Annuity Equation isolating n value HELP

  • #1
Annuity Equation "isolating n value" HELP!

[tex]R =\frac{Pi(1+i)^n}{(1+i)^n-1}[/tex]

1. I need to solve for n.
  • n = number of years
  • P = Investment
  • R = Yearly Savings
  • i = interest rate

3. My Attempt:

[tex]R =\frac{Pi(1+i)^n}{(1+i)^n-1}\rightarrow -1 \cdot R = \frac{Pi(1+i)^n}{(1+i)^n} \cdot -1 [/tex]

Once I get here I'm left with:

[tex]-R =\frac{Pi(1+i)^n}{(1+i)^n}[/tex]

It seems like the quantities of [tex](1+i)^n[/tex] cancel so I'm left with:

[tex]R = Pi[/tex]

I know this isn't correct, and I've tried it many other ways with no luck.

Any help would be greatly appreciated!
Last edited:
  • #2
Welcome to PF!

[tex]R =\frac{Pi(1+i)^n}{(1+i)^n-1}[/tex]

Hi LordofDirT ! Welcome to PF! :smile:

(nice LaTeX, by the way … I think I'll just copy-and-paste it! :rolleyes:)

Hint: [tex]\frac{(1+i)^n}{(1+i)^n-1}\,=\,1\,+\,\frac{1}{(1+i)^n-1}[/tex] :smile:
  • #3
I would be inclined to simplify at first: let 1+ i= u. Then the equation is
[tex]R= \frac{Piu^n}{u^n-1}[/tex]
then rather than use tiny-tim's method (which is perfectly good) I would multiply on both sides by un- 1:
[tex]Ru^n- R= Pi u^n[/tex]
isolate the un term:
[tex]Ru^n- Pi u^n= (R- Pi)u^n= R[/itex]
[tex]u^n= \frac{R}{R- Pi}[/itex]
to solve for n, you will now have to take a logarithm of both sides.

(Don't forget to put 1+ i back in for u in the answer.)
  • #4
the logarithm

So to isolate the n variable in [tex]u^n= \frac{R}{R- Pi}[/tex]

would i have to take the logarithm base "u" on both sides? Or will any logarithm work?
  • #5
Hi LordofDirT! :smile:

Yes … n = (logR - log(R - pi))/logu, for any base of log. :smile:

btw, the reason I suggested my way (instead of HallsofIvy's, which is fine) was to get (1+i)^n just once in the equation … I reckon I'm quite likely to make mistakes, and that lessens the possiblity slightly! :smile:
  • #6

hi everybody!

please help... i need to solve the i value for this formula...

P = the principal = (63,000.00)
R = the amortized payment = (2750.00)
n = the Terms = (36 Months)

P = R((1-(1/(1+i)power of n))/i)

please i really need help to solve for i value...

  • #7
Welcome to PF!

Hi gardzrecah ! Welcome to PF! :smile:

(btw, it's always better to start a new thread … more people will see it)

(and try using the X2 tag just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

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