Question

The half-life of a certain radioactive substance is 13 days. There are 2.5 grand of the substance in totally. Type an expression for the amount of substance as a function of t. When will there be less than 1 g remaining? Please show work

Answer 1

`\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( (1)/(2) \right)^{(t)/(h)}\qquad \begin{cases} A=\textit{current amount}\dotfill &\\ P=\textit{initial amount}\dotfill &2.5\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &13 \end{cases} \\\\\\ ~\hfill {\Large \begin{array}{llll} A=2.5\left( (1)/(2) \right)^{(t)/(13)} \end{array}} ~\hfill`

well, let's instead find when will there be 1 gram first off, then after that it's all downhill, so A = 1

`\stackrel{A}{1}=2.5\left( (1)/(2) \right)^{(t)/(13)}\implies \cfrac{1}{2.5}=\left( (1)/(2) \right)^{(t)/(13)}\implies \log\left( \cfrac{1}{2.5} \right)=\log\left[ \left( (1)/(2) \right)^{(t)/(13)} \right]`

`\log\left( \cfrac{1}{2.5} \right)=t\log\left[ \left( (1)/(2) \right)^{(1)/(13)} \right]\implies \cfrac{\log\left( (1)/(2.5) \right)}{\log\left[ \left( (1)/(2) \right)^{(1)/(13)} \right]}=t\stackrel{\textit{after about 17 days and 5 hours}}{\implies 17.19\approx t ~\hfill }`