Four million $E$. coli bacteria are present in a laboratory culture. An antibacterial agent is introduced and the population of bacteria $P(t)$ decreases by half every $6 \mathrm{hr}$. The population can be represented by
\[
P(t)=4,000,000\left(\frac{1}{2}\right)^{t / 6}
\]
Part: $0 / 2$
Part 1 of 2
(a) Convert the given model to an exponential function using base $\boldsymbol{e}$. Round the value of $k$ to five decimal places.
\[
P(t)=
\]

Step 1 ：The given function is in the form of an exponential decay function. We can convert it to the form \(P(t) = P_0 e^{kt}\), where \(P_0\) is the initial population, \(k\) is the decay constant, and \(t\) is the time.

Step 2 ：We know that \(P_0 = 4000000\) and the population decreases by half every 6 hours. So, we can write the given function as \(P(t) = P_0 \left(\frac{1}{2}\right)^{t/6}\).

Step 3 ：To convert this to the form \(P(t) = P_0 e^{kt}\), we can use the fact that \(e^{ln(a)} = a\). So, we can write \(\left(\frac{1}{2}\right)^{t/6}\) as \(e^{ln\left(\left(\frac{1}{2}\right)^{t/6}\right)}\).

Step 4 ：This simplifies to \(e^{t/6 \cdot ln\left(\frac{1}{2}\right)}\). So, \(k = \frac{1}{6} ln\left(\frac{1}{2}\right)\).

Step 5 ：Let's calculate the value of \(k\).

Step 6 ：\(k = -0.11552453009332421\)

Step 7 ：Final Answer: The value of \(k\) is \(\boxed{-0.11552}\).