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by **themainadam** on Tue Feb 04, 2014 11:36 pm

Hi all,

Is it possible to rearrange this equation in terms of n

an = b * ( (1-r^n) / (1-r) )

Its just beyond me!

a is always much greater than b in this case.

Thanks
Adam

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by **stapel_eliz** on Wed Feb 05, 2014 12:20 pm

themainadam wrote:Is it possible to rearrange this equation in terms of n

an = b * ( (1-r^n) / (1-r) )

"Possible"? Yes. "Reasonably feasible"? No.

According to Wolfram Alpha, the "solution" is as follows:

. . . . . $n\, =\, \frac{a\,W\left(-\frac{b\,r^{\frac{b}{a(1-r)}}\,\log(r)}{a(r\,-\, 1)}\right)\, -\, a\,r\,W\left(-\frac{b\,r^{\frac{b}{a(1-r)}}\,\log(r)}{a(r\,-\, 1)}\right)\, -\, b\,\log(r)}{a\,(r\, -\, 1)\,\log(r)}$

...where "W(x)" is the Lambert W-function. :wink:

by **themainadam** on Wed Feb 05, 2014 2:47 pm

wow!
I tried logs, differentiation, integration and all the other tricks I could barely remember from my school days 20 years ago - but there is no way I could have got there.
Thank you so much for taking the time to answer my question. It's interesting to learn something new about maths - I've never heard of the Lambert W-Function before.

Thanks again,
Adam

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