document.write( "Question 136366: Do you know compounding interest; Pe rt; find the \"r\" rate of interest
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document.write( "At what interest rate must 4800 be compounded annually to equal 9051.12 after 13 years, trying to find a formula like this in one of my 6 books, and I could find P, T, A but not \"r\",
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document.write( "raysecrets @hotmail.com, I have been working on this homework problem for 2 days now. Any help will be great. \n" );
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Algebra.Com's Answer #99888 by scott8148(6628)![]() ![]() You can put this solution on YOUR website! FV=P(1+r)^n \n" ); document.write( "__ FV is the future value \n" ); document.write( "__ P is the starting amount (principal) \n" ); document.write( "__ r is the interest rate per compounding period (as a decimal) \n" ); document.write( "__ n is the number of compounding periods\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the \"Pe rt\" is for continuous compounding __ e is the base for natural logs ( Euler's number)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "9051.12=4800(1+r)^13 __ dividing by 4800 __ 9051.12/4800=(1+r)^13\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "taking the log __ log(9051.12/4800)=13(log(1+r))\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "dividing by 13 __ (log(9051.12/4800))/13=log(1+r)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "taking antilog __ 10^((log(9051.12/4800))/13)=1+r\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "subtracting 1 __ (10^((log(9051.12/4800))/13))-1=r\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |