🚀 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?
>
>

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?
> >
> >
>
>

No worries, I've given up for the mo anyways. I was writing a program for my
Graphics Calculator (I've written one for Simple Interest with and Unknown
Value, Optimisation and various others) but my mathing abilities haven't
reached this far yet.

But if you could point me in the direction of some places to get info on how
to do this kind of thing, it would be great.

 -Chris

"Gags" <drgagnon@ozemail.com.au> wrote in message
news:y7QF7.169$E81.11992@ozemail.com.au...
> Chris......I'm pretty sure my ans for D is wrong.....soz
>
>
> "Gags" <drgagnon@ozemail.com.au> wrote in message
> news:hJaF7.4349$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?
> > >
> > >
> >
> >
>
>

"Chris Harding" <pouchcotato@iinet.net.au> writes:

> 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=...
> 

Starting with y=a(b/100d + 1)^c/d

First step by step for b:

divide both sides by a: y/a = (b/100d + 1)^c/d
raise both to d/c     : (y/a)^d/c = b/100d + 1
subtract 1            : (y/a)^d/c -1 = b/100d
multiply by 100d      : 100d((y/a)^d/c -1) = b

All you need to do this is a good grasp of algebra.

Now step by step for c:

divide both sides by a       : y/a = (b/100d + 1)^c/d
take logs                    : log(y/a) = (c/d) log(b/100d + 1)
multiply by d/log(b/100d + 1): d log(y/a)/log(b/100d + 1) = c 

Here you also need to know a little about logs.

d is much harder ... I'm not sure it is possible to solve for d
algebraically. I can't, at any rate.

Glen.