document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=177896>Question 177896</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n=f(t)=100(2 to the t/3 power). When will the population reach 50,000? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #132887 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=132887\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n=f(t)=100(2 to the t/3 power). When will the population reach 50,000?<BR>\n" );
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document.write( "50,000 = 100(2)^(t/3)<BR>\n" );
document.write( "2^(t/3) = 500<BR>\n" );
document.write( "Take the log of both sides to get:<BR>\n" );
document.write( "(t/3)log2 = log500<BR>\n" );
document.write( "t/3 = log(500)/log(2)<BR>\n" );
document.write( "t/3 = 8.9657..<BR>\n" );
document.write( "t = 26.9 hours<BR>\n" );
document.write( "===================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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