We can start off here by acknowledging that we know the initial amount of the radioactive element is 100 and so \({Q_0} = 100\). Therefore the equation is then,

\[Q\left( t \right) = 100{{\bf{e}}^{k\,t}}\]

Now, we also know that \(Q\left( {1250} \right) = 80\) and plugging this into the equation above gives,

\[80 = Q\left( {1250} \right) = 100{{\bf{e}}^{1250k}}\]

We can use techniques from the Solve Logarithm Equations section to determine the value of \(k\).

\[\begin{align*}80 & = 100{{\bf{e}}^{1250k}}\\ \frac{{80}}{{100}} & = {{\bf{e}}^{1250k}}\\ \ln \left( {\frac{4}{5}} \right) & = 1250k\\ k = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right) & = - 0.000178515\end{align*}\]

Depending upon your preferences we can use either the exact value or the decimal value. Note however that because \(k\) is in the exponent of an exponential function we’ll need to use quite a few decimal places to avoid potentially large differences in the value that we’d get if we rounded off too much.

Putting all of this together the exponential decay equation for this population is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{Q = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}}}\]

What we’re really being asked to do here is to solve the equation,

\[50 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\]

and we know from the Solve Logarithm Equations section how to do that. Here is the solution work for this part.

\[\begin{align*}\frac{{50}}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\\ \ln \left( {\frac{1}{2}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {\frac{1}{2}} \right)}}{{\ln \left( {\frac{4}{5}} \right)}} = 3882.8546}}\end{align*}\]

It will take 3882.8546 years for half of the element to decay. On a side note this time is called the half-life of the element.

In this part we’re being asked to solve the equation,

\[1 = Q\left( t \right) = 100{{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\]

The solution process for this part is the same as that for the previous part. Here is the solution work for this part.

\[\begin{align*}\frac{1}{{100}} & = {{\bf{e}}^{\frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t}}\\ \ln \left( {\frac{1}{{100}}} \right) & = \frac{1}{{1250}}\ln \left( {\frac{4}{5}} \right)\,\,t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{1250\ln \left( {\frac{1}{{100}}} \right)}}{{\ln \left( {\frac{4}{5}} \right)}} = 25797.1279}}\end{align*}\]

There will only be 1 gram of the element left after 25,797.1279 years.