document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171745>Question 1171745</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Joe decides  to invest USD 5400 in an account with a 2.4%  nominal annual interest rate.<BR>\n" );
document.write( "Given that the interest is compounded monthly, find the minimum number of years that it will take for Joe 's investment to triple . Give your answer correct to the nearest year.<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 #796670 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=796670\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There is a rule of 110 for tripling that would suggest 110/2.4=45.83 years.<BR>\n" );
document.write( "That comes from ln 3=1.0986 which is 110 when converted to per cent.\r<BR>\n" );
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document.write( "P=Po(1+0.024/12)^12n<BR>\n" );
document.write( "P/Po=3<BR>\n" );
document.write( "so 3=1.002^12n<BR>\n" );
document.write( "ln 3=12n ln 1.002<BR>\n" );
document.write( "12n=549.86. That is the number of months<BR>\n" );
document.write( "n=45.82 years in close agreement with the above.<BR>\n" );
document.write( "Answer is 46 years. </SPAN>\n" );
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