document.write( "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)\r
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document.write( "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 \n" );
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Algebra.Com's Answer #373478 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A=P(1+r/12)^t. \n" ); document.write( " A= amount of money total i paid for the car, \n" ); document.write( " P= amount of loan i needed) \n" ); document.write( " r= interest rate and \n" ); document.write( " t= length of loan(in months) \n" ); document.write( ": \n" ); document.write( "For my particular case, I have: A= 26641.12, P= 21000, r= .0349 (3.49%), and t= 48. \n" ); document.write( ": \n" ); document.write( "you give t=48, but isn't that what we are trying to find here? \n" ); document.write( "And 1.25 * 21000 = 26250 \n" ); document.write( ": \n" ); document.write( "21000[1+(.0349/12))^t] = 26250 \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "Using common logs \n" ); document.write( " \n" ); document.write( "The log equiv of exponents \n" ); document.write( " \n" ); document.write( "find the logs \n" ); document.write( "t * .00126 = .09691 \n" ); document.write( "t = \n" ); document.write( "t = 76.9, pay it off in 76 months to pay less that $26250 \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this, enter in you calc: 21000(1+(.0349/12))^76 results: $26,186.26 is what you'll pay in 76 months \n" ); document.write( " \n" ); document.write( " |