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Given that $F = A{ (1+i)^n - 1 \over i}$

How can you solve for $i$ or $n$?

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    and for n, just solve for $(1+i)^n$ and use the logarithm with base $1+i$, which you can then convert to natural logarithms if you prefer them.2012-10-05

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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.

$\begin{align*} &\frac{Fi}{A} = (1+i)^n - 1 \\ \Rightarrow &(1+i)^n = 1 + \frac{Fi}{A} \\ \Rightarrow &n \ln (1+i) = \ln(1 + \frac{Fi}{A}) \\ \Rightarrow &n = \frac{\ln(1 + \frac{Fi}{A})}{\ln(1+i)} \end{align*}$

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    @PerManne Sure, good point. So, if you want to achieve some future value $F$ and you end up with a value of $n$ that is not an integer, you would round *up* to figure out how many payments of $A$ are needed. But, at the same time, another way to think about it would be to have the final payment be at time $n$, even if $n$ is not an integer. In that case, the overall future value would be very close to $F$ at that time $n$.2012-10-05
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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 $n$.