06fe1cdd-7f47-4625-a043-9ea1ebeca625.html

<h4>Activity (15 minutes)</h4>
<p>In this activity, students use a description and table of values to write an equation that represents them. Then,
  they interpret the equation for negative values of the exponent and produce a graph. They also interpret the numbers
  in the equation in terms of the context.</p>
<p>For the final question, only an estimate is possible. Expect different answers that range between 2 and 3 hours
  before the medicine was measured because a more precise answer cannot be given right now. Look for students who use
  the graph and try to extend it to values between integer hours. Invite them to share during the discussion.</p>
<h4>Launch</h4>
<p>If students continue to use graphing technology while working on this task, they are likely to enter their equation
  and see a smooth curve. It is not the intention that they try to sketch a curve at this time. Draw their attention to
  the instructions that say to plot the points.</p>
  <div class="os-raise-extrasupport">
  <div class="os-raise-extrasupport-header">
    <p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
    <p class="os-raise-extrasupport-name"> Action and Expression: Provide Access for Physical Action </p>
  </div>
  <div class="os-raise-extrasupport-body">
      <p>Provide access to graph paper or to tools and assistive technologies, such as graphing calculators or graphing
    software.</p>
    <p class="os-raise-text-italicize">Supports accessibility for: Visual-spatial processing; Conceptual processing; Organization</p>
</div>
</div>
<br>
<h4>Student Activity</h4>
<p>A person took some medicine but does not remember how much. Concerned that she took too much, she has a blood test
  every hour for several hours.</p>
  <table class="os-raise-midsizetable">
    <thead>
      <tr>
        <th scope="col">\(t\), time (hours)</th>
        <th scope="col">\(m\), medicine (mg)</th>
      </tr>
    </thead>
    <tbody>
      <tr>
      <td>
        &ndash;3
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
        &ndash;1
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
        0
      </td>
      <td>
        100
      </td>
    </tr>
    <tr>
      <td>
        1
      </td>
      <td>
        50
      </td>
    </tr>
    <tr>
      <td>
        2
      </td>
      <td>
        25
      </td>
    </tr>
  </tbody>
</table>
<br>
<p>Answer the following questions. Double click on mathematical expressions/equations to enlarge.
</p>
<ol class="os-raise-noindent">
  <li> Time \(t\)&nbsp;is measured in hours since the blood test, and amount of medicine in her body, \(m\), is measured
    in milligrams. What is the decay factor? That is, what is \(b\)&nbsp;in an equation of the form \(m=a\cdot
    b^t\)?<br>
    <strong>Answer:</strong> \(b=\frac12\)
  </li>
</ol>
<ol class="os-raise-noindent" start="2">
  <li> What is \(a\)?<br>
    <strong>Answer:</strong> \(a=100\)
  </li>
  </li>
</ol>
<ol class="os-raise-noindent" start="3">
  <li>Find the amounts of medicine in the patient&rsquo;s body when \(t\)&nbsp;is &ndash;1.<br>
    <strong>Answer:</strong> When \(t\)&nbsp;is &ndash;1, \(m\)&nbsp;is 200.
  </li>
</ol>
<ol class="os-raise-noindent" start="4">
  <li>Find the amounts of medicine in the patient&rsquo;s body when \(t\)&nbsp;is &ndash;3.
    <br>
    <strong>Answer:</strong> When \(t\)&nbsp;is &ndash;3, \(m\)&nbsp;is 800.
  </li>
</ol>
<ol class="os-raise-noindent" start="5">
  <li>What does \(t=0\) mean in this context?
    <br>
    <strong>Answer:</strong> \(t=0\) means the time the doctor did the first blood test.
  </li>
</ol>
<ol class="os-raise-noindent" start="6">
  <li>What does \(t=-3\) mean in this context?
    <br>
    <strong>Answer:</strong> \(t=-3\) means 3 hours before the doctor did the first test.

  </li>
</ol>
<ol class="os-raise-noindent" start="7">
  <li>The medicine was taken when \(t\)&nbsp;is &ndash;5. Assuming the person did not have any of the medication in her
    body beforehand, how much medicine did the patient take?
    <br>
    <strong>Answer:</strong> 3,200 mg. Each hour, this is cut in half. After being halved 5 times, it leaves 100 mg, the
    amount of medicine in the person&rsquo;s body when \(t=0\).
  </li>
</ol>
<ol class="os-raise-noindent" start="8">
  <li>Create a graph by plotting the points whose coordinates are in the table from question 1 above. Make sure to draw
    and label tick marks on the axes.
    <br>
    <strong>Answer:</strong> <br><img alt class="img-fluid atto_image_button_text-bottom" height="220"
      role="presentation"
      src="https://k12.openstax.org/contents/raise/resources/87c0825791421cf128a96a1b9eabd7e5cd7ccd97" width="300">
  </li>
</ol>
<ol class="os-raise-noindent" start="9">
  <li>Based on your graph, when do you think the patient will have: 500 mg of medicine remaining in her body
    <br>
    <strong>Answer:</strong> Between 2 and 3 hours before the patient first had her blood tested
  </li>
</ol>
<ol class="os-raise-noindent" start="10">
  <li>Based on your graph, when do you think the patient will have: No medicine remaining in her body
    <strong>Answer:</strong> <br> Your answer may vary, but here are some samples:
    </p>
    <ul>
      <li>Sometime after 10 hours. The amount of medicine falls below 0.1 mg after about 10 hours and might be too small
        to be detectable after that point.</li>
      <li> Eventually, all of the medicine will be out of the patient&rsquo;s body (but the mathematical model for the
        amount of medicine does not reach 0).</li>
    </ul>
  </li>
</ol>


<h4>Student Facing Extension</h4>
<p>Are you ready for more?</p>
<p>Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.
</p>
<ol class="os-raise-noindent">
  <li>\(\frac1{1-2^{-10}}\)</li><br>
  <p><strong>Answer:</strong> close to 1
  </p>
  <li>\(\frac{2^{10}}{2^{10}+1}\)</li><br>
  <p> <strong>Answer:</strong> close to 1
  </p>
  <li>\(\frac{2^{-10}}{2^{10}+1}\)</li><br>
  <p> <strong>Answer:</strong> close to 0
  </p>
  <li>\(\frac{1-2^{-10}}{2^{10}}\)</li><br>
  <p> <strong>Answer:</strong> close to 0
  </p>
  <li>\(\frac{1+2^{10}}{2^{-10}}\)</li><br>
  <p> <strong>Answer:</strong> much larger than 1
  </p>
</ol>
<h4>Anticipated Misconceptions</h4>
<p>Students may struggle with understanding the negative time values. It may help to give clock values to the numbers.
  For example, the first blood test was done at 5:00 p.m. \((t=0)\). Ask students what \(t\)&nbsp;would be when it is
  6:00 p.m. and 4:00 p.m. It is important throughout this entire lesson to emphasize that time is not typically
  negative, but in this case is emphasizing events that happened prior to initial measurement.</p>
<h4>Activity Synthesis</h4>
<p>Ask students how they labeled the axes on the graph. For the domain, it needs to include times from 3 hours before
  the medicine was measured to 2 hours after. (Students may also choose to include 5 hours before the first
  measurement.) The vertical axis should include 0 up to at least 800 (higher if students include the value for \(t=-5\)
  in the table).</p>
<p>Use the last question about when there was 500 mg of medicine to begin a discussion about values between integer
  hours. Elicit from students how the graph might look if we consider non-integer time values. If students still have
  access to graphing technology for this activity, they can enter their equation and see the curve for themselves.</p>
<p>Consider displaying the following graphs for all to see. Explain that if the amount of medicine in the bloodstream is
  plotted at time intervals smaller than 1 hour, the graph may look like the first graph. We sometimes graph a
  relationship where a quantity changes continuously using a curve, as in the second graph. Students will have more
  opportunities to consider non-integer domain values in upcoming lessons.</p>
<p><img height="300" src="https://k12.openstax.org/contents/raise/resources/bb19fdbfada87ee73b88ddb4310cf35ae3bf398a"
    width="300"></p>
<p>To follow up on the last question, consider discussing: How long could the decrease by half go on? Is there a point
  at which it is no longer practical or reasonable to use the mathematical model?</p>
<h3>5.6.3: Self Check</h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the
      concepts explored in the activity.</em></p>
      <p class="os-raise-text-bold">QUESTION:</p>
        <p>A census of hemlock trees is taken in a national park. The rangers have found that only \(\frac45\) of the
          population has survived each year. The data are shown in the table. Find the number of hemlocks that were in
          the park 2 years before the census was taken.</p>
          <table class="os-raise-midsizetable">
            <thead>
              <tr>
                <th scope="col">\(y\), years since census</th>
                <th scope="col">\(h\), number of hemlocks</th>
              </tr>
            </thead>
            <tbody>
              <tr>          
              <td>
                &ndash;2
              </td>
              <td>&nbsp;</td>
            </tr>
            <tr>
              <td>
                0
              </td>
              <td>
                64,000
              </td>
            </tr>
            <tr>
              <td>
               1
              </td>
              <td>
               51,200
              </td>
            </tr>
            <tr>
              <td>
            3
              </td>
              <td>
                32,768
              </td>
            </tr>
          </tbody>
        </table>
        <br>
        <table class="os-raise-textheavytable">
          <thead>
            <tr>
              <th scope="col">Answers</th>
              <th scope="col">Feedback</th>
            </tr>
          </thead>
          <tbody>
            <tr>
      <td>
        156,250
      </td>
      <td>
       Incorrect. Let&rsquo;s try again a different way: This is the solution for 4 years before the census was
          taken. The equation to represent this situation is
       \(h=64,000\cdot\left(\frac45\right)^y\). The solution is found by substituting &ndash;2 for \(y\). The answer
          is 100,000.
      </td>
    </tr>
    <tr>
      <td>
       125,000
      </td>
      <td>
       Incorrect. Let&rsquo;s try again a different way: This is the solution for 3 years before the census was
          taken. The equation to represent this situation is<br>
        \(h=64,000\cdot\left(\frac45\right)^y\). The solution is found by substituting &ndash;2 for \(y\). The answer
          is 100,000.
      </td>
    </tr>
    <tr>
      <td>
       80,000
      </td>
      <td>
        Incorrect. Let&rsquo;s try again a different way: This is the solution for 1 year before the census was
          taken. Find the solution 2 years before the census was taken. The equation to represent this situation is<br>
        \(h=64,000\cdot\left(\frac45\right)^y\). The solution is found by substituting &ndash;2 for \(y\). The answer
          is 100,000.
      </td>
    </tr>
    <tr>
      <td>
        100,000
      </td>
      <td>
       That&rsquo;s correct! Check yourself: The equation to represent this situation is
          \(h=64,000\cdot\left(\frac45\right)^y\).
      </td>
    </tr>
  </tbody>
</table>
<br>
<h3>5.6.3: Additional Resources</h3>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on their experience with
      the self check. Students will not automatically have access to this content, so you may wish to share it with
      those who could benefit from it.</em></p>
<h4>Interpreting Negative Exponents in Exponential Decay</h4>
<p>Let&rsquo;s look at a problem involving an exponential decay equation and negative exponents.</p>
<p>A scientist records the population of a bacterial population in a petri dish. The data are shown in the table.</p>
<br>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">\(d\), days since first test</th>
      <th scope="col">\(b\), number of bacteria</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
       &ndash;3
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
       &ndash;1
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
        0
      </td>
      <td>
       40,000
      </td>
    </tr>
    <tr>
      <td>
        1
      </td>
      <td>
       20,000
      </td>
    </tr>
    <tr>
      <td>
      3
      </td>
      <td>
       5,000
      </td>
    </tr>
  </tbody>
</table>
<br>
<p>The number of days \(d\) since the first test corresponds with the number of bacteria, \(b\), estimated to be alive
  in the petri dish.</p>
<ol class="os-raise-noindent">
  <li> What is the decay factor? That is, what is \(f\) in an equation of the form \(b=n\cdot f^d\)? What is the
    starting population, \(n\), on the first day of testing? <br>
    <br>
    Since the number of bacteria found on day 0 is 40,000, then \(n = 40,000\). Since after each day, the number of
    bacteria is cut in half, then the decay factor, \(f\), is \(\frac12\). <br>
    <br>
    The equation to represent this situation is \(b=40,000\cdot\left(\frac12\right)^d\).
  </li>
  <br>
  <li> What do \(d=-1\) and \(d=-3\) mean in this context? <br>
    <br>
    Since the day the first test is taken is defined as \(d=0\), then days with a negative value occured before the
    first test. \(d=-1\) means 1 day before the scientist did the first test. \(d=-3\) means 3 days before the scientist
    did the first test.
  </li>
  <br>
  <li> How many bacteria were present 1 day before the first test? <br>
    <br>
    Substitute &ndash;1 for \(d\) in the equation \(b=40,000\cdot\left(\frac12\right)^d\).<br>
    <br>
    \(b=40,000\cdot\left(\frac12\right)^d\) <br>
    \(b=40,000\cdot\left(\frac12\right)^{-1}\) <br>
    \(b=40,000\cdot 2\) <br>
    \(b=80,000\) <br>
    <br>
    There were 80,000 bacteria in the petri dish 1 day before the test.
  </li>
  <br>
  <li> Assume the peak of the bacterial population was 3 days before the first test. How many bacteria were present at
    the peak of the population? <br>
    <br>
    Substitute &ndash;3 for \(d\) in the equation \(b=40,000\cdot\left(\frac12\right)^d\). <br>
    <br>
    \(b=40,000\cdot\left(\frac12\right)^d\) <br>
    \(b=40,000\cdot\left(\frac12\right)^{-3}\) <br>
    \(b=40,000\cdot 8\) <br>
    \(b=320,000\) <br>
    <br>
    There were 320,000 bacteria in the petri dish 3 days before the test at the peak of the
    population.
  </li>
</ol>
<h4>Try It: Interpreting Negative Exponents in Exponential Decay</h4>
<p>An investor has tracked the value of a particular investment since 2016 that has depreciated over time. The data are
  shown in the table.</p>
<br>
<table class="os-raise-midsizetable">
  <thead>
    <tr>
      <th scope="col">\(y\), years since 2016</th>
      <th scope="col">\(v\), value of investment ($)</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>
       &ndash;2
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
      &ndash;1
      </td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>
        0
      </td>
      <td>
        36,000
      </td>
    </tr>
    <tr>
      <td>
       1
      </td>
      <td>
       12,000
      </td>
    </tr>
    <tr>
      <td>
        2
      </td>
      <td>
        4,000
      </td>
    </tr>
  </tbody>
</table>
<br>
<ol class="os-raise-noindent">
  <li> What is the equation that represents this situation? </li>
</ol>
<ol class="os-raise-noindent" start="2">
  <li> What do \(y=-1\) and \(y=-2\) mean in this context? </li>
</ol>
<ol class="os-raise-noindent" start="3">
  <li> If the value of the investment was at its highest in 2014, what was the maximum value of the investment? </li>
</ol>
<p>Write down your answers. Then select the <strong>solution</strong> button to compare your work.</p>
<p>Here is how to solve this exponential growth problem using negative exponents:</p>
<p><strong>Solution</strong></p>
<ol class="os-raise-noindent">
  <li> The equation that represents this situation is \(v=36,000\cdot\left(\frac13\right)^y\). The value of the
    investment in year 0 is $36,000, and the decay factor is \(\frac13\). </li>
</ol>
<ol class="os-raise-noindent" start="2">
  <li> \(y=-1\) means 1 year before 2016, which is 2015. \(y=-2\) means 2 years before 2016, which is 2014. </li>
</ol>
<ol class="os-raise-noindent" start="3">
  <li> The maximum value of the investment was $324,000. <br>
    <br>
    \(v=36,000\cdot\left(\frac13\right)^y\) <br>
    \(v=36,000\cdot\left(\frac13\right)^{-2}\) <br>
    \(v=36,000\cdot 9\) <br>
    \(v=324,000\)
  </li>
</ol>