document.write( "Question 885241: The doubling period of a baterial population is 15 minutes. At time t = 90 minutes, the baterial population was 50000. Round your answers to at least 1 decimal place.\r
\n" ); document.write( "\n" ); document.write( "What was the initial population at time t = 0 ? \r
\n" ); document.write( "\n" ); document.write( "Find the size of the baterial population after 4 hours
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Algebra.Com's Answer #535041 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
WITHOUT FORMULAS OR CALCULATOR (but with understanding):
\n" ); document.write( "The population of that kind of bacteria doubles every 15 minutes, and after 90 minutes there's 50,000 bacteria.
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\n" ); document.write( "At time t=90 minutes, \"90%2F15=6\" doublings have happened since \"t=0\"
\n" ); document.write( "The population at t=90 minutes is \"2%5E6=64\" times the population at \"t=0\" ,
\n" ); document.write( "so the population at t=0 must have been
\n" ); document.write( "\"50000%2F2%5E6=1541.25\"
\n" ); document.write( "I do not need to round that number, because it is an exact calculation.
\n" ); document.write( "Saying that at time t=0 there were \"1541%261%2F4\" bacteria is a little silly.
\n" ); document.write( "I expect the number of bacteria to be an integer.
\n" ); document.write( "However, the problem asks to round to at least 1 decimal place, so let's be silly.
\n" ); document.write( "How did I calculate that result? With pencil and paper.
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\n" ); document.write( "At t=4 hours, \"4%2A%2860min%2F%2215+min%22%29=4%2A4=16\" doubling period have passed since t=0.
\n" ); document.write( "That is \"16-6=10\" doubling periods after t=90 minutes,
\n" ); document.write( "so the number of bacteria should be \"2%5E10\" times what it was at 90 minutes.
\n" ); document.write( "That is \"50000%2A2%5E10=%2251%2C200%2C000%22\" and that is an exact number too.
\n" ); document.write( "Should I say it is 51,200,000.00 to have at least one decimal place expressly stated.
\n" ); document.write( "How did I calculate that result? With pencil and paper.
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\n" ); document.write( "WITH FORMULAS AND CALCULATOR:
\n" ); document.write( "We are having a case of exponential growth.
\n" ); document.write( "\"P%28t%29\"= population at t=t minutes
\n" ); document.write( "\"P%28t%29=P%280%29%2A2%5E%28%22t+%2F+15%22%29\"<-->\"P%28t%29%2FP%280%29=2%5E%28%22t+%2F+15%22%29\"
\n" ); document.write( "Since calculators allow us to calculate natural logarithms, and base 10 logarithms,
\n" ); document.write( "we can also write that as
\n" ); document.write( "\"ln%28P%28t%29%2FP%280%29%29=%28t%2F15%29%2Aln%282%29\" or \"+log%28P%28t%29%2FP%280%29%29=%28t%2F15%29%2Alog%282%29\"
\n" ); document.write( "Going from the natural logarithms version to exponentials on base \"e\" we get
\n" ); document.write( "\"P%28t%29%2FP%280%29=e%5E%28%22ln%282%29t+%2F+15%22%29\"<--->\"P%28t%29=P%280%29%2Ae%5E%28%22ln%282%29t+%2F+15%22%29\"
\n" ); document.write( "Those are very popular forms. The approximation \"ln%282%29=0.693\" is often used.
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\n" ); document.write( "At \"t=90\" ,
\n" ); document.write( "\"P%2890%29%2FP%280%29=e%5E%28%22ln%282%2990%2F+15%22%29\"
\n" ); document.write( "\"50000%2FP%280%29=e%5E%286ln%282%29%29\"-->\"P%280%29%2F50000=e%5E%28-6ln%282%29%29\"-->\"P%280%29=50000%2Ae%5E%28-+6ln%282%29%29\"-->\"P%280%29=50000%2Ae%5E%28-+6ln%282%29%29\"-->\"P%280%29=781.25\"
\n" ); document.write( "}}} (using the calculator's value for \"ln%282%29\" )
\n" ); document.write( "(Using 0.693 will give you a different decimal part).
\n" ); document.write( "At \"t=4hours=240minutes\" ,
\n" ); document.write( "\"P%28240%29%2FP%280%29=e%5E%28%22ln%282%29240%2F+15%22%29\"
\n" ); document.write( "\"P%28240%29%2F781.25=e%5E%2816%2Aln%282%29%29\"-->\"P%28240%29=781.25%2Ae%5E%2816%2Aln%282%29%29\"-->\"P%28240%29=%2251%2C200%2C000%22\"
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