We can represent the exponential decay function as
[tex]f(t)=f(0)e^{kt}[/tex]
where f(0) is the initial value of the function and k is a constant.
Since we know the half-life of that liquid is 2.2 hours, f(2.2) equals f(0)/2. So, we have:
[tex]\begin{gathered} \frac{f\mleft(0\mright)}{2}=f(0)e^{2.2k} \\ \\ \frac{1}{2}=e^{2.2k} \\ \\ \ln \frac{1}{2}=\ln (e^{2.2k}) \\ \\ \ln 0.5=2.2k \\ \\ k=\frac{\ln 0.5}{2.2} \end{gathered}[/tex]
Thus, we can use this result and f(0) = 2.25 (ml) to determine the function f(x):
[tex]f(t)=2.25\cdot e^{\frac{\ln0.5}{2.2}t}[/tex]
Thus, to find the number of hours passed after the liquid was administrated when it decreases to 0.41 ml, we need to make f(t) = 0.41 and find t:
[tex]\begin{gathered} 0.41=2.25\cdot e^{\frac{\ln0.5}{2.2}t} \\ \\ \frac{0.41}{2.25}=e^{\frac{\ln0.5}{2.2}t} \\ \\ \ln \frac{0.41}{2.25}=\ln e^{\frac{\ln0.5}{2.2}t} \\ \\ \frac{\ln 0.5}{2.2}t=\ln \frac{0.41}{2.25} \\ \\ t=\frac{2.2\cdot\ln \frac{0.41}{2.25}}{\ln 0.5} \\ \\ t\cong5.40370 \end{gathered}[/tex]
So approximately 5.404 hours passed since the liquid was administrated at 7:00 am. Then, to find the hour and minutes, let's convert:
[tex]5.404h=5h+0.404h=5h+0.404\cdot60\min \cong5h+24\min [/tex]
Thus, adding this result to 7:00 am, we find:
[tex]7am+5h+24\min =12pm+24\min =12\colon24pm[/tex]
