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 $y$ be the number of bacteria $attimet$

It is given that

$\frac{dy}{at}\alpha y$

$\frac{dy}{dt}=\mu y$

$\frac{dy}{y}=\mu at$

Integrating both sides $[\int \frac{dy}{y}=\log y]$

$\log y=\mu t+c---\left(1\right)$

When $t=0,t=1,00,000$

When $t=2hoursey=1,00,000+\frac{10}{100}\times 1,00,000=1,10,000$

Now when $t=0,y=1,00,000$ put in equatin (1)

$\log 1,00,000=\mu \times 0+c$

$\Rightarrow c=\log 1,00,000$

Now puting $c=\log 1,00,000$ in (1) we have

$\log y=\mu +\log 1,00,000---\left(2\right)$

Now when $t=2,y=1,10,000,$ put in (2)

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

$\log \left(\frac{1,10,000}{1,00,000}\right)=2\mu [\log n-\log m=\log \frac{n}{m}]$

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

Now finding $t$ when bacteri$a=2,00,000$

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

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

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

$t=\frac{2\log 2}{\log \left(\frac{11}{10}\right)}=14.54hours$