# Using This Formula to Figure Out Credit Card Debt

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$0.21$? • What is N supposed to represent? – fleablood Oct 7 '15 at 22:08 • 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. – fleablood Oct 7 '15 at 22:12 • Oh. N = n! The capitals confused me. – fleablood Oct 7 '15 at 22:13 ## 1 Answer 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

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