document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=47892>Question 47892</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The amount of A in an account after t years from an initial principle P invested at an annual rate r compounded continuously is given by A=Pe^rt where r is expressed as a decimal. How many years will it take an initial investment of $1,000 to grow to $1,700 at the rate of 4.42% compunded continuously.\r<BR>\n" );
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document.write( "I came up with 12 years. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #31691 by </B><A HREF=/tutors/aboutme.mpl?userid=Nate><B>Nate(3500)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Nate><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=31691\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A = Pe^rt<BR>\n" );
document.write( "1700 = 1000e^0.0442t<BR>\n" );
document.write( "ln(1.7) = 0.0442t<BR>\n" );
document.write( "ln(1.7)/0.0442 = t<BR>\n" );
document.write( "About 12 days </SPAN>\n" );
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