The population $$P = P(t)$$ at time $$t$$ of a certain species follows the differential equation $$\frac{dP}{dt} = 0.5P - 450$$. If $$P(0) = 850$$, then the time at which population becomes zero is:

We are given the differential equation $$\frac{dP}{dt} = 0.5P - 450$$ with $$P(0) = 850$$, and we need to find the time $$t$$ when $$P = 0$$.

We rewrite the equation as $$\frac{dP}{dt} = 0.5(P - 900)$$.

Separating variables, we get $$\frac{dP}{P - 900} = 0.5 \, dt$$.

Integrating both sides, $$\ln|P - 900| = 0.5t + C$$, where $$C$$ is the constant of integration.

Now we use the initial condition $$P(0) = 850$$. Substituting $$t = 0$$ and $$P = 850$$, we get $$\ln|850 - 900| = C$$, so $$C = \ln 50$$.

So the solution is $$\ln|P - 900| = 0.5t + \ln 50$$.

We need to find $$t$$ when $$P = 0$$. Substituting $$P = 0$$, we get $$\ln|0 - 900| = 0.5t + \ln 50$$, which gives $$\ln 900 = 0.5t + \ln 50$$.

Rearranging, $$0.5t = \ln 900 - \ln 50 = \ln\frac{900}{50} = \ln 18$$.

So $$t = 2\ln 18 = 2\log_e 18$$.

Hence, the correct answer is Option B.