The formula for calculating half-life is as shown below
[tex]N(t)=N_0(\frac{1}{2})^{\frac{t}{t2}}[/tex]Where:
[tex]\begin{gathered} N(t)=quantity\text{ of the substance remaining} \\ N_0=initial\text{ quantity of the substance} \\ t=time\text{ elapsed} \\ t_{\frac{1}{2}}=half\text{ life of the substance} \end{gathered}[/tex]Given that
[tex]\begin{gathered} N(t)=\text{?} \\ N_0=30\text{milligrams} \\ t=12\text{hours} \\ t_{\frac{1}{2}}=36\text{hours} \end{gathered}[/tex]So,
[tex]\begin{gathered} N(t)=30(\frac{1}{2})^{\frac{12}{36}} \\ N(t)=30(\frac{1}{2})^{\frac{1}{3}} \\ N(t)=30\times0.7937 \\ N(t)=23.811 \\ N(t)=23.8(\text{nearest tenth)} \end{gathered}[/tex]Hence, there is approximately 23.8 milligrams of Valium in the patient's blood at noon on the first day.
B. To estimate when the Valium concentration will reach 20% of its initial level, it is observed that
[tex]\begin{gathered} N(t)=20\text{ \% of }N_0,N_0=30 \\ \\ N(t)=\frac{20}{100}\times30 \\ N(t)=0.2\times30=6 \end{gathered}[/tex]Let us substitute into the formula to get t as shown below:
[tex]\begin{gathered} N(t)=N_0(\frac{1}{2})^{\frac{t}{36}} \\ 6=30_{}(0.5)^{\frac{t}{36}} \\ \frac{6}{30}=(0.5)^{\frac{t}{36}} \\ 0.2=0.5^{\frac{t}{36}} \\ ^{} \end{gathered}[/tex][tex]\begin{gathered} In0.2=\frac{t}{36}In0.5 \\ t=\frac{36In0.2}{In0.5} \\ t=\frac{36\times-1.6094}{-0.6931} \\ t=83.5894hours \end{gathered}[/tex]Hence, the valium concentration will reach 205 of its initial level at 83.6 hours
