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)\"\" \"About 
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( "
\n" );