section-3-2.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!-- <mathbook><book> -->

<section xml:id="Integer-Exponents"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Integer Exponents</title><introduction>
<p>
    Recall that a positive integer exponent tells us how many times its base occurs as a factor in an expression. For example,
    <me>4a^3b^2 \text{ means } 4aaabb</me>
</p></introduction>
<subsection><title>Negative Exponents</title><idx>negative exponents</idx>
<p>
    Study the list of powers of <m>2</m> shown in <xref ref="table-positive-exponents" text="type-global"/> and observe the pattern as we move up the list from bottom to top. Each time the exponent increases by 1 we multiply by another factor of <m>2</m>. We can continue up the list as far as we like.
</p>
<p>
    If we move back down the list, we divide by <m>2</m> at each step, until we get to the bottom of the list, <m>2^{1} = 2</m>. What if we continue the list in the same way, dividing by <m>2</m> each time we decrease the exponent? The results are shown in <xref ref="table-integer-exponents" text="type-global"/>.
</p>
<sidebyside>
    <table xml:id="table-positive-exponents">
        <caption></caption>
        <tabular top="minor">
            <row bottom="minor" left="minor">
                <cell right="minor"><image source="images/fig-table-positive-exponents" width="100%"/></cell>
            </row>
        </tabular>
    </table>

    <table xml:id="table-integer-exponents">
        <caption></caption>
        <tabular top="minor">
            <row bottom="minor" left="minor">
                <cell right="minor"><image source="images/fig-table-integer-exponents" width="100%"/></cell>
            </row>
        </tabular>
    </table>
</sidebyside>

<p>
    As we continue to divide by <m>2</m>, we generate fractions whose denominators are powers of 2. In particular,
    <me>2^{-1} = \frac{1}{2}= \frac{1}{2^1} ~ \text{ and } ~ 2^{-2} = \frac{1}{4}= \frac{1}{2^2}</me>
    Based on these observations, we make the following definitions.
</p>

<assemblage><title>Definition of Negative and Zero Exponents</title><p>
<me>
    \begin{aligned}
    a^{-n} \amp = \frac{1}{a^n} \amp\amp (a \ne 0) \\
    a^0 \amp = 1  \amp\amp (a \ne 0)
    \end{aligned}
</me></p>
</assemblage>

<p>
    These definitions tell us that if the base <m>a</m> is not zero, then any number raised to the zero power is <m>1</m>, and that a negative exponent denotes a reciprocal.
</p><example xml:id="example-integer-exponents">


<p><ol label="a">
    <li><p><m>2^{-3} = \dfrac{1}{2^3}= \dfrac{1}{8}</m></p></li>
    <li><p><m>9x^{-2} = 9 \cdot \dfrac{1}{x^2}= \dfrac{9}{x^2}</m></p></li>
</ol></p>
</example>

<warning>
<p><ol>
    <li><p>A negative exponent does <em>not</em> mean that the power is negative! For example,
    <me>2^{-3}\ne -2^3</me></p></li>
    <li><p>In <xref ref="example-integer-exponents" text="type-global" />b, note that 
    <me>9x^{-2} \ne  \frac{1}{9x^2}</me>
    The exponent, <m>-2</m>, applies only to the base <m>x</m>, not to <m>9</m>.
    </p></li> 
    
</ol></p></warning>

<exercise xml:id="exercise-integer-exponents"><statement>
<p>
    Write each expression without using negative exponents.
    <ol label="a">
        <li><p><m>5^{-4}</m></p></li>
        <li><p><m>5x^{-4}</m></p></li>
    </ol>
</p>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><m>\dfrac{1}{5^4}</m> </p></li>
    <li><p><m>\dfrac{5}{x^4} </m></p></li>
</ol></p></answer>
</exercise>

<p>
    In the next example, we see how to evaluate expressions that contain negative exponents and how to solve equations involving negative exponents.
</p>

<example xml:id="example-BMI-index">
<p>
    The body mass index, or BMI, is one measure of a person’s physical fitness. Your body mass index is defined by
    <me>BMI = wh^{-2}</me>
    where <m>w</m> is your weight in kilograms and <m>h</m> is your height in meters. The World Health Organization classifies a person as obese if his or her BMI is <m>25</m> or higher.

<ol label="a">
    <li><p>a. Calculate the BMI for a woman who is <m>1.625</m> meters (<m>64</m> inches) tall and weighs <m>54</m> kilograms (<m>120</m> pounds).</p></li>
    <li><p>For a fixed weight, how does BMI vary with height?</p></li>
    <li><p>The world’s heaviest athlete is the amateur sumo wrestler Emanuel Yarbrough, who weighs <m>319</m> kg (<m>704</m> pounds). What height would Yarbrough have to be to have a BMI under <m>25</m>?</p></li>
</ol></p>
<solution>
<p><ol label="a">
    <li><p><m>BMI = 54(1.625^{-2})= 54(\dfrac{1}{1.625^2})= 20.45</m></p></li>
    <li><p><m>BMI = wh^{-2} = \dfrac{w}{h^2}</m>, so BMI varies inversely with the square of height. That is, for a fixed weight, BMI decreases as height increases.</p></li>
    <li><p>To find the height that gives a BMI of <m>25</m>, we solve the equation <m>25 = 319h^{-2}</m>. Note that the variable <m>h</m> appears in the denominator of a fraction, so we begin by clearing the denominator—in this case we multiply both sides of the equation by <m>h^2</m>.
    \begin{align*}
    25 \amp = \frac{319}{h^2} \amp\amp \text{Multiply both sides by }h^2.\\
    25h^2 \amp = 319 \amp\amp \text{Divide both sides by }25.\\
    h^2 \amp = 12.76 \amp\amp \text{Extract square roots.}\\ 
    h \amp\approx 3.57
    \end{align*}
    To have a BMI under <m>25</m>, Yarbrough would have to be over <m>3.57</m> meters, or <m>11</m> feet <m>8</m> inches tall. (In fact, he is <m>6</m> feet <m>8</m> inches tall.)</p></li>
</ol></p>
</solution>
</example>

<exercise xml:id="solve-power-equation"><statement>
<p>Solve the equation <m>0.2x^{-3} = 1.5</m>.</p>
</statement>
<hint><p>Rewrite without a negative exponent.</p>
<p>Clear the fraction.</p>
<p>Isolate the variable.</p></hint>
<answer><p><m>x=\sqrt[3]{\dfrac{2}{15}}\approx 0.51 </m></p></answer>
</exercise>

</subsection>



<subsection><title>Power Functions</title>
<p>
    The functions that describe direct and inverse variation are part of a larger family of functions called <term>power functions</term>.
</p>    
<assemblage><title>Power Function</title>
    <p>
        A function of the form
        <me>f(x) = kx^p</me>
        where <m>k</m> and <m>p</m> are nonzero constants, is called a <term>power function</term><idx>power function</idx>.
    </p>
</assemblage>

<p>
    Examples of power functions are
    <me>V(r ) = \frac{4}{3}\pi r^3 ~~\text{ and }~~L(T ) = 0.8125T^2</me>
    In addition, the basic functions
    <me>f (x) = \frac{1}{x} ~~\text{ and }~~ g(x) = \frac{1}{x^2}</me>
    which we studied in <xref ref="chap2" text="type-global"/> can be written as
    <me>f (x) = x^{-1} ~~\text{ and }~~ g(x) = x^{-2} </me>
    Their graphs are shown in <xref ref="fig-basic-reciprocal-functions" text="type-global"/>. Note that the domains of power functions with negative exponents do not include zero.
    <figure xml:id="fig-basic-reciprocal-functions"><caption></caption><image source="images/fig-basic-reciprocal-functions"  width="60%"><description>graphs of the two basic reciprocal functions</description></image></figure>
</p>

<example xml:id="example-identify-power-functions">
<p>
    Which of the following are power functions?
<ol label="a">
    <li><p><m>f (x) = \dfrac{1}{3}x^4 + 2</m></p></li>
    <li><p><m>g(x) = \dfrac{1}{3x^4}</m></p></li>
    <li><p><m>h(x) = \dfrac{x + 6}{x^3}</m></p></li>
</ol></p>
<solution>
<p><ol label="a">
    <li><p>This is not a power function, because of the addition of the constant term.</p></li>
    <li><p>We can write <m>g(x) = \frac{1}{3}x^{-4}</m>, so <m>g</m> is a power function.</p></li>
    <li><p>This is not a power function, but it can be treated as the sum of two power functions, because <m>h(x) = x^{-2} + 6x^{-3}</m>.</p></li>
</ol></p>
</solution>
</example>

<exercise xml:id="exercise-write-as-power-functions"><statement><p>
Write each function as a power function in the form <m>y = kx ^p</m>.
<ol label="a" cols="3">
    <li><p><m>f (x) = {12}{x^2}</m></p></li>
    <li><p><m>g(x) = \dfrac{1}{4x}</m></p></li>
    <li><p><m>h(x) = \dfrac{2}{5x^6}</m></p></li>
</ol>
</p></statement>
<answer><p><ol label="a" cols="3">
    <li><p><m>f (x) = 12x^{-2}</m></p></li>
    <li><p><m>g(x)=\dfrac{1}{4}x^{-1} </m></p></li>
    <li><p><m>h(x)=\dfrac{2}{5}x^{-6} </m></p></li>
</ol></p></answer>
</exercise>

<p>
    Most applications are concerned with positive variables only, so many models use only the portion of the graph in the first quadrant.
</p>

<example xml:id="example-trebuchet">
<p>
    In the Middle Ages in Europe, castles were built as defensive strongholds. An attacking force would build a huge catapult called a trebuchet to hurl rocks and scrap metal inside the castle walls. The engineers could adjust its range by varying the mass of the projectiles. The mass, <m>m</m>, of the projectile should be inversely proportional to the square of the distance, <m>d</m>, to the target.

<ol label="a">
    <li><p>Use a negative exponent to write <m>m</m> as a function of <m>d</m>, <m>m = f (d)</m>.</p></li>
    <li><p>The engineers test the trebuchet with a <m>20</m>-kilogram projectile, which lands <m>250</m> meters away. Find the constant of proportionality; then rewrite your formula for <m>m</m>.</p></li>
    <li><p>Graph <m>m = f (d)</m>.</p></li>
    <li><p>The trebuchet is <m>180</m> meters from the courtyard within the castle. What size projectile will hit the target?</p></li>
    <li><p>The attacking force would like to hurl a <m>100</m>-kilogram projectile at the castle. How close must the attackers bring their trebuchet?</p></li>
</ol></p>
<solution>
<p><ol label="a">
    <li><p>If we use <m>k</m> for the constant of proportionality, then <m>m = \dfrac{k}{d^2}</m>. Rewriting this equation with a negative exponent gives <m>m = kd^{-2}</m>.</p></li>
    <li><p>We substitute <m>m = 20</m> and <m>d = 250</m> to obtain
    \begin{align*}
    20 \amp = k(250)^{-2}\amp\amp \text{Multiply both sides by }250^2.\\
    1,250,000 \amp = k
    \end{align*}
    Thus, <m>m = 1,250,000 d^{-2}</m>.</p></li>
    <li><p>We evaluate the function for several values of <m>m</m>, or use a calculator to obtain the graph in <xref ref="fig-trebuchet-graph" text="type-global"/>.
    <figure xml:id="fig-trebuchet-graph"><caption></caption><image source="images/fig-trebuchet-graph"  width="40%"><description>graph of mass vs distance rock was hurled</description></image></figure>
    </p></li>
    <li><p>We substitute <m>d = \alert{180}</m> into the formula:
    \begin{align*}
        m \amp = 1,250,000 (\alert{180} )^{-2} \\
          \amp = \frac{1,250,000}{32,400} \\
          \amp \approx 38.58
    \end{align*}
    The attackers should use a mass of approximately <m>38.6</m> kilograms.</p></li>
    <li><p>We substitute <m>m=\alert{100}</m> into the formula and solve for <m>d</m>.
    \begin{align*}
        \alert{100} \amp = 1,250,000 d^{-2} \amp\amp\text{Multiply by }d^2.\\
        100d^2 \amp = 1,250,000 \amp\amp\text{Divide by }100.\\
        d^2 \amp = 12,500 \amp\amp\text{Take square roots.} \\
        d \amp = \pm \sqrt{12,500}
    \end{align*}
    They must locate the trebuchet <m>\sqrt{12,500}\approx 111.8</m> meters from the castle.
    </p></li>
</ol></p></solution>
</example>

<p>
    The function <m>m = \dfrac{k}{d^2}</m> is an example of an <term>inverse square law</term><idx>inverse square law</idx>, because <m>m</m> varies inversely with the square of <m>d</m>. Such laws are fairly common in physics and its applications, because gravitational and other forces behave in this way. Here is a more modern example of an inverse square law.
</p>

<exercise xml:id="exercise-cellphone-coverage"><statement><p>
    Cell phone towers typically transmit signals at <m>10</m> watts of power. The signal strength varies inversely with the square of distance from the tower, and <m>1</m> kilometer away the signal strength is <m>0.8</m> picowatt. (A picowatt is <m>10^{-12}</m> watt.) Cell phones can receive a signal as small as <m>0.01</m> picowatt. How far can you be from the nearest tower and still hope to have cell phone reception?
</p></statement>
<answer><p>About <m>9</m> km</p></answer>
</exercise>

</subsection>



<subsection><title>Working with Negative Exponents</title>
<p>
    A negative exponent denotes the <em>reciprocal</em> of a power. Thus, to simplify a fraction with a  negative exponent, we compute the positive power of its reciprocal.
</p>

<example xml:id="example-negative-exponents-on-fractions">
<p><ol label="a">
    <li><p><m>\left(\dfrac{3}{5}\right)^{-2}= \dfrac{1}{\left(\dfrac{3}{5}\right)^2}
        = \left(\dfrac{5}{3}\right)^2 =\dfrac{25}{9}</m></p></li>
    <li><p><m>\left(\dfrac{x^3}{4}\right)^{-3}=\left(\dfrac{4}{x^3}\right)^3
        = \dfrac{(4)^3}{\left(x^3\right)^3}= \dfrac{64}{x^9}</m>
    </p></li>
</ol></p>
</example> 

<exercise xml:id="exercise-negative-exponent-on-fraction"><statement><p>
    Simplify <m>\left(\dfrac{2}{x^2} \right)^{-4}</m>.
</p></statement>
<answer><p><m>\dfrac{x^8}{16} </m></p></answer>
</exercise> 

<p>
    Dividing by a power with a negative exponent is equivalent to multiplying by a power with a positive exponent.
</p>

<example xml:id="example-dividing-negative-exponents">
    <p><ol label="a">
        <li><p><m>
            \begin{aligned} \\ \\
            \frac{1}{5^{-3}}\amp= 1 \div 5^{-3} \\
            \amp = 1 \div \frac{1}{5^3} \\
            \amp = 1 × 5^3 = 125
            \end{aligned}</m>
        </p></li>
        <li><p><m>
            \begin{aligned}\\ \\
            \frac{k^2}{m^{-4}} \amp = k^2 \div m^{-4} \\
            \amp = k^2 ÷ \frac{1}{m^4} \\
            \amp = k^2m^4
            \end{aligned}</m>
        </p></li>
    </ol></p>
</example>

<exercise xml:id="exercise-write-without-neg-exponents"><statement>
<p>Write each expression without using negative exponents.
<ol label="*a">
    <li><p><m>\left(\dfrac{3}{b^4}\right)^{-2}</m></p></li>
    <li><p><m>\dfrac{12}{x^{-6}}</m></p></li>
</ol>
</p></statement>
<answer><p><ol label="a" cols="2">
    <li><p><m>\dfrac{b^8}{9} </m></p></li>
    <li><p><m>12x^6</m></p></li>
</ol></p></answer>
</exercise>

</subsection>



<subsection><title>Laws of Exponents</title><idx>laws of exponents</idx>
<p>
    The laws of exponents, reviewed in Algebra Skills Refresher 
    <xref ref="appendix-Laws-of-Exponents" text="type-global"/>, apply to all integer exponents, positive, negative, and zero. When we allow negative exponents, we can simplify the rule for computing quotients of powers.
</p>
<assemblage><title>A Law of Exponents</title><p>
II. <m>\displaystyle{\frac{a^m}{a^n}= a^{m-n}\hphantom{blank} (a \ne 0)}</m>
</p></assemblage>

<p>
    For example, by applying this new version of the law for quotients, we find 
    <me>\frac{x^2}{x^5}= x^{2-5} = x^{-3}</me>
    which is consistent with our previous version of the rule,
    <me>\frac{x^2}{x^5}= \frac{1}{x^{5-2}}= \frac{1}{x^3}</me>
</p>

<p>
    For reference, we restate the laws of exponents below. The laws are valid for all integer exponents <m>m</m> and <m>n</m>, and for <m>a, b \ne 0</m>.
</p>

<assemblage><title>Laws of Exponents</title>
<idx>laws of exponents</idx>
<p><ol label="*I*">
    <li><p><m>\displaystyle{a^m\cdot a^n = a^{m+n}}</m></p></li>
    <li><p><m>\displaystyle{\frac{a^m}{a^n}=a^{m-n}}</m></p></li>
    <li><p><m>\displaystyle{\left(a^m\right)^n=a^{mn}}</m></p></li>
    <li><p><m>\displaystyle{\left(ab\right)^n=a^n b^n}</m></p></li>
    <li><p><m>\displaystyle{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}</m></p></li>
</ol></p>
</assemblage>

<example xml:id="example-laws-of-exponents">
<p>
\begin{align*}
    \text{a}\amp. ~x^3\cdot x^{-5} = x^{3-5} = x^{-2} 
    \amp\amp     \text{Apply the first law: Add exponents.} \\
    \text{b}\amp. ~\frac{8x^{-2}}{4x^{-6}}= \frac{8}{4}x^{-2-(-6)} = 2x^4
    \amp\amp     \text{Apply the second law: Subtract exponents.} \\
    \text{c}\amp. ~\left(5x^{-3}\right)^{-2}= 5^{-2}(x^{-3})^{-2}=\frac{x^6}{25}
    \amp\amp     \text{Apply laws IV and III.}
\end{align*}
</p> 
</example>

<p>
    You can check that each of the calculations in <xref ref="example-laws-of-exponents" text="type-global"/> is shorter when we use negative exponents instead of converting the expressions into algebraic fractions.
</p>

<exercise xml:id="exercise-laws-of-exponents"><statement>
<p>
    Simplify by applying the laws of exponents.
    <ol label="a*a">
        <li><p><m>\left(2a^{-4}\right) \left(-4a^2\right)</m></p></li>
        <li><p><m>\dfrac{(r^2)^{-3}}{3r^{-4}}</m></p></li>
    </ol>
</p></statement>
<answer><p><ol label="a" cols="2">
    <li><p><m>\dfrac{-8}{a^2} </m></p></li>
    <li><p><m>\dfrac{1}{3r^2} </m></p></li>
</ol></p></answer>
</exercise>


<warning><statement>
<p>
    The laws of exponents do not apply to sums or differences of powers. We can add or subtract like terms, that is, powers with the same exponent. For example,
    <me>6x^{-2} + 3x^{-2} = 9x^{-2}</me>
    but <em>we cannot add or subtract terms with different exponents</em>. Thus, for example,
    \begin{align*}
    4x^2 \amp- 3x^{-2} \amp\amp \text{cannot be simplified}\\
    x^{-1} \amp + x^{-3}\amp\amp \text{cannot be simplified}
    \end{align*}
</p></statement></warning>

<p>
    In <xref ref="table-integer-exponents" text="type-global"/>, we saw that <m>2^0 = 1</m>, and in fact <m>a^0 = 1</m> as long as <m>a \ne 0</m>. Now we can see that this definition is consistent with the laws of exponents. The quotient of any (nonzero) number divided by itself is <m>1</m>. But by applying the second law of exponents, we also have
    <me>1 = \frac{a^m}{a^m}= a^{m-m} = a^0</me>
</p>
<p>Thus,</p>
<assemblage><title>Zero as Exponent</title><p>
    <me>a^0 = 1, ~~ \text{ if } a \ne 0</me></p>
</assemblage>
<p>
    For example,
    <me>3^0 = 1, ~~ (-528)^0 = 1, ~~ \text{ and } ~~ (0.024)^0 = 1</me>
</p>



</subsection>
<xi:include href="./summary-3-2.xml" /> <!-- summary  -->
<xi:include href="./section-3-2-exercises.xml" /> <!-- exercises  -->

</section> 
<!-- </book>  </mathbook> -->