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

The rate of growth of bacteria is proportional to the number present, so $$\frac{dN}{dt} = \lambda N$$, where $$\lambda$$ is a constant. The solution is $$N(t) = N(0) e^{\lambda t} = 1000 e^{\lambda t}$$.

Since the bacteria count increases by 20% in 2 hours: $$N(2) = 1200 = 1000 e^{2\lambda}$$, giving $$e^{2\lambda} = \frac{6}{5}$$, so $$\lambda = \frac{1}{2}\ln\left(\frac{6}{5}\right)$$.

We need to find when $$N(t) = 2000$$: $$2000 = 1000 e^{\lambda t}$$, so $$e^{\lambda t} = 2$$, which gives $$t = \frac{\ln 2}{\lambda} = \frac{2\ln 2}{\ln(6/5)}$$.

We are told this time equals $$\frac{k}{\log_e(6/5)}$$, so $$\frac{k}{\ln(6/5)} = \frac{2\ln 2}{\ln(6/5)}$$, giving $$k = 2\ln 2$$.

Therefore $$\left(\frac{k}{\log_e 2}\right)^2 = \left(\frac{2\ln 2}{\ln 2}\right)^2 = 2^2 = 4$$.