document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187689>Question 1187689</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if savings in an account are to be at least tripled over 20 years with an interest rate of 5.55% p.a, how often should the interest be compounded? Give your answer in months.\r<BR>\n" );
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document.write( "The answer from textbook: at least every 4 months\r<BR>\n" );
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document.write( "I calculated my own answer and is 1.0169 something. Can someone please show me how the question should be solved? Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #818717 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=818717\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i'm not sure there's a formula that will let you get the answer directly.<BR>\n" );
document.write( "it appears to be an iterative approach.\r<BR>\n" );
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document.write( "the formula to use is f = p * (1 + r/c) ^ (n*c)\r<BR>\n" );
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document.write( "f is the future value which will be 3.<BR>\n" );
document.write( "p is the present value which will be 1.<BR>\n" );
document.write( "r is the interest rate per year (percent divided by 100).<BR>\n" );
document.write( "c is the number of compounding periods per year.<BR>\n" );
document.write( "n is the number of years.\r<BR>\n" );
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document.write( "if the number of compounding periods per year is 1, then the formula becomes:<BR>\n" );
document.write( "f = 1 * (1 + .0555/1) ^ (20 * 1)<BR>\n" );
document.write( "you get f = 2.945538834.\r<BR>\n" );
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document.write( "if the number of compounding periods per year is 2, then the formula becomes:<BR>\n" );
document.write( "f = 1 * (1 + .0555/2) ^ (20 * 2)<BR>\n" );
document.write( "you get f = 2.988817642.\r<BR>\n" );
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document.write( "if the number of compounding periods per year is 3, then the formula becomes:<BR>\n" );
document.write( "f = 1 * (1 + .0555/3) ^ (20 * 3)<BR>\n" );
document.write( "you get f = 3.003737664.\r<BR>\n" );
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document.write( "if the number of compounding periods per year is 4, then the formula becomes:<BR>\n" );
document.write( "f = 1 * (1 + .0555/4) ^ (20*4)<BR>\n" );
document.write( "you get f = 3.011293958.\r<BR>\n" );
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document.write( "if the number of compounding periods per year is 5, then the formula becomes:<BR>\n" );
document.write( "f = 1 * (1 + .0555/5) ^ (20*5)<BR>\n" );
document.write( "you get f = 3.01585904\r<BR>\n" );
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document.write( "it looks like 3 compounding periods per year is about where it will be.<BR>\n" );
document.write( "that's the closest to f = 3.\r<BR>\n" );
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document.write( "3 compounding periods per year is equal to 4 months each because 3 * 4 = 12.\r<BR>\n" );
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document.write( "i tried to do it directly by getting 3 = 1 * (1 + .0555/x) ^ (20*x), but couldn't solve that by logarithms because x was in the log function itself and also in the exponent.\r<BR>\n" );
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document.write( "i then tried to solve it graphically.<BR>\n" );
document.write( "i got x = 2.668.\r<BR>\n" );
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document.write( "when x = 2.668, the formula becomes y = 1 * (1 + .0555/2.668) ^ (20 * 2.668).<BR>\n" );
document.write( "you get f = 3.000000595.\r<BR>\n" );
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document.write( "that's much closer, and would be even closer if the graphing software didn't rounded the answer to 3 decimal digits.\r<BR>\n" );
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document.write( "here's what the graphing software output looks like.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2021/111307.jpg\" >\r<BR>\n" );
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document.write( "the graphing software i used online was at desmos.com.\r<BR>\n" );
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document.write( "i also used my ti-84 plus graphing software.\r<BR>\n" );
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document.write( "that gave me x = 2.6679529.\r<BR>\n" );
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document.write( "when x = 2.6679529, y = (1 + .0555/2.6679529) ^ (20 * 2.6679529) gave me:<BR>\n" );
document.write( "f = 3.\r<BR>\n" );
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document.write( "there may be a way to get it directly through formula, but i wasn't able to do it. </SPAN>\n" );
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