SOLUTION: A population of bugs double every month. Started with 40, what would the function be of the number of bugs with regards to time. How many bugs are there after 1 year and 5 yea

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