A=P((r(1+r)^(n))/((1+r)^(n)-1))
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!
A=P((r(1+r)^(n))/((1+r)^(n)-1))
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!
$\displaystyle \frac{(1+r)^n}{(1+r)^n - 1} =$
$\displaystyle \frac{(1+r)^n - 1 + 1}{(1+r)^n - 1} = $
$\displaystyle \frac{(1+r)^n - 1}{(1+r)^n - 1} + \frac{1}{(1+r)^n-1} =$
$\displaystyle 1 + \frac{1}{(1+r)^n - 1}$
Hello, iamblackhawk!
$\displaystyle \text{Solve for }n\!:\;\;A \:=\:P\frac{r(1+r)^n}{(1+r)^n-1}$
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) $
. . . . . . . . . . . . . . . $\displaystyle n \;=\;\dfrac{\ln\left(\frac{A}{A-Pr}\right)}{\ln(1+r)}$