document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140188>Question 1140188</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Question Help <BR>\n" );
document.write( "How long does it take for an investment to double in value if it is invested at <BR>\n" );
document.write( "20​% compounded​ continuously?\r<BR>\n" );
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document.write( "At 20% compound continuously, the investment doubles in about __ years. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760680 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=760680\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We use the exponential growth or decay formula for continuously compounded interest<BR>\n" );
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document.write( "A = Pe^rt, where P is the original amount, r is the annual interest rate, t is time in years<BR>\n" );
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document.write( "let x be the original amount, then<BR>\n" );
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document.write( "2x = x * e^(0.20t)<BR>\n" );
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document.write( "e^(0.20t) = 2<BR>\n" );
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document.write( "0.20t = ln(2)<BR>\n" );
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document.write( "t = ln(2)/0.20 = 3.4657<BR>\n" );
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document.write( "At 20% compound continuously, the investment doubles in about 3.5 years<BR>\n" );
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document.write( ": </SPAN>\n" );
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