Given that $F = A{ (1+i)^n - 1 \over i}$
How can you solve for $i$ or $n$?
Given that $F = A{ (1+i)^n - 1 \over i}$
How can you solve for $i$ or $n$?
As far as $i$, numerically. The formula you give is the future value of an annuity that pays $A$ at the end of every time period for $n$ periods. There isn't really a good way to solve for it. But, there is guaranteed to be one unique value for $i$, so you can use a financial calculator to solve for it or use numerical methods like Newton's method.
Solving for $i$ is equivalent to solving for $i$ in the $n$th degree polynomial:
$A(1+i)^n - Fi - A = 0$
Solving for $n$ is possible.
$\begin{align*} &\frac{Fi}{A} = (1+i)^n - 1 \\ \Rightarrow &(1+i)^n = 1 + \frac{Fi}{A} \\ \Rightarrow &n \ln (1+i) = \ln(1 + \frac{Fi}{A}) \\ \Rightarrow &n = \frac{\ln(1 + \frac{Fi}{A})}{\ln(1+i)} \end{align*}$
You can solve for $n$ if you know $i$ with $F = A{ (1+i)^n - 1 \over i}$ $\frac{iF}{A} +1 = (1+i)^n $ $n= \frac{\log\left(\frac{iF}{A} +1\right)}{\log (1+i)}= \frac{\log\left(iF +A\right) - \log(A)}{\log (1+i)}. $
Solving for $i$ if you know $n$ requires numerical methods except for some special cases for $n$.