1. Replacing with t = 9 into the equation, we get:
[tex]\begin{gathered} A(t)=100\cdot e^{-0.087\cdot9} \\ A(t)=100\cdot e^{-0.783} \\ A(t)=100\cdot0.457 \\ A(t)=45.7 \end{gathered}[/tex]There are 45.7 grams of iodine left, after 9 days
2. Replacing with A(t) = 70 into the equation, we get:
[tex]\begin{gathered} 70=100\cdot e^{-0.087t} \\ \frac{70}{100}=e^{-0.087t} \\ \ln (\frac{70}{100})=-0.087\cdot t \\ \frac{-0.356}{-0.087}=t \\ 4\approx t \end{gathered}[/tex]70g of iodine will be left after 4 days
