🚀 go-pugleaf

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?
>
>