Solution:
Given the amount A, in micrograms remaining in that person after t hours is given by the equation;
[tex]A=3e^{-0.046t}[/tex]
(a) The graph of the equation is;
(b) After 7 hours;
[tex]\begin{gathered} t=7 \\ A=3e^{-0.046t} \\ A=3e^{-0.046\left(7\right)} \\ A=3e^{-0.322} \\ A=2.17\mu g \end{gathered}[/tex]
(c) At half life of sodium -24;
[tex]\begin{gathered} A=\frac{3}{2} \\ A=1.5\mu g \\ 1.5=3e^{-0.046t} \\ \frac{1.5}{3}=e^{-0.046t} \\ 0.5=e^{-0.046t} \\ \ln\mleft(0.5\mright)=\ln\mleft(e^{-0.046t}\mright) \\ -0.6931=-0.046t \\ t=-\frac{0.6931}{-0.046} \\ t=15.07h \end{gathered}[/tex]
(d) When the amount of sodium -24 is 1 microgram;
[tex]\begin{gathered} 1=3e^{-0.046t} \\ \frac{1}{3}=e^{-0.046t} \\ 0.3333=e^{-0.046t} \\ \ln\mleft(0.3333\mright)=\ln\mleft(e^{-0.046t}\mright) \\ -1.0986=-0.046t \\ t=-\frac{1.0986}{-0.046} \\ t=23.88h \end{gathered}[/tex]
