The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

[Given log_e 3 = 1.0986, e^2.1972 = 9]

Let the count of bacteria be C at any time t.

According to question,

⇒ where k is a constant

⇒

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|C| = kt + a……(1)

Given, we have C = C₀ when t = 0 sec

Putting the value in equation (1)

∴ log|C| = kt + a

⇒ log| C₀| = 0 + a

⇒ a = log| C₀| ……(2)

Putting the value of a in equation (1) we have,

log|C| = kt + log|100000|

⇒ log|C| – log| C₀| = k t []

⇒ log ( = kt ……(3)

Also, at t = 5 years, C = 3C₀

From equation(3),we have

∴ kt = log (

⇒ k×5 = log (

⇒ k = ……(4)

Now, equation (3) becomes,

Now, let C1 be the number of bacteria present in 10 hours, as

⇒

⇒

⇒

Let the time be t1 for bacteria to be 10 times

⇒

⇒

⇒

∴ The time required for = hours