illusiia

2021-01-04

Use your equation to determine the half-life ofthis type of lodine. 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.

Margot Mill

Skilled2021-01-05Added 106 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

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