Question

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

• A. $\frac{2\log 2}{\log \left(\frac{11}{10}\right)}$
• B. $\frac{\log 2}{\log 11}$
• C. $\frac{\log 2}{\log 10}$
• D. $\frac{\log 2}{\log \frac{11}{10}}$

Answer 1

Answer: (A)

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 $\int \frac{dy}{y}=\int kdt$
$\Rightarrow \log |y|=kt+C\dots$ (i)
When $t=0,y=100000$ (initially), then
$\log 100000=C....$(ii)
and when $t=2,y=110000$, then
$\log 110000=2k+C....$(iii)
On subtracting Eqs. (ii) from Eqs. (iii), we get
$\log 110000-\log 100000=2k$
$\Rightarrow \log \frac{110000}{100000}-2k$
$\Rightarrow k-\frac{1}{2}\log \left(\frac{11}{10}\right)$
On substituting the value of $k$ and $C$ in Eq. (i), we get
$\log y=\frac{1}{2}\log \left(\frac{11}{10}\right)t+\log 100000$
When $y=200000$, then
$\log 200000=\frac{1}{2}\log \left(\frac{11}{10}\right)t+\log 100000$
$\Rightarrow \log \left(\frac{200000}{100000}\right)=\frac{1}{2}\log \left(\frac{11}{10}\right)t$
$\Rightarrow 2\log (2)=\log \left(\frac{11}{10}\right)t$
$\Rightarrow t=\frac{2\log 2}{\log \left(\frac{11}{10}\right)}$
Hence, in $\frac{2\log 2}{\log \left(\frac{11}{10}\right)}$ hours the number of bacteria increases from $100000$ to $200000$ .