document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=747333>Question 747333</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A bacteria culture initially contains 3000 bacteria and doubles every half hour. The formula for the population is P(t)=3000e^(kt) for some constant k. How do I solve for k? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #454831 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><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=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=454831\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The formula P(t) = P(o)e^kt  t is in hours<BR>\n" );
document.write( " P(o) - original amount of bacteria.\r<BR>\n" );
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document.write( "working on the principle that the culture<BR>\n" );
document.write( " doubles every half an hour. The amount <BR>\n" );
document.write( "after 1 hour would = 12000 bacteria\r<BR>\n" );
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document.write( "12,000 = 3,000 e^ k x 1<BR>\n" );
document.write( "12,000 = 3,000  e^k<BR>\n" );
document.write( "12000/3000 = e^k<BR>\n" );
document.write( "       4  = e^k<BR>\n" );
document.write( "    lge 4 = k loge e<BR>\n" );
document.write( "        k = loge 4<BR>\n" );
document.write( "        k = 1.38629<BR>\n" );
document.write( " </SPAN>\n" );
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