[tex]\bf \textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\\ t=\textit{elapsed time}\dotfill &120\\ h=\textit{half-life}\dotfill &20 \end{cases} \\\\\\ A=P\left( \frac{1}{2} \right)^{\frac{120}{20}}\implies A=P\left( \frac{1}{2} \right)^6\implies A=P\left( \cfrac{1^6}{2^6} \right)\implies A=\cfrac{1}{64}P[/tex]
