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It's been years since I've had to do math, and to be honest, I'm not sure where to begin with this formula.

Can someone refresh my brain or where should I start?

Questions are below.

The formula is: $N = -(1/30) \cdot \ln(1+b/p(1-(1+i)^{30})) / \ln(1+i)$

  • $n=$ months
  • $b=$ credit card balance
  • $p=$ monthly payment
  • $i=$ daily interest rate (annual interest rate/365)

The question are:

1) What APR value will allow Alice to pay off a $\$7,500.00$ balance in $40$ months if she pays $\$250.00$ per month?

2) What monthly payment amount will allow Alice to pay off a $\$7,500.00$ balance in $40$ months if the APR value is $0.21$?

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  • $\begingroup$ What is N supposed to represent? $\endgroup$ – fleablood Oct 7 '15 at 22:08
  • $\begingroup$ For number 1 we plug in n = 40; b=7500; p=250; i = APR/365 So we get N = -(1/30).ln(1 + 7500/250(1 - (1 + APR/365)^30))/ln(1 + APR/365). And we solve for APR but I need to know what N is. $\endgroup$ – fleablood Oct 7 '15 at 22:12
  • $\begingroup$ Oh. N = n! The capitals confused me. $\endgroup$ – fleablood Oct 7 '15 at 22:13
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Okay:

Formula: $n = -(1/30) \cdot \ln(1+b/p(1-(1+i)^{30})) / \ln(1+i)$

First problem

n = 40

b = 7500

p = 250

i = APR/365; we will solve for APR

plug those in

$ 40 = -(1/30) \cdot \ln(1+7500/250(1-(1+APR/365)^{30})) / \ln(1+APR/365) =-(1/30) \cdot \ln(1+30(1-(1+APR/365)^{30}))/ \ln(1+APR/365)$

multiply both side by 30 and then by $ \ln(1+APR/365)$

$1200 = -\ln(1+30(1-(1+APR/365)^{30}))/ \ln(1+APR/365)$

$1200\ln(1+APR/365) = \ln(1+30(1-(1+APR/365)^{30}))$

We get rid of the $ln$s by rising e to both powers. (Trust me, the e's will vanish.)

$e^{1200\ln(1+APR/365)} = e^{-\ln(1+30(1-(1+APR/365)^{30}))} $

$(e^{\ln(1+APR/365)})^{1200} = (e^{\ln(1+30(1-(1+APR/365)^{30}))})^{-1} $

$(1+APR/365)^{1200} = (1+30(1-(1+APR/365)^{30}))^{-1}= \frac{1}{(1+30(1-(1+APR/365)^{30}))} $

Okay, there's no way in heck I'm going to deal with 1200 roots. So I'm going to cheat.

By the binomial theorem we know $(1 + x)^n = 1 + nx + n(n-1)/2*x^2 + ....$ and if x is very small all the x^i terms will become so small as to be insignificant. So I can estimate $$(1 + x)^n = 1 + nx$ (but only for very small values of x.) Now APR/365 will be a very small number. So I will approximate $(1 + APR/365)^{1200} = 1 + 1200APR365$ and $(1 + APR/365)^{30} = 1 + 30APR365$. So:

$1+1200APR/365 = \frac{1}{(1+30(1-(1+30APR/365)))} = \frac{1}{1+30(-30APR/365)}= \frac{1}{1-900APR/365}$

Multiply both sides by $(1-900APR/365)$

$(1+1200APR/365)(1-900APR/365) = 1$

Pour yourself a stiff drink and expand:

$1 + 1200APR/365 - 900APR/365 - 1200*900(APR/365)^2 = 1$

$1200*900(APR/265)^2 = 300APR/365$

We will assume APR isn't 0 so we divide both sides by $300APR/365$ to get

$1200*900(APR/365)^2/(300APR/365) = 1$

$360000APR/365 = 1$

so

APR = 365/360000 = 0.00101388888888888888888888888889 = 1.01%

Unless I made an error which I almost certainly did. SHEESH

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  • $\begingroup$ I did make a mistake . APR = .1 percent which is clearly way to low. $\endgroup$ – fleablood Oct 8 '15 at 1:11

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