bacteria reproduces from 1 to 2. under ideal conditions bacteria reproduce every 20 minutes. find the constant k

Question

bacteria divides from 1 to 2 . some reproduce every 20 minutes. find the constant k

Answer 1

If I have the number of bacteria represented by the expression

 

Ae^kt 

 

where, here, A will be the initial amount (one bacterium), k will be the constant we're trying to find, and t will be the amount of time that has elapsed in minutes,

 

I can rewrite this expression as 

 

e^kt

 

and use it to find the number of bacteria I will have at any given time.

 

I know that after 20 minutes, the one bacterium I got doubled, so

 

2 = e^20k

 

Then 20k = log(2)

 

So k = log(2)/20 = about 0.035

 

(I'm using natural log btw...)

 

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If you're not using that exponential to approximate the number of bacteria, I guess we could think of another way we can represent that number:

 

t = 0       1 bacterium

t = 20     2 bacteria

t = 40     2² bacteria

t = 60     2³ bacteria

t = 80     2⁴ bacteria

etc.

 

If we're representing the number of bacteria at any time as 2^kt, k would have to be 1/20 or 0.05. (Notice how the time is related to the exponent in each of the lines above. You have to divide it by 20 to get the exponent.)

 

It's more often that I see people working with that exponential growth model though that uses e.

 

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Let's look at the model that uses e again:

 

Notice that I should get the same result as in the case when we used 2 if I plug increments of 20 in for t:

 

f(t) = e^(log(2)/20)t

 

f(20) = e^{(log(2)/20)20} = e^log(2) = 2

 

f(40) = e^{(log(2)/20)40} = e^log(2)*2 = 2² 

 

f(60) = e^{(log(2)/20)60} = e^log(2)*3 = 2³

 

and so on...

Answer 2

Let, time t be the interval between successive reproduction of bacteria 

Thus bacteria becomes                                    1 to 2 in t    or                 2⁰ to 2¹ in time interval t

     each of 2 bacteria reproduces and becomes 2 to 4 in next t interval or  2¹ to 2² in time interval 2t

     each of 2² bacteria reproduces and becomes 4 to 8 in next t interval or 2² to 2³ in time interval 3t

     each of 2³ bacteria reproduces and becomes 2³ to 2⁴ in 4t intervals 

     each of 2⁴  bacteria reproduces and becomes 2⁴ to 2⁵ in 5t intervals 

thus each of 2^x-1 bacteria reproduces and becomes 2^x-1 to 2^x in xt intervals , where x is a positive integer

Thus after[1+2+3+.......+ x+] t time, 2⁰ bacterium grows into 2^x bacteria, 

 

Hence, amount of increase, y = 2^x -2⁰

Rate of increase, k=(2^x-2⁰)/[t.(x/2)(x+1)] = (2^x+1-2)/[tx(x+1)], which remains constant over all other constraints being changed.

when t=20 minutes, and x=2, k becomes = (2²⁺¹-2)/[20.2(2+1)]=(2³-2)/[40.3]=(8-2)/120=6/120₂₀=1/20=0.05

 

Therefore the constant k, the rate of increase is 0.05.