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 ${k \over {{{\log }_e}\left( {{6 \over 5}} \right)}}$ hours, then ${\left( {{k \over {{{\log }_e}2}}} \right)^2}$ is equal to :

${{dx} \over {dt}} \propto x$<br><br>${{dx} \over {dt}} = \lambda x$<br><br>$\int\limits_{1000}^x {{{dx} \over x} = \int\limits_0^t {\lambda dt} }$<br><br>$\ln x - \ln 1000 = \lambda t$<br><br>$\ln \left( {{x \over {1000}}} \right) = \lambda t$<br><br>Put t = 2, x = 1200<br><br>$$\ln \left( {{{12} \over {10}}} \right) = 2\lambda \Rightarrow \lambda = {1 \over 2}\ln {6 \over 5}$$<br><br>Now, $$\ln \left( {{x \over {1000}}} \right) = {t \over 2}\ln \left( {{6 \over 5}} \right)$$<br><br>$x = 1000{e^{{t \over 2}\ln \left( {{6 \over 5}} \right)}}$<br><br>$x = 2000$ at $t = {k \over {\ln \left( {{6 \over 5}} \right)}}$<br><br>$\Rightarrow 2000 = 1000{e^{{k \over {2\ln (6/5)}} \times \ln (6/5)}}$<br><br>$\Rightarrow 2 = {e^{k/2}}$<br><br>$\Rightarrow \ln 2 = {k \over 2}$<br><br>$\Rightarrow {k \over {\ln 2}} = 2$<br><br>$\Rightarrow {\left( {{k \over {\ln 2}}} \right)^2} = 4$