document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=480929>Question 480929</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A lake is stocked with 500 fish, and the fish population P increases according to the following model P=10,000/1+19e^-t/5, t greater or equal than 0, where t in the time in months. After how many months will the population reach 2000?\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #329300 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=329300\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P increases according to the following model P=10,000/1+19e^-t/5, t greater or equal than 0, where t in the time in months. After how many months will the population reach 2000?<BR>\n" );
document.write( "------------<BR>\n" );
document.write( "P=10,000/1+19e^-t/5<BR>\n" );
document.write( "2000 = 10000/[1 + 19e^(-t/5)]<BR>\n" );
document.write( "----<BR>\n" );
document.write( "1/5 = 1/[1 + 19e^(-t/5)]<BR>\n" );
document.write( "---<BR>\n" );
document.write( "Invert both sides to get:<BR>\n" );
document.write( "1 + 19e^(-t/5) = 5<BR>\n" );
document.write( "---<BR>\n" );
document.write( "e^(-t/5) = 4/19<BR>\n" );
document.write( "---<BR>\n" );
document.write( "-t/5 = ln(4/19)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "-t = 5*-1.5581<BR>\n" );
document.write( "t = 7.79 months<BR>\n" );
document.write( "=================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );