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…

Given $\frac{dB}{dt}\propto B\Rightarrow \frac{dB}{dt}=\lambda B,(\lambda \text{is proportionility constant})$dtdB​∝B⇒dtdB​=λB,(λis proportionility constant)

$$\Rightarrow \frac{dB}{B}=\lambda dt$$⇒BdB​=λdt

Integrate both sides with proper limit we get,

$${\int }_{{B}_{1}}^{{B}_{2}}\frac{dB}{B}={\int }_{{t}_{1}}^{{t}_{2}}\lambda dt$$∫B1​B2​​BdB​=∫t1​t2​​λdt

$$\Rightarrow \mathrm{ln}\frac{{B}_{2}}{{B}_{1}}=\lambda ({t}_{2}-{t}_{1})...(i)$$⇒lnB1​B2​​=λ(t2​−t1​)...(i)

Given that, at t=0,B=1000&t=20%\text{increase}=1200

$$\therefore \mathrm{ln}\frac{1200}{1000}=2\lambda \Rightarrow \lambda =\frac{1}{2}\mathrm{ln}\frac{6}{5}$$∴ln10001200​=2λ⇒λ=21​ln56​

Now an equation $(i)$(i)becomes,

$$\mathrm{ln}\frac{{B}_{2}}{{B}_{1}}=\frac{1}{2}\mathrm{ln}\frac{6}{5}({t}_{2}-{t}_{1})...(ii)$$lnB1​B2​​=21​ln56​(t2​−t1​)...(ii)

Given that, at {t}_{2}=\frac{k}{\mathrm{ln}2},{B}_{2}=2000&t=0,{B}_{1}=1000

From $(ii),\mathrm{ln}\frac{2000}{1000}=\frac{1}{2}\mathrm{ln}\frac{6}{5}(\frac{k}{\mathrm{ln}\frac{6}{5}}-0)=\frac{1}{2}k$(ii),ln10002000​=21​ln56​(ln56​k​−0)=21​k

$$\Rightarrow k=2\mathrm{ln}2$$⇒k=2ln2

$$\therefore {(\frac{k}{\mathrm{ln}2})}^{2}={(\frac{2\mathrm{ln}2}{\mathrm{ln}2})}^{2}=4$$∴(ln2k​)2=(ln22ln2​)2=4