Is it possible to transform this equation to give R? $$y=x\left[\frac{\left(1+\frac{R}{12}\right)^{12\times{25}}}{\frac{R}{12}}-1\right]$$
Formula transformation
-
0I've edited your question. Please check whether it is what you meant to write. I appreciated your handwork, but please use LaTeX in future :) – 2012-10-09
-
0For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex). – 2012-10-09
-
0related: http://math.stackexchange.com/questions/207864/how-to-solve-for-i-and-n-in-compound-interest-formula – 2012-10-09
2 Answers
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.
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.