To find the half life of iodine
(i.e) \(t =\ ?\ when\ P=\frac{P_{0}}{2}=\ \frac{10}{2}=5\) grams of iodine
\(P = 10_{e}^{−0.086t}\)
substitute \(P = 5\) in the above equation
\(5 = 10_{e}^{−0.086t}\)
Dividing both sides by 10 we get,
\(\frac{5}{10}=e^{-0.086t}\)

\(\Rightarrow\ e^{-0.086t}=\ \frac{1}{2}\)

\(e^{0.086t} = 2\) Taking log on both sides we get, \(0.086t = \log_{e} 2\)

\(\Rightarrow\ t=\ \frac{\log_{e}2}{0.086}=\ \frac{0.69310}{0.086}=8.0598\ \approx\ 8\) Therefore it took 8 days for the iodine reduces to 5 grams

\(\Rightarrow\ e^{-0.086t}=\ \frac{1}{2}\)

\(e^{0.086t} = 2\) Taking log on both sides we get, \(0.086t = \log_{e} 2\)

\(\Rightarrow\ t=\ \frac{\log_{e}2}{0.086}=\ \frac{0.69310}{0.086}=8.0598\ \approx\ 8\) Therefore it took 8 days for the iodine reduces to 5 grams