Question

In a culture, the bacteria count is $1,00,000$. The number is increased by $10\%$ in $2$ hours. In how many hours will the count reach $2,00,000$, if the rate of growth of bacteria is proportional to the number present?

Answer 1

Let the number of bacteria at time $t$ be $y$

Given that rate growth of bacteria is proportional to the number present

$\frac{dy}{dt}\propto y$

$\frac{dy}{dt}=ky$

$\frac{dy}{y}=kdt$

Integrating both sides,

$\int \frac{dy}{y}=k\int dt$

$\Rightarrow$ $\log y=kt+C$ ---- ( 1 )

Now according to question,

The bacteria court is $1,00,000$. The number is increased by $10\%$ in $2$ hours, in how many hours will the count reach $2,00,000.$

Putting $t=0$ and $y=1,00,000$ in ( 1 )

$\log 1,00,000=k\times 0+C$

$C=\log 1,00,000$

Putting value of $C$ in ( 1 )

$\log y=kt+C$

$\Rightarrow$ $\log y=kt+\log 1,00,000$ ---- ( 2 )

Now,

Putting $t=2$ and $y=1,10,000$ in ( 2 )

$\log 1,10,000=2k+\log 1,00,000$

$\log 1,10,000-\log 1,00,000=2k$

$\log \left(\frac{1,10,000}{1,00,000}\right)=2k$

$\frac{1}{2}\log \left(\frac{11}{10}\right)=k$

Putting value of $k$ in ( 2 )

$\log y=kt+\log 1,00,000$

$\Rightarrow$ $\log y=\frac{1}{2}\log \left(\frac{11}{10}\right)t+\log 1,00,000$ ---- ( 3 )

Now, if bacterial $=2,00,000$ in ( 3 )

$\log 2,00,000=\frac{1}{2}\log \left(\frac{11}{10}\right)t+\log (1,00,000)$

$\log 2,00,000-\log 1,00,000=\frac{1}{2}\log \left(\frac{11}{10}\right)t$

$\log \left(\frac{2,00,000}{1,00,000}\right)=\frac{1}{2}\log \left(\frac{11}{10}\right)t$

$\log 2=\frac{1}{2}\log \left(\frac{11}{10}\right)t$

$\Rightarrow$ $t=\frac{2\log 2}{\log \left(\frac{11}{10}\right)}$