The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in 6 hours is

Step-by-step solution

Let b be the number of bacteria. We have db dt b db b = Kdt b = Kt + c ...(1) Let b₀ be the initial number of bacteria. At t=0, b=b₀ b ₀= K (0)+ c c = b ₀ ( b b₀ )= Kt ...(2) When t=3, b=2 b₀ ( 2 b₀ b₀ )=3 K K= 1 3 ( 2) Thus b= 1 3 ( 2) t+ b₀ When t =6 ( b b₀ )=2 2= 4 b b₀ =4 b=4 b₀ This problem can also be solved as follows : The number of bacteria doubles in 3 hours. Let initial number of bacteria = N . After 3 hours, number of bacteria =2 ~N . After 6 hours, number of bacteria =4 ~N .