Given the mass of a radioactive substance after time t days is given by
[tex]m(t)=Ae^{-kt}[/tex]
The initial mass of the substance is 120 milligrams, which means m(0)=120. So,
[tex]\begin{gathered} m(0)=120 \\ \Rightarrow Ae^0=120 \\ \Rightarrow A=120 \end{gathered}[/tex]
The mass of the substance is 90 milligrams after 10 days, which means m(10)=90. So,
[tex]\begin{gathered} m(10)=90 \\ \Rightarrow120e^{-10k}=90 \\ \Rightarrow e^{-10k}=\frac{3}{4} \end{gathered}[/tex]
Now, take natural logarithm on both the side and use the property
[tex]\ln (e^n)=n[/tex]
So, the above equation will become
[tex]\begin{gathered} \ln (e^{-10k})=\ln (\frac{3}{4}) \\ \Rightarrow-10k=-0.2877 \\ \Rightarrow k=\frac{-0.2877}{-10} \\ \Rightarrow k=0.02877 \end{gathered}[/tex]
Therefore, the relation between the mass of the substance at time t days is given by t
[tex]m(t)=120e^{-0.02877t}[/tex]
The mass of the substance after 16 days will be
[tex]m(16)=120e^{-0.02877(16)}=120(0.631)=75.729\text{ milligrams}[/tex]
The day during which the mass of the substance reaches 50 milligrams can be obtained as follows:
[tex]\begin{gathered} m(t)=50 \\ 120e^{\mleft\{-0.02877t\mright\}}=50 \\ \Rightarrow e^{-0.02877t}=\frac{50}{120} \\ \Rightarrow-0.02877t=\ln (\frac{50}{120}) \\ \Rightarrow t=\frac{-0.8754}{-0.02877} \\ \Rightarrow t=30.43\text{ days} \end{gathered}[/tex]
The graph that shows the relationship between m and t is given below:
