The rate of growth of bacteria in a culture is proportional to the number of bacteria pres…

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

Accepted Answer

$${{dx} \over {dt}} \propto x$$

$${{dx} \over {dt}} = \lambda x$$

$$\int\limits_{1000}^x {{{dx} \over x} = \int\limits_0^t {\lambda dt} } $$

$$\ln x - \ln 1000 = \lambda t$$

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

Put t = 2, x = 1200

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

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

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

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

$$ \Rightarrow 2000 = 1000{e^{{k \over {2\ln (6/5)}} \times \ln (6/5)}}$$

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

$$ \Rightarrow \ln 2 = {k \over 2}$$

$$ \Rightarrow {k \over {\ln 2}} = 2$$

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

Solutions · 1