Answer
a)
The exponential decay rate $k$ is approximately equal to 0.341, and the equation for the exponential decay function is $P\left( t \right)=15.5{{e}^{-0.341t}}$.
b)
5.6 micrograms/milliliter.
c)
4 hours.
d)
2 hours.
Work Step by Step
a)
Consider the exponential decay function, $P\left( t \right)={{P}_{0}}{{e}^{-kt}}$ …… (1)
Now, put 11 at the place of $P\left( t \right)$ and 1 at the place of $t$ in equation-1:
$11={{P}_{0}}{{e}^{-k\left( 1 \right)}}$ …… (2)
Again, put 2 at the place of $P\left( t \right)$ and 6 at the place of $t$ in equation-1:
$2={{P}_{0}}{{e}^{-k\left( 6 \right)}}$ …… (3)
Here, the number of equations is 2, and also the number of unknowns is 2, so solve these two equations and find out the unknowns of the equation as,
$\begin{align}
& 11={{P}_{0}}{{e}^{-k\left( 1 \right)}} \\
& 11={{P}_{0}}{{e}^{-k}} \\
& 11{{e}^{k}}={{P}_{0}}
\end{align}$
And
$\begin{align}
& 2={{P}_{0}}{{e}^{-k\left( 6 \right)}} \\
& 2={{P}_{0}}{{e}^{-6k}} \\
& 2{{e}^{6k}}={{P}_{0}}
\end{align}$
Now, use substitution and solve for $k$ as,
$11{{e}^{k}}=2{{e}^{6k}}$
$\begin{align}
& 11{{e}^{k}}=2{{e}^{6k}} \\
& \ln 11{{e}^{k}}=\ln 2{{e}^{6k}} \\
& \ln 11+\ln {{e}^{k}}=\ln 2+\ln {{e}^{6k}} \\
& \ln 11+k=\ln 2+6k
\end{align}$
Further simplified,
$\begin{align}
& \ln 11+k=\ln 2+6k \\
& \ln 11-\ln 2=6k-k \\
& \ln 11-\ln =5k \\
& \frac{\ln 11-\ln }{5}=k
\end{align}$
Further simplified,
$\begin{align}
& \frac{\ln 11-\ln 2}{5}=k \\
& \frac{2.397-0.693}{5}=k \\
& \frac{1.704}{5}=k \\
& 0.341\approx k
\end{align}$
Then, the exponential decay rate is $k\approx 0.341$.
$\begin{align}
& 11={{P}_{0}}{{e}^{-0.341\left( 1 \right)}} \\
& 11={{P}_{0}}{{e}^{-0.341}} \\
& 11{{e}^{0.341}}={{P}_{0}} \\
& 15.5\approx {{P}_{0}}
\end{align}$
Now, put the values of $k$ and ${{P}_{0}}$ in exponential decay function.
Thus, the exponential decay function is $P\left( t \right)=15.5{{e}^{-0.341t}}$.
Where $t$ is in hours.
b)
From part-(a), it is clear that the exponential decay function is $P\left( t \right)=15.5{{e}^{-0.341t}}$. Thus:
$\begin{align}
& P\left( t \right)=15.5{{e}^{-0.341t}} \\
& P\left( 3 \right)=15.5{{e}^{-0.341\left( 3 \right)}} \\
& P\left( 3 \right)=15.5{{e}^{-1.023}} \\
& P\left( 3 \right)=15.5\left( 0.359 \right)
\end{align}$
Further simplified,
$\begin{align}
& P\left( 3 \right)=15.5\left( 0.359 \right) \\
& P\left( 3 \right)=5.572 \\
& P\left( 3 \right)\approx 5.6
\end{align}$
c)
From part-(a), it is clear that the exponential decay function is $P\left( t \right)=15.5{{e}^{-0.341t}}$, so put 4 at a place of $P\left( t \right)$ and calculate the value of $t$:
$\begin{align}
& P\left( t \right)=15.5{{e}^{-0.341t}} \\
& \left( 4 \right)=15.5{{e}^{-0.341t}} \\
& \frac{4}{15.5}={{e}^{-0.341t}}
\end{align}$
$\begin{align}
& \frac{4}{15.5}={{e}^{-0.341t}} \\
& \ln \frac{4}{15.5}=\ln {{e}^{-0.341t}} \\
& \ln 0.258=-0.341t \\
& -1.354=-0.341t
\end{align}$
$\begin{align}
& -1.354=-0.341t \\
& 1.354=0.341t \\
& \frac{1.354}{0.341}=t \\
& 4\approx t
\end{align}$
d)
$\begin{align}
& P\left( t \right)=15.5{{e}^{-0.341t}} \\
& \frac{1}{2}\left( 15.5 \right)=15.5{{e}^{-0.341t}} \\
& \frac{7.75}{15.5}={{e}^{-0.341t}} \\
\end{align}$
$\begin{align}
& \frac{7.75}{15.5}={{e}^{-0.341t}} \\
& \ln \frac{7.75}{15.5}=\ln {{e}^{-0.341t}} \\
& -0.693=-0.341t \\
& 0.693=0.341t
\end{align}$
Further simplified,
$\begin{align}
& 0.693=0.341t \\
& \frac{0.693}{0.341}=t \\
& 2\approx t
\end{align}$
