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Is it possible to transform this equation to give $R$? $y=x\left[\frac{\left(1+\frac{R}{12}\right)^{12\times{25}}}{\frac{R}{12}}-1\right]$

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    related: http://math.stackexchange.com/questions/207864/how-to-solve-for-i-and-n-in-compound-interest-formula2012-10-09

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Letting $w = \frac y x + 1$ and $r = \frac R {12}$, we are left with inverting $ w = \frac {(1+r)^{300}} r $ $ w = \frac {(1+r)^{300}} {(r^{1/300})^{300}} $ $ w^{1/300} = r^{-1/300} + r^{299/300} $ Letting $z = r^{1/300}$ and multiplying by $z$ we have $ z^{300} - w^{1/300}z + 1 = 0$

This is a trinomial equation of degree 300 in a form similar to Glasser's form, which you can read about here. (You can get Glasser's form exactly by substituting $z$ with $cz$ with an appropriate constant $c$, dependent on $w$ of course.)

The answer's not pretty. You might be better off solving it numerically.

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You could rewrite like $\frac{R}{12}\left(\frac yx+1\right)-\sum_{k=0}^{12\times 25}\binom{12\times 25}{k}\left(\frac R{12}\right)^k=0$, but to my knownledge, there is no general way to solve that other than numerical.