OpenEuphoria

Re: Algebra problem

• Posted by Alexander Toresson <alexander.toresson at gmail.com> Jul 07, 2005
• 795 views

Derek Parnell wrote:
> 
> My high school algebra has deserted me 
> 
>  Given the formula 
> 
>    A^b + Cb = d
> 
> how do I solve for 'b'?
> 
> -- 
> Derek Parnell
> Melbourne, Australia
> 

I dunno really if this is solvable. Anyway, here are my attempts:

-----------------------

A^b + Cb = d

log(A^b) = log(d - Cb)

b * log(A) = log(d - Cb)

log(b * log(A)) = log(log(d - Cb))

log(b) + log(log(A)) = log(log(d - Cb))

e^(log(b) + log(log(A))) = log(d - Cb)

e^(e^(log(b) + log(log(A)))) = d - Cb

e^(e^(log(b)) + e^(log(log(A)))) = d - Cb

e^(b + log(A)) = d - Cb

--------------------

A^b + Cb = d

A^b = d - Cb

-A^b + d = Cb

(-A^b + d)/C = b

log((-A^b + d)/C) = log(b)

log(-A^b + d) - log(C) = log(b)

--------------------

A^b + Cb = d

A = (d - Cb)^(1/b)

log(A) = log((d - Cb)^(1/b))

log(A) = (1/b) * log(d - Cb)

log(A) * b = log(d - Cb)


Regards, Alexander Toresson