section-4-3.xml

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

<section xml:id="Logarithms"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Logarithms</title>
<introduction>
<p>
	In this section, we introduce a new mathematical tool called a <term>logarithm</term>, which will help us solve exponential equations.
</p>
<p>
	Suppose that a colony of bacteria doubles in size every day. If the colony starts with <m>50</m> bacteria, how long will it be before there are <m>800</m> bacteria? We answered questions of this type in <xref ref="Exponential-Functions" text="type-global"/> by writing and solving an exponential equation. The function
	<me>P(t) = 50 \cdot 2^t</me>
	gives the number of bacteria present on day <m>t</m>, so we must solve the equation
	<me>800 = 50 \cdot 2^t</me>
	Dividing both sides by 50 yields
	<me>16 = 2^t</me>
	The solution of this equation is the answer to the following question: To what power must we raise <m>2</m> in order to get <m>16</m>?
</p>
<p>
	The value of <m>t</m> that solves the equation is called the base <m>2</m> <term>logarithm</term><idx>logarithm</idx> of <m>16</m>. Since <m>2^4 = 16</m>, the base <m>2</m> logarithm of <m>16</m> is <m>4</m>. We write this as
	<me>\log_{2}16 = 4</me>
	In other words, we solve an exponential equation by computing a logarithm. You can check that <m>t = \alert{4}</m>
	solves the problem stated above:
	<me>P(\alert{4}) = 50 \cdot 2^\alert{4}= 800</me>
</p>
<p>
	Thus, the unknown exponent is called a logarithm. In general, for positive values of <m>b</m> and <m>x</m>, we make the following definition.
</p>
<assemblage><title>Definition of Logarithm</title>
<p>
	For <m>b\gt 0, b\ne 1</m>, the <idx><term>base <m>b</m> logarithm of <m>x</m></term></idx>, written <m>\log_{b} x</m>, is the exponent to which <m>b</m> must be raised in order to yield <m>x</m>.
</p></assemblage>
<p>
	It will help to keep in mind that a logarithm is just an exponent. Some logarithms, like some square roots, are easy to evaluate, while others require a calculator. We will start with the easy ones.
</p>

<example xml:id="example-easy-logarithm">
<p>
	Compute the logarithms.
	<ol label="a">
		<li><p><m>\log_3 9</m></p></li>
		<li><p><m>\log_5 125 </m></p></li>
		<li><p><m>\log_4 \dfrac{1}{16}</m></p></li>
		<li><p><m>\log_5 \sqrt{5}</m></p></li>
	</ol>
</p>
<solution>
<ol label="a">
	<li><p>
		To evaluate <m>\log_3 9</m>, we ask what exponent on base <m>3</m> will produce <m>9</m>. In symbols, we want to fill in the blank in the equation <m>3^{\underline{?}} = 9</m>. The exponent we need is <m>2</m>, so
		<me>\log_3 9 = 2 \text{ because } 3^2 = 9</me>
		We use similar reasoning to compute the other logarithms.
	</p></li>
	<li><p><m>\log_5{125} = 3 \text{ because } 5^3 = 125</m></p></li>
	<li><p><m>\log_4{\dfrac{1}{16}}= -2 \text{ because } 4^{-2} = \dfrac{1}{16}</m></p></li>
	<li><p><m>\log_5{\sqrt{5}} = \dfrac{1}{2} \text{ because } 5^{1/2} =\sqrt{5}</m></p></li>
</ol>
</solution>
</example>

<exercise xml:id="exercise-easy-logarithms"><statement>
	Find each logarithm.
	<ol label="a">
		<li><p><m>\log_{3}{81}</m></p></li>
		<li><p><m>\log_{10}{\dfrac{1}{1000}}</m></p></li>
	</ol>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>4</m></p></li>
	<li><p><m>-3</m></p></li>
</ol></p></answer>
</exercise>

<p>
	From the definition of a logarithm and the examples above, we see that the following two statements are equivalent.
</p>

<assemblage><title>Logarithms and Exponents: Conversion Equations</title>
<p>
	If <m>b \gt 0</m>, <m>b\ne 1</m>, and <m>x \gt 0</m>,
	<me>y = \log_b x \text{ if and only if } x = b^y</me>
</p></assemblage>

<p>
	In other words, the logarithm <m>y</m>, is the same as the <em>exponent</em> in <m>x = b^y</m>. We see again that <em>a logarithm is an exponent</em>; it is the exponent to which <m>b</m> must be raised to yield <m>x</m>.
</p> 
<p>
	These equations allow us to convert from logarithmic to exponential form, or vice versa. You should memorize the conversion equations, because we will use them frequently. 
</p>
<p>
	As special cases of the equivalence in (1), we can compute the following useful logarithms.
For any base <m>b \gt 0, b\ne 1</m>,
</p>

<assemblage><title>Some Useful Logarithms</title>
<p>
	\begin{align*}
	\log_b b \amp = 1 \text{ because } b^1 = b \\
	\log_b 1 \amp = 0  \text{ because } b^0 = 1 \\
	\log_b{bx} \amp = x \text{ because } b^x = b^x
	\end{align*}
</p></assemblage>

<example xml:id="example-useful-logarithms">
	<ol label="a">
		<li><p><m>\log_{2}{2} = 1</m></p></li>
		<li><p><m>\log_{5}{1} = 0</m></p></li>
		<li><p><m>\log_{3}{3^4} = 4</m></p></li>
	</ol>
</example>

<exercise xml:id="exercise-useful-logarithms"><statement>
	Find each logarithm.
	<ol label="a">
		<li><p><m>\log_{n}{1}</m></p></li>
		<li><p><m>\log_{n}{n^3}</m></p></li>
	</ol>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>0</m></p></li>
	<li><p><m>3</m></p></li>
</ol></p></answer>
</exercise>
</introduction>
<subsection><title>Using the Conversion Equations</title>
<p>
	We use logarithms to solve exponential equations, just as we use square roots to solve quadratic equations. Consider the two equations
	<me>x^2 = 25 ~ \text{ and } ~ 2^x = 8</me>
	We solve the first equation by taking a square root, and we solve the second equation by computing a logarithm:
	<me>x = \pm\sqrt{25} = \pm 5 ~ \text{ and } ~ x = \log_{2}{8} = 3</me>
	The operation of taking a base <m>b</m> logarithm is the inverse operation for raising the base <m>b</m> to a power, just as extracting square roots is the inverse of squaring a number.
</p>
<p>
	Every exponential equation can be rewritten in logarithmic form by using the conversion equations. Thus,
	<me>3 = \log_{2}{8} \text{ and } 8 = 2^3</me>
	are equivalent statements, just as
	<me>5 = \sqrt{25} \text{ and } 25 = 5^2</me>
	are equivalent statements. Rewriting an equation in logarithmic form is a basic  strategy for finding its solution.
</p>

<example xml:id="example-exponential-to-log-form">
<p>
	Rewrite each equation in logarithmic form.
	<ol label="a">
		<li><p><m>2^{-1} = \dfrac{1}{2}</m></p></li>
		<li><p><m>a^{1/5} = 2.8</m></p></li>
		<li><p><m>6^{1.5} = T</m></p></li>
		<li><p><m>M^v = 3K</m></p></li>
	</ol>
</p>
<solution>
<p>
	First identify the base <m>b</m>, and then the exponent or logarithm <m>y</m>. Use the conversion equations to rewrite <m>b^y = x</m> in the form <m>\log_{b}{x} = y</m>.
	<ol label="a">
		<li><p>The base is <m>2</m> and the exponent is <m>-1</m>. Thus, <m>\log_{2}{\dfrac{1}{2}}= -1</m>.</p></li>
		<li><p>The base is <m>a</m> and the exponent is <m>\dfrac{1}{5}</m>. Thus, <m>\log_{a}{2.8} = \dfrac{1}{5}</m>.</p></li>
		<li><p>The base is <m>6</m> and the exponent is <m>1.5</m>. Thus, <m>\log_{6}{T} = 1.5</m>.</p></li>
		<li><p>The base is <m>M</m> and the exponent is <m>v</m>. Thus, <m>\log_{M}{3K} = v</m>.</p></li>
	</ol>
</p>
</solution>
</example>

<exercise xml:id="exercise-exponential-to-log-form"><statement>
<p>
	Rewrite each equation in logarithmic form.
	<ol label="a">
		<li><p><m>8^{-1/3} = \dfrac{1}{2}</m></p></li>
		<li><p><m>5^x = 46</m></p></li>
	</ol>
</p>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>\log_8 \left(\dfrac{1}{2} \right)=\dfrac{-1}{3} </m></p></li>
	<li><p><m>\log_{5} 46=x </m></p></li>
</ol></p></answer>
</exercise>

</subsection>



<subsection><title>Approximating Logarithms</title>
<p>
	Suppose we would like to solve the equation
	<me>2^x = 26</me>
	The solution of this equation is <m>x = \log_{2}{26}</m>, but can we find a decimal approximation for this value? There is no integer power of <m>2</m> that equals <m>26</m>, because
	\begin{align*}
	2^4 \amp = 16 \\
	\text{and } 2^5 \amp = 32
	\end{align*} 
	Thus, <m>\log_{2}{26}</m> must be between <m>4</m> and <m>5</m>. We can use trial and error to find the value of <m>\log_{2}{26}</m> to the nearest tenth. Use your calculator to make a table of values for <m>y = 2^x</m>, starting with <m>x = 4</m> and using increments of <m>0.1</m>.
</p>



<table xml:id="table-decimal-powers-of-two"><caption></caption>
    <tabular halign="center" right="minor" left="minor" bottom="minor">
        <col top="minor" />
        <col top="minor" />
        <col />
        <col top="minor"/>
        <col top="minor"/>
                
        <row bottom="major">
            <cell><m>x</m></cell>
            <cell><m>2^x</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>x</m></cell>
            <cell><m>2^x</m></cell>
        </row>
        <row> 
            <cell><m>4</m></cell>
            <cell><m>2^4=16</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>4.5</m></cell>
            <cell><m>2^{4.5}=22.627</m></cell>
        </row>
        <row> 
            <cell><m>4.1</m></cell>
            <cell><m>2^{4.1}=17.148</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>4.6</m></cell>
            <cell><m>2^{4.6}=24.251</m></cell>
        </row>
        <row> 
            <cell><m>4.2</m></cell>
            <cell><m>2^{4.2}=18.379</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>\alert{4.7}</m></cell>
            <cell><m>2^{\alert{4.7}}=25.992</m></cell>
        </row>
        <row> 
            <cell><m>4.3</m></cell>
            <cell><m>2^{4.3}=19.698</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>\alert{4.8}</m></cell>
            <cell><m>2^{\alert{4.8}}=27.858</m></cell>
        </row>
        <row> 
            <cell><m>4.4</m></cell>
            <cell><m>2^{4.4}=21.112</m></cell>
            <cell top="none"  bottom="none"></cell>
            <cell><m>4.9</m></cell>
            <cell><m>2^{4.9}=29.857</m></cell>
        </row>
     </tabular>
</table>
<p>
	From <xref ref="table-decimal-powers-of-two" text="type-global"/>, we see that <m>26</m> is between <m>24.7</m> and <m>24.8</m>, and is closer to <m>24.7</m>. To the nearest tenth, <m>\log_{2}{26} \approx 4.7</m>.
</p>
<p>
	Trial and error can be a time-consuming process. In Example 4, we illustrate a graphical method for estimating the value of a logarithm.
</p>

<example xml:id="example-approximate-log">
<p>
	Approximate <m>\log_{3}{7}</m> to the nearest hundredth.
</p>
<solution>
<p>
	If <m>\log_{3}{7}=x</m>, then <m>3^x = 7</m>. We will use the graph of <m>y = 3^x</m> to approximate a solution to <m>3^x = 7</m>. Graph <m>Y_1 =3</m><c>^</c> X and <m>Y_2 = 7</m> in the standard window (<c>ZOOM</c> 6) to obtain the graph shown in <xref ref="fig-GC-approx-log" text="type-global"/>. Activate the intersect feature to find that the two graphs intersect at the point <m>(1.7712437, 7)</m>. Because this point lies on the graph of <m>y = 3^x</m> , we know that
	<me>31.7712437 \approx 7\text{, or } \log_{3}{7} \approx 1.7712437</me>
	To the nearest hundredth, <m>\log_{3}{7} \approx 1.77</m>.
</p>
<figure xml:id="fig-GC-approx-log"><caption></caption><image source="images/fig-GC-approx-log"  width="50%"><description>GC intersection of expnential curve and horizontal line</description></image></figure>
</solution>
</example>
<exercise xml:id="exercise-approximate-log"><statement>
	<ol label="a">
		<li><p>Rewrite the equation <m>3^x = 90</m> in logarithmic form.</p></li>
		<li><p>Use a graph to approximate the solution to the equation in part (a). Round your answer to three decimal places.</p></li>
	</ol>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>\log_8(90)=x </m></p></li>
	<li><p><m>x\approx 4.096 </m></p></li>
</ol></p></answer>
</exercise>

</subsection>



<subsection><title>Base 10 Logarithms</title>
<p>
	Some logarithms are used so frequently in applications that their values are programmed into scientific and graphing calculators. These are the base <m>10</m> logarithms, such as 
	<me>\log_{10}{1000} = 3 ~\text{ and }~ \log_{10}{0.01} = -2</me>
	Base <m>10</m> logarithms are called <term>common logarithms</term><idx>common logarithms</idx>, and the subscript <m>10</m> is often omitted, so that <m>\log x</m> is understood to mean <m>\log_{10}{x}</m>.
</p>
<p>
	To evaluate a base <m>10</m> logarithm, we use the <c>LOG</c> key on a calculator. Many logarithms are irrational numbers, and the calculator gives as many digits as its display allows. We can then round off to the desired accuracy.
</p>

<example xml:id="example-calculator-log">
	<p>
		Approximate the following logarithms to <m>2</m> decimal places.
		<ol label="a">
			<li><p><m>\log{6.5}</m></p></li>
			<li><p><m>\log{256}</m></p></li>
		</ol>
	</p>
<solution>
	<ol label="a">
		<li><p>
			The keying sequence <c>LOG</c> <m>6.5</m> <c>)</c><c>ENTER</c> produces the display</p><p>
			<sidebyside><tabular>
				<col halign="left"/><col /><col halign="right"/>
				<row>
					<cell><m>\log {(6.5)}</m></cell>
					<cell><m></m></cell>
					<cell><m></m></cell>
				</row>
				<row>
					<cell><m></m></cell>
					<cell><m></m></cell>
					<cell><m>.812913566</m></cell>
				</row>
			</tabular></sidebyside>
		so <m>\log {6.5}\approx 0.81</m>.
		</p></li>
		<li><p>
			The keying sequence <c>LOG</c> <m>256</m> <c>)</c> <c>ENTER</c> yields <m>2.408239965</m>, so <m>\log {256} \approx 2.41</m>.
		</p></li>
	</ol>
</solution>
</example>

<p>
	We can check the approximations found in <xref ref="example-calculator-log" text="type-global"/> with our conversion equations. Remember that a logarithm is an exponent, and in this example the base is <m>10</m>. We find that
	\begin{align*}
	\amp\amp 10^{0.81} \amp\approx 6.45654229 \\
	\text{and} \amp\amp 10^{2.41} \amp\approx 257.0395783
	\end{align*}
	so our approximations are reasonable, although you can see that rounding a logarithm to <m>2</m> decimal places does lose some accuracy.</p>
	<p> For this reason, <em>rounding logarithms to <m>4</m> decimal places is customary.</em>
</p>

<exercise xml:id="exercise-calculator-log"><statement>
	<ol label="a">
		<li><p>Evaluate <m>\log{250}</m>, and round your answer to two decimal places. Check your answer using the conversion equations.</p></li>
		<li><p>Evaluate <m>\log{250}</m>, and round your answer to four decimal places. Check your answer using the conversion equations.</p></li>
	</ol>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>2.40</m></p></li>
	<li><p><m>2.3979</m></p></li>
</ol></p></answer>
</exercise>

</subsection>



<subsection><title>Solving Exponential Equations</title>
<p>
	We can now solve any exponential equation with base <m>10</m>. For instance, to solve the equation
	<m>16 \cdot 10^t = 360</m>, 
	we first divide both sides by <m>16</m> to obtain
	<me>10^t = 22.5</me>
	Then we convert the equation to logarithmic form and evaluate:
	<me>t = \log_{10}{22.5} \approx 1.352182518</me>
	To <m>4</m> decimal places, the solution is <m>1.3522</m>.
</p>
<p>
	To solve exponential equations involving powers of 10, we can use the following steps.
</p>
<assemblage><title>Steps for Solving Base 10 Exponential Equations</title><p>
<ol>
	<li><p>Isolate the power on one side of the equation.</p></li>
	<li><p>Rewrite the equation in logarithmic form.</p></li>
	<li><p>Use a calculator, if necessary, to evaluate the logarithm.</p></li>
	<li><p>Solve for the variable.</p></li>
</ol></p>
</assemblage>

<example xml:id="example-solve-exponential">
<p>
	Solve the equation <m>38 = 95 - 15 \cdot 10^{0.4x}</m>.
</p>
<solution>
<p>
	First, we isolate the power of <m>10</m>: We subtract <m>95</m> from both sides of the equation and divide by <m>-15</m> to obtain
	\begin{align*}
	-57 \amp = -15 \cdot 10^{0.4x} \amp\amp \text{Divide by }-15. \\
	3.8 \amp = 10^{0.4x}
	\end{align*}
	Next, we convert the equation to logarithmic form as
	<me>\log_{10}{3.8} = 0.4x</me>
	Solving for <m>x</m> yields
	<me>\frac{\log_{10}{3.8}}{0.4}= x</me>
	We can evaluate this expression on the calculator by entering
</p>
<p>
	<c>LOG</c> <m>3.8</m> <c>)</c> <c>÷</c> <m>0.4</m> <c>ENTER</c>
</p>
<p>
	which yields <m>1.449458992</m>. Thus, to four decimal places, <m>x \approx 1.4495</m>.
</p>
</solution>
</example>

<warning><statement>
<p>
		Be careful when using a calculator to evaluate expressions involving logs. We can evaluate a single logarithm like <m>\log{3.8}</m> by entering <c>LOG</c> <m>3.8</m><c>ENTER</c> without an ending parenthesis, so that the calculator shows</p><p>
		<sidebyside><tabular>
			<col halign="left"/><col /><col halign="right"/>
			<row>
				<cell><m>\log {(3.8}</m></cell>
				<cell><m></m></cell>
				<cell><m></m></cell>
			</row>
			<row>
				<cell><m></m></cell>
				<cell><m></m></cell>
				<cell><m>.5795835966</m></cell>
			</row>
		</tabular></sidebyside>
		But if we want to evaluate <m>\dfrac{\log{3.8}}{0.4}</m>, we must enclose <m>3.8</m> in parentheses, as shown in <xref ref="example-solve-exponential" text="type-global"/>. If we omit the parenthesis after <m>3.8</m>, the calculator will interpret the expression as <m>\log{\left(\dfrac{3.8}{0.4}\right)}</m>, which is not the expression we wanted.
</p>
</statement>
</warning>

<exercise xml:id="exercise-solve-exponential"><statement>
	<p>
		Solve <m>12 - 30(10^{-0.2x} ) = 11.25</m>.
	</p>
</statement>
<answer><p><m>8.01</m></p></answer>
</exercise>

</subsection>




<subsection><title>Application to Exponential Models</title>
<p>
	We have seen that exponential functions are used to describe some applications of growth and decay, <m>P(t) = P_0b^t</m>. There are two common questions that arise in connection with exponential models:
	<ol>
		<li><p>Given a value of <m>t</m>, what is the corresponding value of <m>P(t)</m>?</p></li>
		<li><p>Given a value of <m>P(t)</m>, what is find the corresponding value of <m>t</m>?</p></li>
	</ol>
	To answer the first question, we evaluate the function <m>P(t)</m> at the appropriate value. To answer the second question, we must solve an exponential equation, and this usually involves logarithms.
</p>

<example xml:id="example-tractor-depreciates">
<p>
	The value of a large tractor originally worth <dollar/><m>30,000</m> depreciates exponentially according to the formula <m>V(t) = 30,000(10)^{-0.04t}</m>, where <m>t</m> is in years. When will the tractor be worth half its original value?
</p>
<solution>
<p>
	We want to find the value of <m>t</m> for which <m>V(t) = 15,000</m>. That is, we want to solve the equation
	<me>15,000 = 30,000(10)^{-0.04t}</me>
	We divide both sides by 30,000 to obtain
	<me>0.5 = 10^{-0.04t}</me>
	We convert the equation to logarithmic form as
	<me>\log_{10}{0.5} = -0.04t</me>
	and divide by <m>-0.04</m> to obtain
	<me>\frac{\log_{10}{0.5}}{-0.04} = t </me>
	To evaluate this expression, we key in
</p>
<p>
	<c>LOG</c> <m>0.5</m> <c>)</c> <c>÷</c> <c>(-)</c> <m>0.04</m> <c>ENTER</c>
</p>
<p>
	to find <m>t \approx 7.525749892</m>. The tractor will be worth <dollar/>15,000 in approximately <m>7\frac{1}{2}</m> years.
</p>
</solution>
</example>

<exercise xml:id="exercise-home-computers"><statement>
<p>
	The percentage of American homes with computers grew exponentially from 1994 to 1999. For <m>t = 0</m> in 1994, the growth law was <m>P(t) = 25.85(10)^{0.052t}</m>. [Source: Los Angeles Times, August 20, 1999]
	<ol label="a">
		<li><p>What percent of American homes had computers in 1994?</p></li>
		<li><p>If the percentage of homes with computers continued to grow at the same rate, when did <m>90</m><percent/> of American homes have a computer?</p></li>
		<li><p>Do you think that the function <m>P(t)</m> will continue to model the percentage of American homes with computers? Why or why not?</p></li>
	</ol>
</p>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>25.85\%</m></p></li>
	<li><p><m>t\approx 10.4</m> (year 2004)</p></li>
	<li><p>No, the percent of homes with computers cannot exceed <m>100\%</m>.</p></li>
</ol></p></answer>
</exercise>

<p>
	At this stage, it seems we will only be able to solve exponential equations in which the base is <m>10</m>. However, we will see in <xref ref="Properties-of-Logarithms" text="type-global"/> how the properties of logarithms enable us to solve exponential equations with any base.
</p>


</subsection>
<xi:include href="./summary-4-3.xml" /> <!-- summary  -->
<xi:include href="./section-4-3-exercises.xml" /> <!-- exercises  -->

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