document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1081518>Question 1081518</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find out how long it takes a $3500 investment to double of it is invested at 8% compunded semiannuakky. Round to the nearest ten of a 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 #695721 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=695721\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If the money is invested at 8%,compounded semiannually, then the formula for its value after n years is:<BR>\n" );
document.write( "fv=3500(1+.08/2)^2n<BR>\n" );
document.write( "So:<BR>\n" );
document.write( "7000=3500(1.04)^2n<BR>\n" );
document.write( "2=1.04^2n<BR>\n" );
document.write( "ln 2=ln 1.04^2n<BR>\n" );
document.write( "ln 2=2n ln 1.04<BR>\n" );
document.write( "ln 2/2 ln 1.04=n<BR>\n" );
document.write( "n=8.83 years. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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