document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140160>Question 1140160</A>: <I><SPAN STYLE=\"word-break: break-all;\"> john invested $20 000 on 1/06/2009 at 9% p.a. interest compounded monthly. <BR>\n" );
document.write( "a) what is the value of the investment by 1/06/2019<BR>\n" );
document.write( "b) If he only takes the interest earned on this investment on this date and invests this at 10.5% p.a. (compounded continuously), how much can he expect back by 1/06/2025 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760891 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=760891\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it will be 20000(1+.09/12)^1120=20000*(2.45 but round at end)=$49027.14 by 1/6/19\r<BR>\n" );
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document.write( "the interest is $29027.14 and that is multiplied by e^(.105*6)or e^0.735 =$60535.58 by 1/6/25<BR>\n" );
document.write( "That formula is P=Poe^rt<BR>\n" );
document.write( "Rule of 72 would say the first would double in 8 years and be about 2.5 times<BR>\n" );
document.write( "The second would double in under 7 years, which it has. </SPAN>\n" );
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