Question

How do I find the initial number of bacteria if after 3 hours there are 400 bacteria present and after 10 hours there are 2000 bacteria present?

Answer 1

Initial number of bacteria is `200`

Explanation:

Growth of bacteria takes place exponentially and if initial population is `p_0`, then population after `t` seconds it would be `p_t=p_0e^(kt)`, where `k` is a constant.

Here we have to find initial number of bacteria i.e. at `t=0`, which is `p_0`. As population after `3` hours is `400` and after `10` hours is `2000`, we have

`p_3=400=p_0e^(3k)` (A)

and `p_10=2000=p_0e^(10k)` (B)

and dividing (B) by (A) we get `e^(10k)/e^(3k)=2000/400=5`

i.e. `e^(7k)=5` i.e. `7k=ln5` and `k=1/7ln5`

putting this in (A), we get `400=p_0e^(3/7ln5)`

or `400=p_0(e^(ln5))^(3/7)=p_0 5^(3/7)`

and `p_0=400xx5^(-3/7)=400xx0.5017~=200`