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?

: Let $P$ be the number of bacteria present in the culture at any time $t$.

$\therefore$ $\cfrac{{dP}}{{dt}} \propto P \Rightarrow \cfrac{{dP}}{{dt}} = kP \Rightarrow \cfrac{{dP}}{P} = kdt$

…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{dP}}{P}} = \int k dt \Rightarrow \log P = kt + C$

…(2)
When $t = 0$ hours, $P = 1,00,000$

Substituting the values of $P$ and $t$ in (2),

we get
$\log 1,00,000 = C$

Equation (2) becomes $\log P = kt + \log 1,00,000$

…(3)
When $t = 2$ hours, $P = \;1,00,000 + 10\%$ of $1,00,000 = 1,10,000$

Substituting the values of $P$ and $t$ in (3),

we get
$\log 1,10,000 = 2k + \log 1,00,000$
$\Rightarrow 2k = \log \left( {\cfrac{{11}}{{10}}} \right) \Rightarrow k = \cfrac{1}{2}\log \left( {\cfrac{{11}}{{10}}} \right)$

$\therefore$ $\log P = t \cdot \cfrac{1}{2}\log \left( {\cfrac{{11}}{{10}}} \right) + \log 1,00,000$

…(4)
Now, again when $P = 2,00,000$ we are required to find $t$

$\therefore$ $\log 2,00,000 = t \cdot \cfrac{1}{2}\log \cfrac{{11}}{{10}} + \log 1,00,000$

$\Rightarrow t \cdot \cfrac{1}{2}\log \cfrac{{11}}{{10}} = \log 2 \Rightarrow t = \cfrac{{2\log 2}}{{\log \left( {\cfrac{{11}}{{10}}} \right)}}$hours