Here's the full problem:
If a bacterial culture starts with 7000 bacteria and doubles every 20 minutes, how many minutes will it take the population to reach 22000?
Is there a specific formula I can use to solve this and problems like it?
If a bacterial culture starts with 7000 bacteria and doubles every 20 minutes......?
2016-05-11 5:44 am
回答 (2)
2016-05-11 5:54 am
In 20 min, the number of bacteria is multiplied by 2.
In 40 min, the number of bacteria is multiplied by 2².
In 60 min, the number of bacteria is multiplied by 2³.
......
Hence, in 20n min, the number of bacteria is multiplied by 2ⁿ.
22000 = 7000 × 2ⁿ
2ⁿ = 22000/7000
2ⁿ = 22/7
log(2ⁿ) = log(22/7)
n log(2) = log(22/7)
n = log(22/7) / log(2)
Time taken
= 20n min
= 20 × [log(22/7) / log(2)] min
= 33.04 min (to 2 decimal places)
In 40 min, the number of bacteria is multiplied by 2².
In 60 min, the number of bacteria is multiplied by 2³.
......
Hence, in 20n min, the number of bacteria is multiplied by 2ⁿ.
22000 = 7000 × 2ⁿ
2ⁿ = 22000/7000
2ⁿ = 22/7
log(2ⁿ) = log(22/7)
n log(2) = log(22/7)
n = log(22/7) / log(2)
Time taken
= 20n min
= 20 × [log(22/7) / log(2)] min
= 33.04 min (to 2 decimal places)
2016-05-11 6:03 am
P = Po [ 2^(t/3) ] , t = hours.....P = Po [2^(z/20)] , z = minutes
......................22000 = 7000 2^(z/20) ---> z = 20 ln(22/7) / ln 2....≈ 33 minutes
......................22000 = 7000 2^(z/20) ---> z = 20 ln(22/7) / ln 2....≈ 33 minutes
收錄日期: 2021-04-18 14:49:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160510214427AAL9AXZ