document.write( "Question 702020: The bacteria in a certain culture double every 7.1 hours. The culture has 6,500 bacteria at the start.
\n" ); document.write( "How many bacteria will the culture contain after 4 hours? \n" ); document.write( "
Algebra.Com's Answer #432762 by nerdybill(7384)\"\" \"About 
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The bacteria in a certain culture double every 7.1 hours. The culture has 6,500 bacteria at the start. How many bacteria will the culture contain after 4 hours?
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\n" ); document.write( "General exponential growth/decay equation:
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( "where
\n" ); document.write( "A is amount after time t
\n" ); document.write( "P is initial amount
\n" ); document.write( "r is growth/decay rate
\n" ); document.write( "t is time
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\n" ); document.write( "From: \"The bacteria in a certain culture double every 7.1 hours.\" we can determine r:
\n" ); document.write( "P/2 = Pe^(r*7.1)
\n" ); document.write( "1/2 = e^(r*7.1)
\n" ); document.write( ".5 = e^(r*7.1)
\n" ); document.write( "ln(.5) = r*7.1
\n" ); document.write( "ln(.5)/7.1 = r
\n" ); document.write( "-0.09763 = r
\n" ); document.write( ".
\n" ); document.write( "Now, we can answer:
\n" ); document.write( "How many bacteria will the culture contain after 4 hours?
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( "A = 6500e^(-0.09763*4)
\n" ); document.write( "A = 6500e^(-0.3905055)
\n" ); document.write( "A = 4398.65 (answer)\r
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