Given differential equation

$\frac{dP}{dt}=0.5P-450$

with initial condition

$P(0)=850$

We need the time when population becomes zero, i.e. when $P=0$.

Rewrite the equation

$\frac{dP}{dt}=\frac12 P-450=\frac12(P-900)$

So,

$\frac{dP}{P-900}=\frac12 dt$

Integrate both sides

$\int \frac{dP}{P-900}=\int \frac12 dt$

$\ln|P-900|=\frac t2 + C$

Hence,

$P-900=Ce^{t/2}$

so

$P=900+Ce^{t/2}$

Use the initial condition

Since $P(0)=850$,

$850=900+C$

$C=-50$

Therefore,

$P(t)=900-50e^{t/2}$

Find when population becomes zero

Set $P(t)=0$:

$0=900-50e^{t/2}$

$50e^{t/2}=900$

$e^{t/2}=18$

Taking natural log,

$\frac t2=\ln 18$

$t=2\ln 18$

Match with options

$t=2\log_e 18$

This is Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.