document.write( "Question 602310: At the beginning of an experiment, a culture contains 1000 bacteria . Five hours later, there are 7600 bacteria. Assuming that the bacteria grow exponentially, how many bacteria will there be after 24 hours? \n" ); document.write( "
Algebra.Com's Answer #380199 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! At the beginning of an experiment, a culture contains 1000 bacteria . Five hours later, there are 7600 bacteria. Assuming that the bacteria grow exponentially, how many bacteria will there be after 24 hours? \n" ); document.write( ". \n" ); document.write( "Exponential growth equation: \n" ); document.write( "p = e^(kt) \n" ); document.write( "where \n" ); document.write( "p is amount after time t \n" ); document.write( "k is a constant \n" ); document.write( "t is time \n" ); document.write( ". \n" ); document.write( "From:\"At the beginning of an experiment, a culture contains 1000 bacteria . Five hours later, there are 7600 bacteria.\" we get: \n" ); document.write( "7600 = 1000e^(5k) \n" ); document.write( "7600/1000 = e^(5k) \n" ); document.write( "7.600 = e^(5k) \n" ); document.write( "ln(7.600) = 5k \n" ); document.write( "ln(7.600)/5 = k \n" ); document.write( "0.40563 = k \n" ); document.write( ". \n" ); document.write( "Our \"general equation\" for growth then is: \n" ); document.write( "p = e^(0.40563t) \n" ); document.write( "which we use to solve: \n" ); document.write( "how many bacteria will there be after 24 hours? \n" ); document.write( "p = e^(0.40563*24) \n" ); document.write( "p = 16900 (answer) \n" ); document.write( " \n" ); document.write( " |