SOLUTION: Okay so for a car project I have this problem and I need to find out how long it would take for me to pay for the car to pay less than 125% of what it is worth.- A=P(1+r/12)^t. A=

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Okay so for a car project I have this problem and I need to find out how long it would take for me to pay for the car to pay less than 125% of what it is worth.- A=P(1+r/12)^t. A=       Log On


   

Question 585645: Okay so for a car project I have this problem and I need to find out how long it would take for me to pay for the car to pay less than 125% of what it is worth.- A=P(1+r/12)^t. A= amount of money total i paid for the car,P= amount of loan i needed) r= interest rate and t= length of loan(in months)
For my particular case, I have: A= 26641.12, P= 21000, r= .0349 (3.49%), and t= 48. I know I have to use logarithms to get T down to find the time, but I'm stuck
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A=P(1+r/12)^t.
A= amount of money total i paid for the car,
P= amount of loan i needed)
r= interest rate and
t= length of loan(in months)
:
For my particular case, I have: A= 26641.12, P= 21000, r= .0349 (3.49%), and t= 48.
:
you give t=48, but isn't that what we are trying to find here?
And 1.25 * 21000 = 26250
:
21000[1+(.0349/12))^t] = 26250
:
1.0029083%5Et = 26250%2F21000
Using common logs
log%28%281.0029083%5Et%29%29 = log%28%2826250%2F21000%29%29
The log equiv of exponents
t%2Alog%28%281.0029083%29%29 = log%28%2826250%2F21000%29%29
find the logs
t * .00126 = .09691
t = .09691%2F.00126
t = 76.9, pay it off in 76 months to pay less that $26250
:
:
Check this, enter in you calc: 21000(1+(.0349/12))^76 results: $26,186.26 is what you'll pay in 76 months