
> Gags (3rd Year Elec Eng Student)
> "Chris Harding" <pouchcotato@iinet.net.au> wrote in message
> news:3be51fc1$0$29765@echo-01.iinet.net.au...
> > I have this equation that I need to be changed around, but have no idea
> how
> > to do it. In letters, the equation is (sorry for the numerous brackets,
> but
> > it's hard without pen and paper abilities on here):
> >
> > Y = A � (((B�100)�D) + 1)^(C�D)
> >
> > And what I need is for the equation to become like this:
> >
> > A= Y � (((B�100) � D) +1)^(C�D)
> >
> > but for B=... C=... and D=...
> >
> > the ^ means to the power of, and I can't for the life of me figure out
how
> > to move them around. The equation is for Compound interest, and I do
> > Discrete maths.
> >
> > Your help would be greatly appreciated, and if you could email it to me,
> > that would also be great.
> >
> > Thanks for your help
> >
> >  -Chris
> >
> > P.S-If anyone knows of the names of some good math books for this kind
of
> > thing, could you tell me that too?
> >
> >
>
>