Rui Baldwin

2020-11-30

Use your equation to determine the half-life ofthis type of Fodine, That is, find
out how many days it will take for half of the original amount to be left. Show an
algebraic solution using logs

Viktor Wiley

Skilled2020-12-01Added 84 answers

To find the half life of iodine
(i.e) $t=\text{}?\text{}when\text{}P=\frac{{P}_{0}}{2}=\text{}\frac{10}{2}=5$ 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,
$\frac{5}{10}={e}^{-0.086t}$

$\Rightarrow \text{}{e}^{-0.086t}=\text{}\frac{1}{2}$

${e}^{0.086t}=2$
Taking log on both sides we get,
$0.086t={\mathrm{log}}_{e}2$

$\Rightarrow \text{}t=\text{}\frac{{\mathrm{log}}_{e}2}{0.086}=\text{}\frac{0.69310}{0.086}=8.0598\text{}\approx \text{}8$
Therefore it took 8 days for the iodine reduces to 5 grams