068a5358-d013-4544-9566-5358e3677fc9.html

<h4>Analyzing Exponential Decay</h4>
<p>At the beginning of April, a colony of ants had a population of 5,000.</p>
<ol class="os-raise-noindent">
  <li> The colony decreases by \(\frac15\) during April. Write an expression that represents the ant population at the end of April. </li>
</ol>
<ol class="os-raise-noindent" start="2">
  <li> During May, the colony decreases again by \(\frac15\) of its size. Write an expression that represents the ant population at the end of May. </li>
</ol>
<ol class="os-raise-noindent" start="3">
  <li> The colony continues to decrease by \(\frac15\) of its size each month. Write an expression that represents the ant population after 6 months. </li>
</ol>
<ol class="os-raise-noindent" start="4">
  <li> Write an equation to represent the population of ants, \(p\), after \(m\) months. </li>
</ol>
<p>When you are analyzing exponential decay and are given the fraction or percent by which it decreases, you need to remember that one minus that fraction or percent is the growth rate. In the case of the ant colony, because it is decreasing by \(\frac15\), that means that the growth factor is \(1-\frac15\) or \(\frac45\).</p>
<p>Part 1 asks for an expression to represent the ant population at the end of April. There are several options:</p>
<ul>
  <li>\(
    5000-(\frac{1}{5} \cdot 5000)\) </li>
  <li> \(5000 \cdot \frac{4}{5}\) </li>
  <li>\(
    4000\) </li>
</ul>
<p>Each line is the simplification of the line above it. Part 2 has a few more options.</p>
<p>Part 2 is asking you to find \(\frac45\) of April&rsquo;s total. There are several expressions to represent that:</p>
<ul>
  <li> \(5000 \cdot \frac{4}{5} \cdot \frac{4}{5}\) </li>
  <li> \(4000 \cdot \frac{4}{5}\) </li>
  <li>\(
    4000-(\frac{1}{5} \cdot 4000)\) </li>
  <li>\(3200\) </li>
</ul>
<p>Basically, you take the answer from April and multiply by \(\frac45\) again since the growth factor is \(\frac45\). You could also look at it as taking the initial value times \(\frac45\) twice.</p>
<p>Part 3 asks you to extend the pattern to 6 months after the initial population was measured. After the 2nd month, you multiplied the initial value by the growth factor twice, so for this situation you would need to multiply the initial value by the growth factor 6 times, or raise it to the sixth power. The possible answers would be \(5000 \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5}\) or \(5000 \cdot (\frac{4}{5})^6\), which means that after 6 months the population would be about 1,311.</p>
<p>Part 4 wants the general equation to represent the situation. If you use the pattern from part 3, you can see that it can be written in the exponential form \(y=ab^x\), where \(a\) is the initial value and \(b\) is the growth factor. Using the variables suggested, the equation would be \(p = 5000( \frac{4}{5})^m\).</p>
<h4>Try It: Analyzing Exponential Decay</h4>
<p>A doctor prescribes 125 milligrams of a therapeutic drug that decreases in effectiveness by \(\frac{3}{10}\) each hour.</p>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
  <div class="os-raise-ib-cta-content">
    <ol class="os-raise-noindent" type="a">
      <li> Write an expression to represent how much is left after 1 hour. </li>
      <li> Write an expression to represent how much is left after 3 hours. </li>
      <li> Write an equation that represents the amount of drug, \(d\), left after \(h\) hours.&nbsp;</li>
    </ol>
  </div>
  <div class="os-raise-ib-cta-prompt">
    <p>Write down your answer, then select the <strong>solution</strong> button to compare your work. </p>
  </div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
  <p>Compare your answers:</p>
  <p>Here is how to analyze the exponential decay situation:</p>
  <p>In part a, you need to multiply the initial amount by \(\frac{7}{10}\), or \(125 \cdot \frac{7}{10}\), or 87.5 mg.</p>
  <p>In part b, since it&rsquo;s 3 hours later, it would be \(125 \cdot \frac{7}{10} \cdot \frac{7}{10} \cdot \frac{7}{10}\), or \(125 \cdot (\frac{7}{10})^3\), or 42.875 mg.</p>
  <p>The equation that represents the situation for part c is \(d =125 \cdot (\frac{7}{10})^h\).</p>
</div>