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Section 1-9 : Exponential And Logarithm Equations

Exponential Growth/Decay. Many quantities in the world can be modeled (at least for a short time) by the exponential growth/decay equation.

\[Q = {Q_0}{{\bf{e}}^{k\,t}}\]

If \(k\) is positive we will get exponential growth and if \(k\) is negative we will get exponential decay.

16. We initially have 100 grams of a radioactive element and in 1250 years there will be 80 grams left.

  1. Determine the exponential decay equation for this element.
  2. How long will it take for half of the element to decay?
  3. How long will it take until there is only 1 gram of the element left?

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Hint : We have an equation with two unknowns and two values of the amount of the element left at two times so use these values to find the two unknowns.
a Determine the exponential decay equation for this element. Show Solution

We can start off here by acknowledging that we know,

\[Q\left( 0 \right) = 100\hspace{0.5in}{\rm{and}}\hspace{0.5in}Q\left( {1250} \right) = 80\]

If we use the first condition in the equation we get,

\[100 = Q\left( 0 \right) = {Q_0}{{\bf{e}}^{k\left( 0 \right)}} = {Q_0}\hspace{0.5in} \to \hspace{0.5in}{Q_0} = 100\]

We now know the first unknown in the equation. Plugging this as well as the second condition into the equation gives us,

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

We can use techniques from earlier problems in this 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,

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

b How long will it take for half of the element to decay? Show Solution

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 earlier problems in this section how to do that. Here is the solution work for this part.

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

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.


c How long will it take until there is only 1 gram of the element left? Show Solution

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

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

and we know from earlier problems in this section how to do that. Here is the solution work for this part.

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

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