document.write( "Question 202641: A certain bacterial infection has 110,000 organisms at 12:00pm. At 12:30 pm the population was 140,000. If this growth rate has been steady, when did the infection start? \n" ); document.write( "
Algebra.Com's Answer #152830 by stanbon(75887) ![]() You can put this solution on YOUR website! A certain bacterial infection has 110,000 organisms at 12:00pm. At 12:30 pm the population was 140,000. If this growth rate has been steady, when did the infection start? \n" ); document.write( "-------------------------------- \n" ); document.write( "A(t) = Ao*(r)^t \n" ); document.write( "--- \n" ); document.write( "r is the rate of growth \n" ); document.write( "t is the time in minutes \n" ); document.write( "--- \n" ); document.write( "140,000 = 110,000*r^(30) \n" ); document.write( "14/11 = r^(30) \n" ); document.write( "Take the log: \n" ); document.write( "30*log(r) = log(14/11) \n" ); document.write( "log(r) = [log(14/11)]/30 \n" ); document.write( "log(r) = 0.0034911784 \n" ); document.write( "r = 10^0.0034911784 \n" ); document.write( "r = 1.008071133 per minute \n" ); document.write( "------------------------------- \n" ); document.write( "1 = 110,000(1.008071133)^t \n" ); document.write( "1.008071133)^t = 110,000^-1 \n" ); document.write( "t*log(1.998071133) = -log(110,000) \n" ); document.write( "t = -16.7705 minutes \n" ); document.write( "---------------------------- \n" ); document.write( "12:00 - 16.77 minutes is approximately 11:43 am \n" ); document.write( "=================================================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |