Question

In a culture the bacteria count is 100000. The number is increases by 10% in 2h. In how many hours will the count reach 200000, if the ratio of growth of bacteria is proportional to the number present?

Answer 1

Let y be the count of the bacteria at the end of t hours, then
$\frac{dy}{dt}\propto y\Rightarrow \frac{dy}{dt}=ky,$
where, k is constant of proportionality.
Now, separating the variables, we get $\frac{dy}{y}=kdt$
On integrating, we get $\int \frac{dy}{y}=\int kdt\Rightarrow log|y|=kt+C$ ...(i)
When t=0, y=100000 then log 100000=C ...(ii)
and went t=2, y=110000, then log 110000=2k+C ...(iii)
On substracting Eq. (ii) from Eq. (iii), we get
log 110000-log 100000=2k $\Rightarrow log\frac{110000}{100000}=2k\Rightarrow k=\frac{1}{2}log\left(\frac{10}{11}\right)$
On substracting the value of k and C in Eq. (i), we get
$logy=\frac{1}{2}log\left(\frac{10}{11}\right)t+log100000$
When y=200000, then log 200000= $\frac{1}{2}log\left(\frac{11}{10}\right)t+log100000$
$\Rightarrow log\left(\frac{200000}{100000}\right)=\frac{1}{2}log\left(\frac{11}{10}\right)t\Rightarrow 2log\left(2\right)=log\left(\frac{11}{10}\right)t\Rightarrow t=\frac{2log2}{log\left(\frac{11}{10}\right)}$
Hence, in $\frac{2log2}{log\left(\frac{11}{10}\right)}h$ the number of bacteria increases from 100000 to 200000.