take the log to any base of both sides
for example base 10
log 1000 = d log 2
d = 3/.301 = 9.96
2 day periods
or 20 days
check my reply to your earlier efficiency problem
(I skipped the lengthy work problem background as it is irrelevant)
The number of bacteria needed to break down an oil spill is 1000000 per millitre of oil. The bacteria double in number every two days.The starting concentration of bacteria is 1000 bacteria per millimetre.The situation can be modelled by the equation C=1000(2^d) where C is estimated concentration of bacteria and d is the number of 2-day periods the bacteria grow.Approximately how long would it take for the bacteria to reach the required concentartion?
C= 1000(2^d)
1000000=1000(2^d)
1000=2^d
The only part that confuses me is this: how do I go on from there?
6 answers
O.o We haven't even come close to logarithms. What grade level is this?
hmmm
well, you need 2^something = 1000
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = about 1000 so d = about 10
so 2*10 = 20 days. It said approximate :)
well, you need 2^something = 1000
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = about 1000 so d = about 10
so 2*10 = 20 days. It said approximate :)
These problems are pretty advanced for class that has not reached logarithms.
This was one of the advanced problems in my textbook. I think the author intended to solve the problem using the second method.
Thank you,once again for helping me!
Thank you,once again for helping me!
You are welcome :)
1 answer