document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=38435>Question 38435</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If $1000 is invested at 7.5% compounded annually, find the future value that results in 30 years to the nearest $1. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #23952 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=23952\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I think the equation here is<BR>\n" );
document.write( "P = P0 * e^(rt)<BR>\n" );
document.write( "so plugging in we get<BR>\n" );
document.write( "P = 1000 * e^(.075 * 30)<BR>\n" );
document.write( "P = 9487.74 or $9488 to the nearest dollar<BR>\n" );
document.write( "I've had second thoughts on this one...the formula I gave you was when the monies are compounded continuously, not annually as the problem poses...<BR>\n" );
document.write( "Thus I think the correct formula is<BR>\n" );
document.write( "P = P0 * (1 + r)^t<BR>\n" );
document.write( "P = 1000 * (1.075)^30<BR>\n" );
document.write( "P = 8754.96 or $8755 to the nearest dollar </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );