document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1153513>Question 1153513</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Edgar accumulated $5,000 in credit card debt. If the interest rate is 30% per year, and he does not make any payments for 3 years, how much will he owe (in dollars) on this debt in 3 years by each method of compounding? (Simplify your answers completely. Round your answers to the nearest cent.)<BR>\n" );
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document.write( "compound quarterly<BR>\n" );
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document.write( "(b)<BR>\n" );
document.write( "compound monthly<BR>\n" );
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document.write( "(c)<BR>\n" );
document.write( "compound continuously<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #775749 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=775749\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the discrete compounding formula is f = p * (1 + r) ^ n<BR>\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 then number of time periods\r<BR>\n" );
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document.write( "in your problem, you are given:<BR>\n" );
document.write( "f = what you want to find<BR>\n" );
document.write( "p = 5000<BR>\n" );
document.write( "r = 30% per year / 100 = .3 per year (percent /  100  = rate).<BR>\n" );
document.write( "n = 3 years\r<BR>\n" );
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document.write( "if you compound annually, the formula becomes:\r<BR>\n" );
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document.write( "f = 5000 * (1 + .3) ^ 3 = 10985\r<BR>\n" );
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document.write( "if you compound quarterly, the formula becomes:\r<BR>\n" );
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document.write( "f = 5000 * (1 + .3 / 4) ^ (3 * 4) = 11908.898\r<BR>\n" );
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document.write( "if you compound monthly, the formula becomes:\r<BR>\n" );
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document.write( "f = 5000 * (1 + .3 / 12) ^ (3 * 12) = 12162.67658\r<BR>\n" );
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document.write( "if you compound continuously, a different formula is used.\r<BR>\n" );
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document.write( "that formula is f = p * e ^ (r * n)<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "e is the scientific constant of 2.718281828.......<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" );
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document.write( "with this formula, you leave the time periods in terms of years.<BR>\n" );
document.write( "it will make no difference what time periods and compounding periods you use, the answer will be the same.<BR>\n" );
document.write( "most of the time you will just give it the interest rate per year and the number of years.\r<BR>\n" );
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document.write( "the reason is as follows:\r<BR>\n" );
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document.write( "r * n = .3 * 3 = .9 when giving it rate and time in terms of years.<BR>\n" );
document.write( "r * n = .3 / 4 * 3 * 4 = .9 when giving it rate and time in terms of quarters.<BR>\n" );
document.write( "r * n = .3 / 12 * 3 * 12 = .9 when giving it rate and time in terms of months.\r<BR>\n" );
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document.write( "in your problem, the formula becomes f = 5000 * e ^ (.3 * 3) = 12298.01556.\r<BR>\n" );
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document.write( "the more compounding periods per year, the higher the future value.<BR>\n" );
document.write( "the highest is when you compound continuously.\r<BR>\n" );
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document.write( "this is apparent from the data.\r<BR>\n" );
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document.write( "5000                              future value in 3 years\r\n" );
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document.write( "compound annually                       10985\r\n" );
document.write( "compound quarterly                      11908.898\r\n" );
document.write( "compound monthly                        12162.67658\r\n" );
document.write( "compound continuously                   12298.01556\r\n" );
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