Help me find the mistake for the spread of rumour logistic equation Question
Let us suppose that 20 people have heard a rumour by 8am in a town of 9,000 people. Assume that the spread of the rumour follows the logistic equation,
$$\frac{dP}{dt}=kp\left(1-\frac{P}{C}\right)$$
Given that 500 people have heard the rumour by 12 noon, determine how long it takes for half the population to have heard the rumour. Enter your answer in hours rounded to exactly one decimal place.

My Answer
$$\frac{dP}{dt}=kp\left(\frac{C-P}{C}\right)$$
$$\frac{1}{P(C-P)} dP=\frac{k}{C} dt$$
$$\left[\frac{\ln P-\ln (P-C)}{C}\right]_{20}^P=\left[\frac{kt}{C}\right]_8^t$$
$$\ln P - \ln\left(\frac{20}{20-C}\right)=kt-k(8)$$
$$\ln \left(\frac{P(20-C)}{20}\right)=e^{k(t-8)}$$
Substitute P=20 when t=8
$$20-C=e^{k(8-8)}$$
$$\implies C=19$$
Substitute P=500 when t=12
$$25=e^4$$
$$k=0.8047$$
So finally since we want to find when the population is halfed so we take P=600 to find t
$$225=e^{0.8047(t-8)}$$
$$t=14.7$$
However this answer is unfortunately wrong, and I dont know why. Will really appreciate if anyone can shed some light to what I am doing wrong here.

Assume that the population of the town is $C=9000$ and that at 8am, $t=0$.
At $t=0, P=200$ (given).
Also, at $t=4$ (i.e. 12 noon), $P=500$ (given).
$$\begin{align}
\left[\ln \frac P{C-P}\right]_{20}^{500}&=\bigg[kt\bigg]_0^4\qquad (C=9000)\\
\left[\ln \frac 1{\frac {9000}P-1}\right]_{20}^{500}&=4k\\
\ln(449)-\ln(17)&=4k&\cdots (1)
\end{align}$$
When rumour has spread to half of the population, $P=\frac 12 C=4500$, $t=T$.
$$\begin{align}
\left[\ln \frac 1{\frac {9000}P-1}\right]_{20}^{4500}&=Tk\\
\ln(449)&=Tk&\qquad\qquad \qquad\cdots (2)\\
\end{align}$$
$(2)\div (1)$:
$$T=\frac {4\ln449}{\ln449-\ln17}=7.46\approx 7.5$$
Time = $8$am + $7.5$ hrs = $\color{red}{3.30\text{pm}}$.
