document.write( "Question 821723: A bacteria culture started with 6500 bacteria and increased by 35% in the first 2 hours.\r
\n" ); document.write( "\n" ); document.write( "How long does it take the population to triple in size? \r
\n" ); document.write( "\n" ); document.write( "Use Q(t)=6500e^1.35/2(t)
\n" ); document.write( "

Algebra.Com's Answer #494366 by TimothyLamb(4379)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "f(t)= P*e^(tk)
\n" ); document.write( "where:
\n" ); document.write( "P = initial value
\n" ); document.write( "k = growth or decay factor per unit of time
\n" ); document.write( "t = time
\n" ); document.write( "---
\n" ); document.write( "f(0) = P = 6500
\n" ); document.write( "f(t)= 6500*e^(tk)
\n" ); document.write( "---
\n" ); document.write( "f(2) = 6500*1.35 = 8775
\n" ); document.write( "8775 = 6500*e^(2k)
\n" ); document.write( "8775/6500 = e^(2k)
\n" ); document.write( "ln(8775/6500) = ln( e^(2k) )
\n" ); document.write( "ln(8775/6500) = 2k
\n" ); document.write( "k = ln(8775/6500)/2
\n" ); document.write( "k = 0.1500523 growth factor per hour
\n" ); document.write( "---
\n" ); document.write( "time to triple the population:
\n" ); document.write( "f(t)= 3*6500 = 6500*e^(tk)
\n" ); document.write( "3 = e^(tk)
\n" ); document.write( "ln(3) = ln( e^(tk) )
\n" ); document.write( "ln(3) = tk
\n" ); document.write( "t = ln(3)/k
\n" ); document.write( "t = 7.3215293
\n" ); document.write( "---
\n" ); document.write( "answer:
\n" ); document.write( "time to triple the population = 7.3 hours = 7 hours and 19.3 minutes
\n" ); document.write( "---
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