Annuity Equation isolating n value HELP

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Homework Help Overview

The discussion revolves around isolating the variable n in the annuity equation, specifically in the context of financial mathematics. Participants are exploring the manipulation of the equation to solve for n, which represents the number of years in an annuity investment scenario.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate the annuity equation but expresses confusion about the correctness of their approach. Some participants suggest alternative methods for isolating n, including the use of logarithms. Others question the appropriateness of the logarithm base to use in the calculations.

Discussion Status

Participants are actively engaging with the problem, offering hints and alternative methods. There is a mix of approaches being discussed, with no explicit consensus on a single method yet. Some guidance has been provided regarding the use of logarithms and simplifications of the equation.

Contextual Notes

One participant introduces a different equation related to solving for the interest rate i, indicating that multiple related problems are being discussed within the thread. There is a suggestion that starting a new thread might be beneficial for clarity.

LordofDirT
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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:
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Welcome to PF!

LordofDirT said:
[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:
 
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]<br /> [tex]u^n= \frac{R}{R- Pi}[/itex]<br /> to solve for n, you will now have to take a logarithm of both sides.<br /> <br /> (Don't forget to put 1+ i back in for u in the answer.)[/tex][/tex]
 
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?
 
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:
 


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...


thanks,...
 
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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