[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 &75\\ P=\textit{initial amount}\dotfill &2400\\ t=seconds\\ h=\textit{half-life}\dotfill &13.8 \end{cases} \\\\\\ 75=2400\left( \frac{1}{2} \right)^{\frac{t}{13.8}}\implies \cfrac{75}{2400}=\left( \frac{1}{2} \right)^{\frac{t}{13.8}}\implies \cfrac{1}{32}=\left( \frac{1}{2} \right)^{\frac{1}{13.8} t}[/tex]
[tex]\log\left( \cfrac{1}{32} \right)=\log\left[ \left( \frac{1}{2} \right)^{\frac{1}{13.8} t} \right]\implies \log\left( \cfrac{1}{32} \right)=t\log\left[ \left( \frac{1}{2} \right)^{\frac{1}{13.8}} \right] \\\\\\ \cfrac{\log\left( \frac{1}{32} \right)}{\log\left[ \left( \frac{1}{2} \right)^{\frac{1}{13.8}} \right]}=t\implies \cfrac{\log\left( \frac{1}{32} \right)}{\log\left[ \sqrt[13.8]{\frac{1}{2}} \ \right]}=t\implies 69\approx t[/tex]
