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 loge 3 = 1.0986, e2.1972 = 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