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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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I've edited your question. Please check whether it is what you meant to write. I appreciated your handwork, but please use LaTeX in future :) –  Johnny Westerling Oct 9 '12 at 10:06
    
For some basic information about writing math at this site see e.g. here, here, here and here. –  Américo Tavares Oct 9 '12 at 10:21
    
related: math.stackexchange.com/questions/207864/… –  Henry Oct 9 '12 at 11:37

2 Answers 2

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.

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