document.write( "Question 684145: A population of bugs double every month.\r
\n" ); document.write( "\n" ); document.write( "Started with 40, what would the function be of the number of bugs with regards to time.
\n" ); document.write( "How many bugs are there after 1 year and 5 years. When will there be 10,000 bugs. \n" ); document.write( "
Algebra.Com's Answer #423935 by ankor@dixie-net.com(22740)\"\" \"About 
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A population of bugs double every month.
\n" ); document.write( "Started with 40, what would the function be of the number of bugs with regards to time.
\n" ); document.write( "m = no. of months
\n" ); document.write( "f(m) = 40*2^m
\n" ); document.write( ":
\n" ); document.write( "How many bugs are there after 1 year?
\n" ); document.write( "f(12) = 40*2^12
\n" ); document.write( "f(12) = 40 * 4096
\n" ); document.write( "f(12) = 163,840 bugs in 1 year
\n" ); document.write( "and 5 years?
\n" ); document.write( "f(60) = 40*2^60
\n" ); document.write( "f(60) = 40*1.15(10^18)
\n" ); document.write( "f(60) = 4.61(10^19)
\n" ); document.write( ":
\n" ); document.write( "When will there be 10,000 bugs.
\n" ); document.write( "40*2^m = 10000
\n" ); document.write( "2^m = 10000/40
\n" ); document.write( "2^m = 250
\n" ); document.write( "using logs
\n" ); document.write( "m*log(2) = log(250)
\n" ); document.write( ".301m = 2.4
\n" ); document.write( "m = 2.4/.301
\n" ); document.write( "m ~ 8 months to have over 10000 bugs
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