# Question #2ac56

Dec 3, 2014

I think the answer is 7.2 secons.

An exponential decay function can be easily written as

$N \left(t\right) = {N}_{0} \left(t\right) \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$ , where

$N \left(t\right)$ - the quantity that remains and has not yet decayed after a time t;
${N}_{0} \left(t\right)$ - the initial quantity of the substance that will decay;
${t}_{\frac{1}{2}}$ - the half-life of the decaying quantity;

We have $N \left(t\right) = 12.5$ grams, and ${N}_{0} \left(t\right) = 100$ grams, therefore

$12.5 = 100 \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$ ; we also know that the requested decay process took 21.6 seconds, so $t = 21.6$ seconds.

$\frac{12.5}{100} = {\left(\frac{1}{2}\right)}^{\frac{21.6}{t} _ \left(\frac{1}{2}\right)}$

$\frac{21.6}{t} _ \left(\frac{1}{2}\right) = {\log}_{\frac{1}{2}} \left(0.125\right)$

$\frac{21.6}{t} _ \left(\frac{1}{2}\right) = 3$, which gives us ${t}_{\frac{1}{2}} = 7.2$ seconds.