Research shows that the radioactive isotope Americium-241 has a half-life of 432.2 years Use the following to construct a function that will model the amount of Americium-241 remaining after $t$ years, from an initial amount of $15 \mathrm{~kg}$. \[ Q(t)=P e^{r t} \] Where $Q(t)$ describes the amount of Americium-241 remaining after $t$ years from an initial quantity of $P \mathrm{~kg}$. 1. $Q(t)=$ 2. How long (in years) will it take for the amount of Americium-241 remaining to reach $6 \mathrm{~kg}$ ?

Step 1 ：Given that the half-life of Americium-241 is 432.2 years, we can set up the equation \(P e^{r \cdot 432.2} = \frac{P}{2}\) to solve for the decay rate \(r\).

Step 2 ：Dividing both sides by \(P\) gives \(e^{r \cdot 432.2} = \frac{1}{2}\).

Step 3 ：Taking the natural logarithm of both sides gives \(r \cdot 432.2 = \ln\left(\frac{1}{2}\right)\).

Step 4 ：Solving for \(r\) gives \(r = \frac{\ln\left(\frac{1}{2}\right)}{432.2}\).

Step 5 ：So, the function \(Q(t)\) that models the amount of Americium-241 remaining after \(t\) years from an initial quantity of \(15 \mathrm{~kg}\) is \(Q(t) = 15 e^{\frac{\ln\left(\frac{1}{2}\right)}{432.2} \cdot t}\).

Step 6 ：To find how long it will take for the amount of Americium-241 remaining to reach \(6 \mathrm{~kg}\), we can set up the equation \(6 = 15 e^{\frac{\ln\left(\frac{1}{2}\right)}{432.2} \cdot t}\) and solve for \(t\).

Step 7 ：Dividing both sides by 15 gives \(e^{\frac{\ln\left(\frac{1}{2}\right)}{432.2} \cdot t} = \frac{6}{15}\).

Step 8 ：Taking the natural logarithm of both sides gives \(\frac{\ln\left(\frac{1}{2}\right)}{432.2} \cdot t = \ln\left(\frac{6}{15}\right)\).

Step 9 ：Solving for \(t\) gives \(t = \frac{\ln\left(\frac{6}{15}\right)}{\frac{\ln\left(\frac{1}{2}\right)}{432.2}}\).

Step 10 ：Calculating this gives \(t \approx 720.3\) years. So, it will take approximately 720.3 years for the amount of Americium-241 remaining to reach \(6 \mathrm{~kg}\).

Step 11 ：\(\boxed{t \approx 720.3}\) years is the final answer.