document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1136388>Question 1136388</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose $2,900 is invested in an account at an annual interest rate of 3.5% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #754174 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=754174\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the continuous compounding formula is f = p * e ^ (r * n)\r<BR>\n" );
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document.write( "f if 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 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 = 2900<BR>\n" );
document.write( "f = 2 * 2900<BR>\n" );
document.write( "r = 3.5% / 100 = .035 per year.<BR>\n" );
document.write( "n = the number of years you want to find.\r<BR>\n" );
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document.write( "the formula becomes:\r<BR>\n" );
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document.write( "2 * 2900 = 2900 * e ^ (.035 * n)\r<BR>\n" );
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document.write( "simplify this to get:\r<BR>\n" );
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document.write( "5800 = 2900 * e ^ (.035 * n)\r<BR>\n" );
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document.write( "divide both sides of this equation by 2900 to get:\r<BR>\n" );
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document.write( "2 = e ^ (.035 * n)\r<BR>\n" );
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document.write( "take the natural log of both sides of this equation to get:\r<BR>\n" );
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document.write( "ln(2) = ln(e ^ (.035 * n)\r<BR>\n" );
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document.write( "since ln(a^b) = b*ln(a), this equation becomes:\r<BR>\n" );
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document.write( "ln(2) = .035 * n * ln(e).\r<BR>\n" );
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document.write( "since ln(e) = 1, this equation becomes:\r<BR>\n" );
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document.write( "ln(2) = .035 * n\r<BR>\n" );
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document.write( "divide both sides of this equation by .035 to get:\r<BR>\n" );
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document.write( "ln(2) / .035 = n\r<BR>\n" );
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document.write( "solve for n to get:\r<BR>\n" );
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document.write( "n = ln(2) / .035 = 19.80420516.\r<BR>\n" );
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document.write( "confirm by replacing n in the original equation of 5800 = 2900 * e ^ (.035 * n) to get:\r<BR>\n" );
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document.write( "5800 = 2900 * e ^ (.035 * 19.80420516) which becomes:\r<BR>\n" );
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document.write( "5800 = 5800\r<BR>\n" );
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document.write( "this confirms the solution is correct.\r<BR>\n" );
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document.write( "the solution is that investment will double in 19.8 years rounded to the nearest tenth of a year.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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