Question

Use the given formula to solve the following problem: When the amount financed is $15,480, the monthly payment is $621.33, and the monthly interest rate is 0.00583, how long do you have to pay?

A. 18 months
B. 23 months
C. 27 months
D. 32 months
@dan815

Answer 1

Are you using a financial calculator, or are you doing this by hand?

Answer 2

By hand

Answer 3

use a calculator

Answer 4

I tried that its not working

Answer 5

I cant figure it out :(

Answer 6

ok hang on

Answer 7

K

Answer 8

http://www.wolframalpha.com/input/?i=log%28621.33%2F%28621.33+-+15480+*+0.00583%29%29+%2F+log%281.00583%29

Answer 9

Arm thanks but that helps me out none at this point cause I dont knownthe order

Answer 10

P = $621.33
A = $15,480
r = 0.00583
Plug these three numbers into the formula.

Answer 11

Dan just pm me

Answer 12

@dan815 ^^^

Answer 13

I figured it out

Answer 14

This is the equation i found
its kinda long, bear with me
P=15480
a=621.33
r=0.00583

Answer 15

Its ok Ps I still haven't event got to the q

Answer 16

So hence y I said pm me @dan815

Answer 17

ok sry internet crashed

Answer 18

P=Payment to be made in total
A=Monthly installements for the payment
R=rate of interset per month, that is adde to the money that is left over to pay

Answer 19

1st month before the interest rate kicks in she will pay A, so P-A left over to pay
P-A
2 nd month, now there is an interest on the payment left over and then A is subtracted
(P-A)+r(P-A) - A,
=(P-A)*(1+r) -A
=y*k +y-p, if P-A=y , 1+r =k
3rd Month
=(yk+y-p)*k +y-p
=k^2y+yk+y-kp-p
4th month
=k^3y+k^2y+ky+y-k^2p=kp-p
...
n+1 months
=k^ny+k^n-1y....k^0y-k^(n-1)p-..-k^0p
=y((k^(n+1)-1)/(k-1))-p((k^n-1)/(k-1) =0 <--- equal to zero before after n months we have 0 payment left