The rate of increase of bacteria in a culture is proportional to the number of bacteria present, and it is found that the number doubles in 6 hours. Prove that the bacteria becomes eight times at the end of 18 hours.
Let the count of bacteria be C at any time t.
According to the 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| C0|
⇒ log|C| – log| C0| = k t [
]
⇒ log (
= kt ……(3)
Also, at t = 6 years, C = 2C0
From equation(3),we have
∴ kt = log (
⇒ k×6 = log (
⇒ k = 
……(4)
Now, equation (3) becomes,
Now, C = 8C0
⇒ 
⇒ 
⇒ 
⇒ t = 18
∴ The time required = 18 hours