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We have a simple calculation that we are having troubles solving for n.

A=P((r(1+r)^(n))/((1+r)^(n)-1))

http://www.mathway.com/math_image.as...03)?p=127?p=42

solve for n please?

Is this possible? I've tried for a couple of hours but have failed and I need it for some software I'm writing. Any help would be welcomed!

Re: We have a simple calculation that we are having troubles solving for n.

$\displaystyle A = Pr(\frac{(1+r)^n}{(1+r)^n - 1})$

$\displaystyle \frac{A}{Pr} = \frac{(1+r)^n}{(1+r)^n - 1} = 1 + \frac{1}{(1+r)^n - 1}$

Can you take it from here?

Re: We have a simple calculation that we are having troubles solving for n.

great thanks, yes that helped a lot. How did you get to there? The rest I've got.

Re: We have a simple calculation that we are having troubles solving for n.

$\displaystyle \frac{(1+r)^n}{(1+r)^n - 1} =$

$\displaystyle \frac{(1+r)^n - 1 + 1}{(1+r)^n - 1} = $

$\displaystyle \frac{(1+r)^n - 1}{(1+r)^n - 1} + \frac{1}{(1+r)^n-1} =$

$\displaystyle 1 + \frac{1}{(1+r)^n - 1}$

Re: We have a simple calculation that we are having troubles solving for n.

Hello, iamblackhawk!

Quote:

$\displaystyle \text{Solve for }n\!:\;\;A \:=\:P\frac{r(1+r)^n}{(1+r)^n-1}$

Multiply by $\displaystyle (1+r)^n-1\!:$

. . . . . . . $\displaystyle A\big[(1+r)^n-1\big] \;=\;Pr(1+r)^n$

. . . . . . . .$\displaystyle A(1+r)^n - A \;=\;Pr(1+r)^n$

m $\displaystyle A(1+r)^n - Pr(1+r)^n \;=\;A$

Factor:.$\displaystyle (A - Pr)(1+r)^n \;=\;A$

. . . . . . . . . . . .$\displaystyle (1+r)^n \;=\;\frac{A}{A-Pr}$

Take logs: n . $\displaystyle \ln(1+r)^n \;=\;\ln\left(\frac{A}{A-Pr}\right) $

. . . . . . . . n $\displaystyle n\ln(1+r) \;=\;\ln\left(\frac{A}{A-Pr}\right) $

. . . . . . . . . . . . . . . $\displaystyle n \;=\;\dfrac{\ln\left(\frac{A}{A-Pr}\right)}{\ln(1+r)}$