The rate of growth of bacteria in a culture is

Question:

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time $t=0$. The number of bacteria is increased by $20 \%$ in 2 hours. If the population of bacteria is 2000 after $\frac{\mathrm{k}}{\log _{\mathrm{e}}\left(\frac{6}{5}\right)}$ hours, then $\left(\frac{\mathrm{k}}{\log _{\mathrm{e}} 2}\right)^{2}$ is equal to

1. (1) 4

2. (2) 2

3. (3) 16

4. (4) 8

Correct Option: 1

Solution:

$\frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{X}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\lambda \mathrm{x}$

$\int_{1000}^{\mathrm{x}} \frac{\mathrm{dx}}{\mathrm{x}}=\int_{0}^{\mathrm{t}} \lambda \mathrm{dt}$

$\ln \mathrm{x}-\ln 1000=\lambda \mathrm{t}$

$\ln \left(\frac{\mathrm{x}}{1000}\right)=\lambda \mathrm{t}$

$\ln \left(\frac{\mathrm{x}}{1000}\right)=\lambda \mathrm{t}$

Put $t=2, x=1200$

$\ln \left(\frac{12}{10}\right)=2 \lambda \Rightarrow \lambda=\frac{1}{2} \ln \frac{6}{5}$

Now $\ln \left(\frac{x}{1000}\right)=\frac{t}{2} \ln \left(\frac{6}{5}\right)$

$x=1000 e^{\frac{t}{2} \ln \left(\frac{6}{5}\right)}$

$x=2000$ at $t=\frac{k}{\ln \left(\frac{6}{5}\right)}$

$\Rightarrow \quad 2000=1000 e^{\frac{1}{2 \ln (6 / 5)}} \times \ln (6 / 5)$

$\Rightarrow 2=e^{k / 2}$

$\Rightarrow \quad \ln 2=\frac{k}{2}$

$\Rightarrow \quad \frac{k}{\ln 2}=2$

$\Rightarrow\left(\frac{k}{\ln 2}\right)^{2}=4$