document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1109513>Question 1109513</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How long will it take to triple your money at a nominal interest rate<BR>\n" );
document.write( "j1 = 12% if simple interest is allowed for part of a year?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #724553 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=724553\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> for simple interest, the formula would be:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = p * r * n + p\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "f is the future value.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you are interested in years, then your interest rate has to be per year.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you are interested in months, then your interest rate has to be per month.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you always have to adjust your interest rate to be applicable to the time period you are interested in.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "your interest rate was given per year.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you want to solve for years, then the formula would become:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = p * .12 * n + p\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you assume p = 1000, and you want to triple your money, then f = 3000 and your formula would become:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3000 = 1000 * .12 * n + 1000\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would then solve this formula for n.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would get n = (3000 - 1000) / (1000 * .12) = 2000 / 120 = 16.666666667 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you wanted to calculate in months, then you would divide .12 by 12 to get .01 per month.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n would then be in months.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the formula would become 3000 = 1000 * .01 * n + 1000\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would solve for n to get n = (3000 - 1000) / (1000 * .01) = 2000 / 10 = 200 months.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "with simple interest, the answer would be the same, whether used interest rate per year or interest rate per month.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "200 months divided by 12 results in 16.6666666667 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's simple interest.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "compound interest is a different story.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "with compound interest, the more compounding periods per year, the greater the future value, or the less time it takes to get the same future value.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the compound interest formula is f = p * (1 + r) ^ (n)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you start with 1000 and want to triple it, and your interest rate is .12 per year, and you want to compound once a year, then the formula becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3000 = 1000 * (1 + .12) ^ n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would then divide both sides of this equation by 1000 to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3 = (1 + .12) ^ n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "to solve for n, you would take the log of both sides of this equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) = log((1 + .12) ^ n)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since log(1 + .12) ^ n) is equal to n * log(1 + .12), the formula would become:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) = n * log(1 + .12)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "divide both sides of this equation by log (1 + .12) to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) / log(1 + .12) = n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "solve for n to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n = 9.694035413 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you wanted to find out how long it would take with monthly compounding, then you need to adjust r to be per month and n will be the number of months.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the monthly interest rate is .12 / 12 = .01\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the formula becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3000 = 1000 * (1 + .01) ^ n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the process is the same.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would divide both sides of the equation by 1000 to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "3 = (1 + .01) ^ n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would take the log of both sides of the equation to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) = log((1 + .01) ^ n)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since log(1 + .01) ^ n) is equal to n * log(1 + .01), the formula would become:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) = n * log(1 + .01)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would divide both sides of this equation by log(1 + .01) to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "log(3) / log(1 + .01) = n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would solve for n to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n = 110.409624 months.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "divide this by 12 to translate to years to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "n = 9.200802004 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "with annual compounding, it would take 9.694035413 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "with monthly compounding, it would take 110.409624 months, which is equivalent to 9.200802004 years.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it takes less time to triple your money with monthly compounding than with annual compounding.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the formula for annual compounding would become f = 1000 * (1 + .12) ^ 9.694035413 = 3000.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the formula for monthly compounding would become f = 1000 * (1 + .01) ^ 110.409624 = 2999.99999999\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it doesn't show up as 3000 because there is a small amount of internal rounding being done that doesn't make it come out right on.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if i use an online calculator that displays more digits, then the answer would become n = 110.40962405\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "using that figure in my calculator, i then get f = 1000 * (1 + .01) ^ 110.40962405 and the result becomes f = 3000.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "regardless, 2999.999999 is very close and rounds to 3000 if you need the answer to be within 2 decimal places or so.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );