Algebra problem (deriving a variable in a formula)

Question

How do I derive the $m$ in the formula: $$I=\left(1+\frac{r}{m}\right)^{mn} -1$$

all the values of the variables in the formula except $m$ is given and the question is find $m$. I just don't know how to derive the formula using the knowledge of Algebra I have.

In order to solve equations like these, you need the Lambert-W function. — David H, Jul 26 '14 at 11:26
Also, numerical methods. Could you clarify what you can use (Newton method ?) ? — Claude Leibovici, Jul 26 '14 at 11:29
@DavidH. Are you sure you could ? Just curious. Cheers :) OK, I checked but this leads to amonster, isn't ? — Claude Leibovici, Jul 26 '14 at 11:50
@MhenniBenghorbal. In fact you can if you write $$(I+1)^{\frac{1}{n}}=\left(1+\frac{r}{m}\right)^m$$ — Claude Leibovici, Jul 26 '14 at 11:54
@ClaudeLeibovici The expression I found was a ghastly creature indeed. Not something I'd expect from an algebra-precalc student. — David H, Jul 26 '14 at 12:04
@DavidH. Me neither ! Once I asked if there is a limit to teachers imagination. — Claude Leibovici, Jul 26 '14 at 12:07
$m$ is probably an integer. Try all values from $1$. If you find two values that bracket $I$ without reaching it, you can refine $m$ by dichotomic search. (There are more efficient methods but let's keep it simple for now.) — , Mar 18 '15 at 08:52

score 1 · Answer 1 · answered Mar 18 '15 at 09:30

Using the answer by David, we set $t=\dfrac rm$, and

$$\sqrt[nr]{1+I}=(1+t)^{1/t}=s,$$ or $$1+t=s^t=e^{t\ln(s)}.$$

Then $$(1+t)s=e^{(1+t)\ln(s)},$$ $$-(1+t)\ln(s)\,e^{-(1+t)\ln(s)}=-\frac{\ln(s)}s,$$ and $$-(1+t)\ln(s)=W(-\frac{\ln(s)}s).$$ Hence, $$m=\frac r{-\dfrac{W(-\frac{\ln(s)}s)}{\ln(s)}-1}.$$

David Holden · Answer 2 · 2014-07-26T12:01:29.653

this looks like a compound interest question, so we see that the somewhat esoteric Lambert-W function may soon be part of the technical toolbox of chartered accountants! in order to gain a clearer idea of what is happening, using only elementary algebra, OP may find it useful to make an initial substitution, say $ x = \frac{m}{r} $, so that the equation becomes: $$ I=(1+x^{-1})^{nrx} -1 $$ or $$ \sqrt[nr]{1+I} = (1+x^{-1})^x $$ where the LHS is a constant computable from the problem data

Algebra problem (deriving a variable in a formula)

2 Answers2