[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 &11\\ P=\textit{initial amount}\dotfill &160\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &14 \end{cases} \\\\\\ 11=160\left( \frac{1}{2} \right)^{\frac{t}{14}}\implies \cfrac{11}{160}=\left( \cfrac{1}{2} \right)^{\frac{t}{14}}\implies \log\left( \cfrac{11}{160} \right)=\log\left[ \left( \cfrac{1}{2} \right)^{\frac{t}{14}} \right][/tex]
[tex]\log\left( \cfrac{11}{160} \right)=\cfrac{t}{14}\log\left[ \left( \cfrac{1}{2} \right)\right]\implies \cfrac{\log\left( \frac{11}{160} \right)}{\log \left( \frac{1}{2} \right)}=\cfrac{t}{14} \\\\\\ 14\left( \cfrac{\log\left( \frac{11}{160} \right)}{\log \left( \frac{1}{2} \right)} \right)=t\implies {\LARGE \begin{array}{llll} \stackrel{mins}{54.1}\approx t \end{array}}[/tex]
