Find an exponential function of the form $P(t)=P_{0} n^{\frac{t}{T}}$ that models the situation, and then find the equivalent exponential model of the form $P(t)=P_{0} e^{r t}$. Doubling time of $6 \mathrm{yr}$, initial population of 400 .

Step 1 ：Given that the doubling time is 6 years, we can say that the population doubles every 6 years. This means that after 6 years, the population is twice the initial population. We can write this as $P(6)=2P_{0}$.

Step 2 ：Substitute $P(6)=2P_{0}$ into the exponential function $P(t)=P_{0} n^{\frac{t}{T}}$, we get $2P_{0}=P_{0} n^{\frac{6}{T}}$.

Step 3 ：Divide both sides by $P_{0}$, we get $2=n^{\frac{6}{T}}$.

Step 4 ：Since the base of the exponential function is $n$, we can take the natural logarithm of both sides to solve for $n$. We get $\ln 2=\frac{6}{T} \ln n$.

Step 5 ：Divide both sides by $\frac{6}{T}$, we get $n=e^{\frac{\ln 2}{6/T}}$.

Step 6 ：Substitute $T=6$ into the equation, we get $n=e^{\frac{\ln 2}{6/6}}=e^{\ln 2}=2$.

Step 7 ：Substitute $n=2$ back into the exponential function $P(t)=P_{0} n^{\frac{t}{T}}$, we get $P(t)=P_{0} 2^{\frac{t}{6}}$.

Step 8 ：Now we need to find the equivalent exponential model of the form $P(t)=P_{0} e^{r t}$. We know that $2=e^{ln 2}$, so we can write $P(t)=P_{0} e^{(\ln 2) \frac{t}{6}}$.

Step 9 ：Simplify the exponent, we get $P(t)=P_{0} e^{\frac{t \ln 2}{6}}$.

Step 10 ：Therefore, the equivalent exponential model of the form $P(t)=P_{0} e^{r t}$ is $P(t)=P_{0} e^{\frac{t \ln 2}{6}}$.

Step 11 ：Finally, substitute the initial population $P_{0}=400$ into the equation, we get $P(t)=400 e^{\frac{t \ln 2}{6}}$.

Step 12 ：Check the result: When $t=6$, $P(6)=400 e^{\frac{6 \ln 2}{6}}=400 e^{\ln 2}=400 \times 2=800$, which is twice the initial population. Therefore, the result meets the requirements of the problem.