The rate of growth of bacteria in a culture is proportional to the...

Given $\dfrac{dB}{dt} ∝ B ⇒ \dfrac{dB}{dt} = \lambdaB, \left(\lambda\text{is proportionility constant}\right)$

$⇒ \dfrac{dB}{B} = \lambdadt$

Integrate both sides with proper limit we get,

${∫}_{{B}_{1}}^{{B}_{2}}\dfrac{dB}{B} = {∫}_{{t}_{1}}^{{t}_{2}}\lambdadt$

$⇒ ln\dfrac{{B}_{2}}{{B}_{1}} = \lambda\left({t}_{2} - {t}_{1}\right) . . . \left(i\right)$

Given that, at $t = 0, B = 1000 \text{ and } t = 20$

$∴ ln\dfrac{1200}{1000} = 2\lambda ⇒ \lambda = \dfrac{1}{2}ln\dfrac{6}{5}$

Now an equation \left(i\right)$becomes, </p><p>$ln\dfrac{{B}_{2}}{{B}_{1}} = \dfrac{1}{2}ln\dfrac{6}{5}\left({t}_{2} - {t}_{1}\right) . . . \left(ii\right)$</p><p>Given that, at${t}_{2} = \dfrac{k}{\ln 2}, {B}_{2} = 2000 \text{ and } t = 0, {B}_{1} = 1000$</p><p>From \left(ii\right), ln\dfrac{2000}{1000} = \dfrac{1}{2}ln\dfrac{6}{5}\left(\dfrac{k}{ln\dfrac{6}{5}} - 0\right) = \dfrac{1}{2}k$

$⇒ k = 2\ln 2$

$∴ {\left(\dfrac{k}{\ln 2}\right)}^{2} = {\left(\dfrac{2\ln 2}{\ln 2}\right)}^{2} = 4$