document.write( "Question 237581: three yeast cells are placed in a laboratory dish at 9:00 am. the number of yeast cells doubles in every 5 minutes period. the number of cells in the dish at 9:30 am the same day is \r
\n" ); document.write( "\n" ); document.write( "a) 30 b) 64 c) 192 d) 256 e) 300 \n" ); document.write( "
Algebra.Com's Answer #174698 by nerdybill(7384)\"\" \"About 
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three yeast cells are placed in a laboratory dish at 9:00 am. the number of yeast cells doubles in every 5 minutes period. the number of cells in the dish at 9:30 am the same day is
\n" ); document.write( "a) 30 b) 64 c) 192 d) 256 e) 300
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\n" ); document.write( "This is an \"exponential growth\" problem...apply :
\n" ); document.write( "A(t) = Pe^(rt)
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\n" ); document.write( "For your problem, they give:
\n" ); document.write( "Let P (initial amount) = 3
\n" ); document.write( "then A(t) (amount at time t) = 2(3) = 6
\n" ); document.write( "r is growth ratio -- this is what we're looking for
\n" ); document.write( "t is 5 minutes
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\n" ); document.write( "So,
\n" ); document.write( "A(t) = Pe^(rt)
\n" ); document.write( "6 = 3e^(5r)
\n" ); document.write( "Dividing both sides by 3:
\n" ); document.write( "2 = e^(5r)
\n" ); document.write( "ln 2 = 5r
\n" ); document.write( "ln(2)/5 = r
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\n" ); document.write( "Finally, for t = 30 (9:30 - 9:00)
\n" ); document.write( "we have
\n" ); document.write( "A(t) = 3e^(ln(2)/5t)
\n" ); document.write( "A(30) = 3e^(ln(2)/5(30))
\n" ); document.write( "A(30) = 3e^(ln(2)(6))
\n" ); document.write( "A(30) = 3(64)
\n" ); document.write( "A(30) = 192 cells\r
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