document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1054018>Question 1054018</A>: <I><SPAN STYLE=\"word-break: break-all;\"> suppose that $14018 is invested at an interest rate of 6.3% per year, compounded continuously.<BR>\n" );
document.write( "Find the exponential function that describes the amount in the account after time t, in years.<BR>\n" );
document.write( "what is the doubling time? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #669244 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=669244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the continuoous compounding formula is:\r<BR>\n" );
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document.write( "f = p * e^(r*n)\r<BR>\n" );
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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( "in your problem:\r<BR>\n" );
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document.write( "p = 14018<BR>\n" );
document.write( "r = .063 per year<BR>\n" );
document.write( "n = t years\r<BR>\n" );
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document.write( "the formula becomes f = 14018 * e^(.063 * t)\r<BR>\n" );
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document.write( "you want to know what is the doubling time.\r<BR>\n" );
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document.write( "if 14018 doubles, then it is worth 28036.\r<BR>\n" );
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document.write( "the formula becomes 28036 = 14018 * e^(.063 * t)\r<BR>\n" );
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document.write( "divide both sides of the equation by 14018 to get 28036 / 14018 = e^(.063 * t)\r<BR>\n" );
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document.write( "simplify to get 2 = e^(.063 * t)\r<BR>\n" );
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document.write( "take the natural log of both sides of the equation to get ln(2) = ln(e^(.063 * t).\r<BR>\n" );
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document.write( "since ln(e^x) = x*ln(e), your equation becomes ln(2) = .063 * t * ln(e).\r<BR>\n" );
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document.write( "since ln(e) = 1, your equation becomes ln(2) = .063 * t\r<BR>\n" );
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document.write( "divide both sides of the equation by .063 to get ln(2) / .063 = t.\r<BR>\n" );
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document.write( "solve for t to get t = ln(2) / .063 = 11.0023362\r<BR>\n" );
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document.write( "thqt's the number of years it would take for the money to double at 6.3% per year using continuous compounding.\r<BR>\n" );
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document.write( "14018 * e^(.063 * 11.0023362) is equal to 28036. </SPAN>\n" );
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