## Precalculus (6th Edition) Blitzer

a) The decay model is $A\left( t \right)=900{{e}^{-0.017t}}$. b) The amount is $759.298\,\,\text{grams}$.
(a) Consider that at any time $t$, the amount of the radioactive substance present is $A\left( t \right)$, following the exponential decay model: $A\left( t \right)={{A}_{0}}{{e}^{kt}}$ Initially there are 900 grams of the substance present,$A\left( 0 \right)=900$, hence \begin{align} & 900={{A}_{0}}{{e}^{0\times k}} \\ & {{A}_{0}}=900 \end{align} Therefore $A\left( t \right)=900{{e}^{kt}}$ The substance has a half-life of 40 days; hence for $t=40$, $A\left( 40 \right)=450$ Therefore: \begin{align} & 450=900{{e}^{40k}} \\ & {{e}^{40k}}=\frac{1}{2} \\ & k=\frac{1}{40}{{\log }_{e}}\frac{1}{2} \\ & =-0.017 \end{align} Therefore, the decaying model for the substance is $A\left( t \right)=900{{e}^{-0.017t}}$ (b) The radioactive substance remains after 10 days, that is, for $t=10$ \begin{align} & A\left( 10 \right)=900{{e}^{-0.017\times 10}} \\ & =759.298 \end{align} Thus, the radioactive substance that remains after 10 days is $759.298\,\,\text{grams}$.