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 $\mathit{e}$. 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 $\boxed{{\displaystyle -0.11552}}$.