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

  1. #1
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    Cool We have a simple calculation that we are having troubles solving for n.

    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!


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

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

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

    Can you take it from here?
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    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.
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    Re: We have a simple calculation that we are having troubles solving for n.

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

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

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

    1 + \frac{1}{(1+r)^n - 1}
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  5. #5
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    Re: We have a simple calculation that we are having troubles solving for n.

    Hello, iamblackhawk!

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

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

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

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

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

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

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

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

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

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