A bacteria culture initially contains 2500 bacteria and doubles every half hour..?

Question

A bacteria culture initially contains 2500 bacteria and doubles every half hour.

Find the size of the bacterial population after 20 minutes. 

Find the size of the bacterial population after 7 hours.

Answer 1

The exponential growth equation is in the form of: y(t) = y₀ * e^(k*t), y₀ is the initial size, t is the time, and k is the growth constant.

Using the given information we know that

y₀ = 2500

t_dbl = 30

y(t_dbl) = 2*y₀

n finding the doubling period you are finding where y(t) = 2 * y₀

The longer method would be to set the equation equal to 2*2500 and solve for k:

y(t)= 5000 = 2500 * e^{(30 * k)}

The more elegant solution is knowing that since

y(t) = 2 * y₀ and y(t) = y₀ * e^(k*t)

then

2 * y₀ = y₀ * e^(k*t)

or better yet

2 = e^(k*t)

Solving for t gets:

k = 1/t_dbl * ln(2)

Substitute the known and solved for variables into the full equation:

use k from where you solved above

y(t) = 2500 * e^(k*t)

For population size at 20 minutes:

y(20) = 2500 * e^(k*20)

For population size at 7hrs, you must first convert to whatever units were used for "half an hour". I have used minutes above, so 7hrs must first be converted into minutes:

7hrs = 7*60 minutes

Use k from where you solved above, use t = 7*60

y(20) = 2500 * e^(k*7*60)