document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1128550>Question 1128550</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Alice invests $18,000 at age 30 from the signing bonus of her new job. She hopes the investments will be worth $36,000 when she turns 41. If the interest compounds continuously, approximately what rate (as a percent) of growth will she need to achieve her goal? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #745034 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=745034\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Alice invests $18,000 at age 30 from the signing bonus of her new job. She hopes the investments will be worth $36,000 when she turns 41. If the interest compounds continuously, approximately what rate (as a percent) of growth will she need to achieve her goal? <BR>\n" );
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document.write( "A(t) = P*e^(rt)<BR>\n" );
document.write( "36000 = 18000*e^(r*11)<BR>\n" );
document.write( "e^(11r) = 2<BR>\n" );
document.write( "----<BR>\n" );
document.write( "11r = ln2/lne<BR>\n" );
document.write( "11r = 0.693<BR>\n" );
document.write( "r = 0.063<BR>\n" );
document.write( "Answer:: rate = 6.3%<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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