document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=212782>Question 212782</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0321233832/algebrahomeworkh>Elementary and Intermediate Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Growth of bacteria. The bacteria Escherichia coli (E. coli) are commonly found in the human bladder. Suppose that 3000 of the bacteria are present at time t=0. Then t minutes later, the number of bacteria present is N(t) = 3000(2)^t/20. if 100,000 bacteria accumulate, a bladder infection can occur. If, at 11:00am, a patient's bladder contains 25,000 E. coli bacteria, at what time can infection occur? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #160712 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=160712\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Growth of bacteria. The bacteria Escherichia coli (E. coli) are commonly found in the human bladder. Suppose that 3000 of the bacteria are present at time t=0. Then t minutes later, the number of bacteria present is N(t) = 3000(2)^t/20. if 100,000 bacteria accumulate, a bladder infection can occur. If, at 11:00am, a patient's bladder contains 25,000 E. coli bacteria, at what time can infection occur?<BR>\n" );
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document.write( "100,000 = 3000*2^(t/20)<BR>\n" );
document.write( "2^(t/20) = 100/3<BR>\n" );
document.write( "(t/20)ln2 = ln(100/3)<BR>\n" );
document.write( "t = 20[ln(100/3)]/[ln(2)]<BR>\n" );
document.write( "t = 101.18 minutes (time after time = 0 to reach 100,000)<BR>\n" );
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document.write( "25000 = 3000*2^(t/2)<BR>\n" );
document.write( "2^(t/2) = 25/3<BR>\n" );
document.write( "t = 2[ln(25/3)/ln(2)]<BR>\n" );
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document.write( "t = 6.18 minutes (time after time = 0 to reach 25000)<BR>\n" );
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document.write( "Time from 25000 to 100000: 101.18-6.18 = 95 minutes<BR>\n" );
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document.write( "95 minutes after 11:00 the time will be 12:35 PM<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );