a. The model of the decay equation is given by

\(\displaystyle{P}={P}_{{0}}{e}^{k}{t}\)

Here \(\displaystyle{P}_{{0}}={10}\) grams of iodine

\(\displaystyle{k}=\text{rate of continuous rate}=-{8.6}\%\) { negative sign implies the decay}

Which implies \(\displaystyle{k}=-{0.086}\)

t is measured in days

Therefore, the decay equation for this type of Iodine is

\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)

b. To find the half life of iodine

\(\displaystyle{\left({i}.{e}\right)}{t}=?\ \text{then}\ {P}=\frac{{P}_{{0}}}{{2}}=\frac{10}{{2}}={5}\) grams of iodine

\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)

substitute \(\displaystyle{P}={5}\) in the above equation

\(\displaystyle{5}={10}{e}^{{-{0.086}{t}}}\)

Dividing both sides by 10 we get,

\(\displaystyle\frac{5}{{10}}={e}^{{-{0.086}{t}}}\)

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

\(\displaystyle{e}^{{{0.086}{t}}}={2}\)

Taking log on both sides we get,

\(\displaystyle{0.086}{t}={{\log}_{{e}}{2}}\)

\(\displaystyle\Rightarrow{t}=\frac{{{{\log}_{{e}}{2}}}}{{{0.086}}}=\frac{{{0.6931}}}{{{0.086}}}={8.0598}\approx{8}\)

Therefore it took 8 days for the iodine reduces to 5 grams

\(\displaystyle{P}={P}_{{0}}{e}^{k}{t}\)

Here \(\displaystyle{P}_{{0}}={10}\) grams of iodine

\(\displaystyle{k}=\text{rate of continuous rate}=-{8.6}\%\) { negative sign implies the decay}

Which implies \(\displaystyle{k}=-{0.086}\)

t is measured in days

Therefore, the decay equation for this type of Iodine is

\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)

b. To find the half life of iodine

\(\displaystyle{\left({i}.{e}\right)}{t}=?\ \text{then}\ {P}=\frac{{P}_{{0}}}{{2}}=\frac{10}{{2}}={5}\) grams of iodine

\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)

substitute \(\displaystyle{P}={5}\) in the above equation

\(\displaystyle{5}={10}{e}^{{-{0.086}{t}}}\)

Dividing both sides by 10 we get,

\(\displaystyle\frac{5}{{10}}={e}^{{-{0.086}{t}}}\)

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

\(\displaystyle{e}^{{{0.086}{t}}}={2}\)

Taking log on both sides we get,

\(\displaystyle{0.086}{t}={{\log}_{{e}}{2}}\)

\(\displaystyle\Rightarrow{t}=\frac{{{{\log}_{{e}}{2}}}}{{{0.086}}}=\frac{{{0.6931}}}{{{0.086}}}={8.0598}\approx{8}\)

Therefore it took 8 days for the iodine reduces to 5 grams