a group of bacteria doubles every second,but 1/3 of the population dies before the reproduction after 1 minute their population 2^100, at what time was the population 56.25% of the final value
Answers
Given : a group of bacteria doubles every second,
but 1/3 of the population dies before the reproduction after 1 minute their population 2^100,
To Find : at what time was the population 56.25% of the final value
Solution:
Let say Initial bacteria = 3B
Then Then after 1 sec
= 2( 3B - (1/3)3B)
= 2(3B - B)
= 4B
After 2 secs
= 2( 4B - (1/3)4B)
= 16B/3
Hence 4/3 is the multiplier factor
if initial is
3B Then after n secs
= 3B ( 4/3)ⁿ
and after 1 minutes
= 3B( 4/3)⁶⁰
3B( 4/3)⁶⁰ = 2¹⁰⁰
After n secs = 3B ( 4/3)ⁿ = (56.25/100) 2¹⁰⁰
=> 3B ( 4/3)ⁿ = (56.25/100)3B( 4/3)⁶⁰
=> ( 4/3)ⁿ = (9/16)( 4/3)⁶⁰
=> ( 4/3)⁶⁰⁻ⁿ = 16/9
=> ( 4/3)⁶⁰⁻ⁿ = (4/3)²
=> 60 - n = 2
=> n = 58
At 58 secs population 56.25% of the final value
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