[tex]\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\dotfill &20\\ P=\textit{initial amount}\dotfill &200\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &12 \end{cases} \\\\\\ 20=200\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{20}{200}=\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{1}{10}=\left( \frac{1}{2} \right)^{\frac{t}{12}}[/tex]
[tex]\log\left( \cfrac{1}{10} \right)=\log\left[ \left( \frac{1}{2} \right)^{\frac{t}{12}} \right]\implies \log\left( \cfrac{1}{10} \right)=t\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right] \\\\\\ \cfrac{\log\left( \frac{1}{10} \right)}{\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right]}=t\implies \implies \stackrel{mins}{39.9}\approx t[/tex]
