document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=598014>Question 598014</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many years does it take for an investment to double in value if it is invested at 6%<BR>\n" );
document.write( "if interest is compounded quarterly?\r<BR>\n" );
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document.write( "if interest is compounded continuously?<BR>\n" );
document.write( "thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #378425 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=378425\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How many years does it take for an investment to double in value if it is invested at 6%<BR>\n" );
document.write( "if interest is compounded quarterly? <BR>\n" );
document.write( "if interest is compounded continuously?<BR>\n" );
document.write( "**<BR>\n" );
document.write( "compound interest formula: A=P(1+r)^t, P=initial investment, r=interest rate per period, t=number of periods, A=amount after t periods.<BR>\n" );
document.write( "For continuous compounding: A=Pe^rt<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Quarterly compounding: A/P=2, r=.06/4=.015, t=quarters<BR>\n" );
document.write( "A=P(1+r)^t<BR>\n" );
document.write( "A/P=(1+.06/4)^t<BR>\n" );
document.write( "A/P=(1+.015)^t<BR>\n" );
document.write( "2=(1.015)^t<BR>\n" );
document.write( "take log of both sides<BR>\n" );
document.write( "log(2)=t*log(1.015)<BR>\n" );
document.write( "t=log(2)/log(1.015)<BR>\n" );
document.write( "t&#8776;46.55 qtrs&#8776;11.64 years<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Continuous compounding: A/P=2, r=.06, t=years<BR>\n" );
document.write( "A=Pe^rt<BR>\n" );
document.write( "A/P=e^rt<BR>\n" );
document.write( "2=e^rt<BR>\n" );
document.write( "2=e^.06t<BR>\n" );
document.write( "take log of both sides<BR>\n" );
document.write( "ln2=.06t*lne<BR>\n" );
document.write( "lne=1<BR>\n" );
document.write( ".06t=ln2<BR>\n" );
document.write( "t=ln2/.06<BR>\n" );
document.write( "t&#8776;11.55 years<BR>\n" );
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