Rearrange the formula for the sum of a geometric series to find the value of its common ratio? What I'm attempting to do is to rearrange the formula for the sum of a geometric series so as to find the value of its common ratio $r$. I've tried several different methods, all of which have failed; though I can't understand why.
This was my last attempt using logarithms:
$$ S_n=\frac{a(1-r^n)}{1-r}$$
$$S_n(1-r) = a(1-r^n)$$
$$S_n-S_nr = a - ar^n$$
$$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$
$$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$
$$0 = (\log{r})(n-1)$$
$$r = 10^{\frac{0}{n-1}}$$
$$r=1$$
I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ and $a$, then that would be equally fantastic.
Thanks.
 A: Added: I assume you don't know both $S_n$ and $S_{n+1}$. Also I thought you had a geometric progression, i.e. not an infinite series. If you mean a series then you have to apply limits to $S_n$. 

From 
$$\frac{S_{n}}{a}-1=\frac{S_{n}}{a}r-r^{n}$$
you cannot deduce
$$\log \frac{S_{n}}{a}=\log \frac{S_{n}}{a}+\log r-n\log r$$
because
$$\log (x+y)\neq \log x+\log y.$$
The correct property of the $\log $ function is
$$\log (xy)=\log x+\log y.$$ 
The algebraic identity
$$x^{n}-c^{n}=(x-c)(x^{n-1}+cx^{n-2}+c^{2}x^{n-3}+\cdots +c^{n-2}x+c^{n-1})$$
gives for $c=1$
$$\frac{x^{n}-1}{x-1}=x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x+1.$$
Therefore you have
$$\frac{S_{n}}{a}=\frac{1-r^{n}}{1-r}=r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1.$$
You would have to find $r$ such that
$$r^{n-1}+r^{n-2}+r^{n-3}+\cdots +r+1-\frac{S_{n}}{a}=0.$$
However this polynomial equation has no algebraic solution, in general, for $n\ge 6$. You need a different method (a numerical one).
A: No need to solve the full polynomial as 
$$S_{n+1} - rS_n = a$$ (by def. of geometric series)
(I would not leave this as an own entry if I had enough points to comment everywhere, sorry about that)
A: HINT $\ \ $ Consider $\ S_{n+1} - r\ S_n $
