Under ideal conditions a certain bacteria population is know to double every three hours. Suppose that there are initially 100 bacteria. How would you go about formulating a function for this?

ubwicanyil5

ubwicanyil5

Answered question

2022-09-09

Under ideal conditions a certain bacteria population is know to double every three hours. Suppose that there are initially 100 bacteria. How would you go about formulating a function for this?

Answer & Explanation

Raphael Singleton

Raphael Singleton

Beginner2022-09-10Added 19 answers

The general approach described by Todd Wilcox is the best way to go about things: It is important to become thoroughly familiar with the function e r t
When doubling time is given, there is a quick formula that works. Let d be the doubling time. Then if P(t) is the population at time t, we have
P ( t ) = P ( 0 ) 2 t / d .
So in our case,
( 1 ) P ( t ) = ( 100 ) 2 t / 3 .
The general approach, which you should carry out, will give you after some manipulation
P ( t ) = ( 100 ) e t log 2 / 3 ,
where by log we mean logarithm to the base e (ln on your calculator). This is the same as the answer (1), since
e x log 2 = ( e log 2 ) x = 2 x ,
since e log 2 = 2
Remark: As long as one is willing to believe that the answer has shape P ( t ) = A 2 k t , the formula is easy to prove. For put t=0. Then P ( 0 ) = A 2 ( k ) ( 0 ) = A, so A=P(0). Also, since doubling time is d, we have P(d)=2P(0). But P ( t ) = P ( 0 ) 2 k t . Put t=d. We get 2 P ( 0 ) = P ( 0 ) 2 k d , and therefore kd=1, meaning that k=1/d. This yields the formula P ( t ) = P ( 0 ) 2 t / d mentioned above.

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