section-4-4.xml

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<!-- <mathbook><book> -->

<section xml:id="Properties-of-Logarithms"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Properties of Logarithms</title>
<introduction>
<p>
	Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.

	<ol>
		<li><p>
			To multiply two powers with the same base, add the exponents and leave the base unchanged.
			<me>a^m \cdot a^n = a^{m+n}</me>
		</p></li>
		<li><p>
			To divide two powers with the same base, subtract the exponents and leave the base unchanged.
			<me>\frac{a^m}{a^n} = a^{m-n}</me>
		</p></li>
		<li><p>To raise a power to a power, keep the same base and multiply the exponents.
			<me>\left(a^m\right)^n = a^{mn}</me>
		</p></li>
	</ol>
	Each of these laws corresponds to one of three properties of logarithms.
</p>

<assemblage xml:id="assemblage-Properties-of-Logarithms"><title>Properties of Logarithms</title>
<p>
	If <m>x</m>, <m>y</m>, <m>b \gt 0</m>, and <m>b\ne 1</m>, then
	<ol>
		<li><p><m>\log_{b}{(xy)} = \log_{b}{x} + \log_{b}{y}</m></p></li>
		<li><p><m>\log_{b}\dfrac{x}{y} = \log_b x - \log_b y</m></p></li>
		<li><p><m>\log_b {x^k} = k \log_b x </m></p></li>
	</ol>
</p>
</assemblage>

<p>
	We will consider proofs of the three properties of logarithms in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is an exponent.
</p>
<ol>
	<li><p>
		Property (1):
		\begin{align*}
		\log_{2}{32} \amp=\log_{2}{(4 \cdot 8)} \amp\amp=\log_2 4 +\log_2 8 \amp \text{ because }2^\alert{5}  \amp= 2^\alert{2} \cdot 2^\alert{3} \\
		 \alert{5} 		\amp\amp\amp = \alert{2} + \alert{3}		\amp 32 					\amp = 4 \cdot 8
		\end{align*}
	</p></li>
	<li><p>
		Property (2):
		\begin{align*}
		\log_{2}{8} \amp=\log_{2}{\frac{16}{2}} \amp\amp=\log_2{16}-\log_2 2 \amp \text{ because }2^\alert{3}  \amp= \frac{2^\alert{4}}{2^\alert{1}} \\
		 \alert{3} 		\amp\amp\amp = \alert{4} - \alert{1}		\amp 8 					\amp = \frac{16}{2}
		\end{align*}
	</p></li>
	<li><p>
		Property (3):
		\begin{align*}
		\log_{2}{64} \amp=\log_{2}{(4)^3} \amp\amp=3\log_2 4  \amp \text{ because }\left(2^{\alert{2}}\right)^\alert{3}  \amp= 2^\alert{6}  \\
		 \alert{6} 		\amp\amp\amp = \alert{3}\cdot \alert{2}		\amp (4)^\alert{3} 					\amp = 64
		\end{align*}

	</p></li>
</ol>
</introduction>


<subsection><title>Using the Properties of Logarithms</title>
<p>
	Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In Example 1, we rewrite one log in terms of simpler logs.
</p>

<example xml:id="example-use-log-property">
<p>
	Simplify <m>\log_{b}{xy}</m>.
</p>
<solution>
<p>
	First, we write <m>\sqrt{xy}</m> using a fractional exponent:
	<me>\log_{b}{xy} = \log_{b}{(xy)^{1/2}}</me>
	Then we apply Property (3) to rewrite the exponent as a coefficient:
	<me>\log{b}{(xy)^{1/2}} = \frac{1}{2}\log_{b}{(xy)}</me>
	Finally, by Property (1) we write the log of a product as a sum of logs:
	<me>\frac{1}{2}(\log_{b}{xy}) = \frac{1}{2}(\log_{b}{x} + \log_{b}{y})</me>
	Thus, <m>\log_{b}\sqrt{xy} = \frac{1}{2}(\log_{b}{x} + \log_{b}{y})</m>.
</p>
</solution>
</example>

<exercise xml:id="exercise-use-log-property"><statement>
<p>
	Simplify <m>\log_{b}{xy^2}</m>.
</p>
</statement>
<answer><p><m>\log_b x - 2\log_b y</m></p></answer>
</exercise>

<warning>
<p>
	Be careful when using the properties of logarithms! Compare the statements below:
	<ol>
		<li><p><m>\log_{b}{(2x)} = \log_{b}{2} + \log_{b}{x}    \text{   by Property 1,}</m></p>
		<p>but</p>
		<p><m>\log_{b}{(2 + x)} \ne \log_{b}{2} + \log_{b}{x}</m></p><p></p></li>
		<li><p><m>\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b x - \log_b 5   \text{   by Property 2,}</m></p>
		<p>but</p>
		<p><m>\log_{b}{\left(\dfrac{x}{5}\right)} \ne  \dfrac{\log_b x}{\log_b 5}</m></p></li>
	</ol>
</p>
</warning>

<p>
	We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.
</p>
<example xml:id="example-rewrite-as-single-log">
<p>
	Express <m>3(\log_b x - \log_b y)</m> as a single logarithm with a coefficient of <m>1</m>.
</p>
<solution>
<p>
	We begin by applying Property (2) to combine the logs.
	<me>3(\log_b x - \log_b y) = 3 \log_{b}\left(\frac{x}{y}\right)</me>
	Then, using Property (3), we replace the coefficient <m>3</m> by an exponent <m>3</m>.
	<me>3 \log_{b}\left(\frac{x}{y}\right)=\log_{b}\left(\frac{x}{y}\right)^3</me>
</p>

</solution>
</example>
<exercise xml:id="exercise-rewrite-as-single-log"><statement>
	<p>
		Express <m>2\log_b x + 4\log_{b}{(x + 3)}</m> as a single logarithm with a coefficient of <m>1</m>.
	</p>
</statement>
<answer><p><m>\log_b x^2(x+3)^4 </m></p></answer>
</exercise>
</subsection>



<subsection><title>Solving Exponential Equations</title>
<p>
	By using Property (3), we can now solve exponential equations in which the base is not <m>10</m>. For example, to solve the equation
	<me>5^x = 7</me>
	we could rewrite the equation in logarithmic form to obtain the exact solution
	<me>x = \log_{5}{7}</me>
	However, we cannot evaluate <m>\log_{5}{7}</m>; there is no log base <m>5</m> button on the calculator. If we want a decimal approximation for the solution, we begin by taking the base <m>10</m> logarithm of both sides, even though the base of the power is not <m>10</m>. This gives us
	<me>\log_{10}{(5^x)} = \log_{10}{7}</me>
	Then we use Property (3) to rewrite the left side as
	<me>x \log_{10}{5} = \log_{10}{7}</me>
	Note how using Property (3) allows us to solve the equation: The variable, <m>x</m>, is no longer in the exponent, and it is multiplied by a constant, <m>\log_{10}{5}</m>. To finish the solution, we divide both sides by <m>\log_{10}{5}</m> to get
	<me>x = \frac{\log_{10}{7}}{\log_{10}{5}}</me>
	On your calculator, enter the sequence
</p>
<p>
	<c>LOG</c> <m>7</m> <c>)</c> <c>÷</c> <c>LOG</c> <m>5</m> <c>)</c> <c>ENTER</c>
</p>
<p>
	to find that <m>x \approx 1.2091</m>.
</p>

<warning>
<p>
	Do not confuse the expression <m>\dfrac{\log_{10}{7}}{\log_{10}{5}}</m> with <m>\log_{10}{\left(\dfrac{7}{5}\right)}</m>; they are not the same! Property (2) allows us to simplify <m>\log{\left(\dfrac{x}{y}\right)}</m>, but not <m>\dfrac{\log x}{\log y}</m>. We cannot rewrite <m>\dfrac{\log_{10}{7}}{\log_{10}{5}}</m>, so we must evaluate it as <m>(\log 7)/(\log 5)</m>. You can check on your calculator that
	<me>\dfrac{\log_{10}{7}}{\log_{10}{5}}\ne \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}{1.4}</me>.
</p>
</warning>

<example xml:id="example-solve-exp-equation">
<p>
	Solve <m>1640 = 80 \cdot 6^{0.03x}</m>.
</p>
<solution>
<p>
	First we divide both sides by <m>80</m> to obtain
	<me>20.5 = 6^{0.03x}</me>
	Next, we take the base <m>10</m> logarithm of both sides of the equation and use Property (3) of logarithms to get
	<me>\log_{10}{20.5} = \log_{10}{6^{0.03x}}= 0.03x \log_{10}{6}</me>
	On the right side of the equation, <m>x</m> is multiplied by two constants, <m>0.03</m> and <m>\log_{10}{6}</m>. So, to solve for <m>x</m> we must divide both sides of the equation by <m>0.03 \log_{10}{6}</m>. We use a calculator to evaluate the answer:
	<me>x = \frac{\log_{10}{20.5}}{0.03 \log_{10}{6}}\approx 56.19</me>
	(On your calculator, remember to enclose the denominator, <m>0.03 \log_{10}{6}</m>, in parentheses.)
</p>
</solution>
</example>

<warning>
	<p>
		In <xref ref="example-solve-exp-equation" text="type-global"/>, do not try to simplify 
		<me>80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \text{ Incorrect!}</me>
		Remember that the order of operations tells us to compute the power <m>6^{0.03x}</m> before multiplying by <m>80</m>.
	</p>
</warning>

<p>
	We summarize our method for solving exponential equations as follows.
</p>
<assemblage xml:id="assemblage-Steps-for-Solving-Exponential"><title>Steps for Solving Exponential Equations</title><p>
	<ol>
		<li><p>Isolate the power on one side of the equation.</p></li>
		<li><p>Take the log base <m>10</m> of both sides.</p></li>
		<li><p>Simplify by applying Log Property (3).</p></li>
		<li>Solve for the variable.</li>
	</ol></p>
</assemblage>

<exercise xml:id="exercise-solve-exp-equation"><statement>
	<p>
		Solve <m>5(1.2)^{2.5x} = 77</m>.
	</p>
</statement>
	<hint>
		<p>Divide both sides by 5.</p>
		<p>Take the log of both sides.</p>
		<p>Apply Property (3) to simplify the left side.</p>
		<p>Solve for <m>x</m>.</p>
	</hint>
	<answer><p><m>x=\dfrac{\log 15.4}{2.5\log 1.2}\approx 5.999 </m></p></answer>
</exercise>

</subsection>



<subsection><title>Applications</title>
<p>
	By using the properties of logarithms, we can now solve equations that arise in exponential growth and decay models, no matter what base the exponential function uses.
</p>

<example xml:id="example-population-growth">
<p>
	The population of Silicon City was <m>6500</m> in <m>1990</m> and has been tripling every <m>12</m> years. When will the population reach <m>75,000</m>?
</p>
<solution>
<p>
	The population of Silicon City grows according to the formula
	<me>P(t) = 6500 \cdot 3^{t/12}</me>
	where <m>t</m> is the number of years after <m>1990</m>. We want to find the value of <m>t</m> for which <m>P(t) = 75,000</m>; that is, we want to solve the equation
	\begin{align*}
	6500 \cdot 3^{t/12} \amp = 75,000\amp\amp\text{ Divide both sides by 6500.}\\
	3^{t/12} \amp = \frac{150}{13} 
	\end{align*}
	Now we take the base <m>10</m> logarithm of both sides and solve for <m>t</m>.
	\begin{align*}
	\log_{10}{(3^{t/12})} \amp = \log_{10}\left(\frac{150}{13}\right)
	\amp\amp\text{Apply Property (3).}\\
	\frac{t}{12}\log_{10}{3} \amp= \log_{10}\left(\frac{150}{13}\right)
	\amp\amp\text{Divide by }\log_{10}{3}\text{; multiply by 12.}\\
	t \amp = \frac{12 \left(\log_{10}{\frac{150}{13}}\right)}{\log_{10}{3}}\\
	\amp\approx 26.71
	\end{align*}
	The population of Silicon City will reach <m>75,000</m> about <m>27</m> years after <m>1990</m>, or in <m>2017</m>.
</p>
</solution>
</example>

<exercise xml:id="exercise-traffic-growth"><statement>
<p>
	Traffic on U.S. highways is growing by <m>2.7</m><percent/> per year. (Source: <em>Time</em>, Jan. 25, 1999)
	<ol label="a">
		<li>Write a formula for the volume, <m>V</m>, of traffic as a function of time, using <m>V_0</m> for the current volume.</li>
		<li>How long will it take the volume of traffic to double? <em>Hint:</em> Find the value of <m>t</m> that gives <m>V = 2V_0</m>.</li>
	</ol>
</p>
</statement>
<answer><p><ol label="a" cols="2">
	<li><p><m>V(t) = V_0(1.027)^t</m></p></li>
	<li><p>about <m>26</m> years</p></li>
</ol></p></answer>
</exercise>
</subsection>



<subsection><title>Compound Interest</title>
<p>
	The amount of money in an account that earns interest compounded annually grows exponentially according to the formula
	<me>A(t) = P(1 + r )^t</me>
	(See <xref ref="Exponential-Growth-and-Decay" text="type-global"/> to review compound interest.) Many accounts compound interest more frequently than once a year. If the interest is compounded <m>n</m> times per year, then in <m>t</m> years there will be <m>nt</m> compounding periods, and in each period the account earns interest at a rate of <m>\dfrac{r}{n}</m>. The amount accumulated is given by a generalization of our earlier formula.
</p>
<assemblage xml:id="assemblage-Compound-Interest"><title>Compound Interest</title>
<p>
	The amount <m>A(t)</m> accumulated (principal plus interest) in an account bearing interest compounded <m>n</m> times annually is
	<me>A(t)=P\left(1+ \frac{r}{n}\right)^{nt}</me>
	where  
	<tabular>
		<row>
			<cell></cell><cell></cell>
			<cell><m>P</m></cell>
			<cell>is the principal invested,</cell>
		</row>
		<row>
			<cell></cell><cell></cell>
			<cell><m>r</m></cell>
			<cell>is the interest rate,</cell>
		</row>
		<row>
			<cell></cell><cell></cell>
			<cell><m>t</m></cell><cell>is the time period, in years.</cell>
		</row>
	</tabular>
</p>
</assemblage>

<example xml:id="example-compound-interest">
<p>
	Rashad deposited <dollar/><m>1000</m> in an account that pays <m>4</m><percent/> interest. Calculate the amount in his account after <m>5</m> years if the interest is compounded
	<ol label="a">
		<li><p>semiannually</p></li>
		<li><p>quarterly</p></li>
		<li><p>monthly</p></li>
	</ol>
</p>
<solution>
	<ol label="a">
		<li><p>
			<em>Semiannually</em> means "twice a year,"" so we use the formula for compound interest with <m>P = 1000</m>, <m>r = 0.04</m>, <m>n = 2</m>, and <m>t = 5</m>.
			\begin{align*}
			A(5) \amp = 1000\left(1 + \frac{0.04}{2}\right)^{2(5)} \\
			\amp = 1000(1.02)^{10} = 1218.99
			\end{align*}
			If interest is compounded semiannually, the balance in the account after <m>5</m> years is <dollar/><m>1218.99</m>.
		</p></li>
		<li><p>
			<em>Quarterly</em> means "4 times a year,"" so we use the formula for compound interest with <m>P = 1000</m>, <m>r = 0.04</m>, <m>n = 4</m>, and <m>t = 5</m>.
			\begin{align*}
			A(5) \amp = 1000\left(1 + \frac{0.04}{4}\right)^{4(5)} \\
			\amp = 1000(1.01)^{20} = 1220.19
			\end{align*}
			If interest is compounded quarterly, the balance in the account after <m>5</m> years is <dollar/><m>1220.19</m>.
		</p></li>
		<li><p>
			There are <m>12</m> months in a year, so we use the formula for compound interest with <m>P = 1000</m>, <m>r = 0.04</m>, <m>n = 12</m>, and <m>t = 5</m>.
			\begin{align*}
			A(5) \amp = 1000\left(1 + \frac{0.04}{12}\right)^{12(5)}\\
			\amp = 1000(1.003)^{60} = 1221.00
			\end{align*}
			If interest is compounded monthly, the balance in the account after <m>5</m> years is <dollar/><m>1221</m>.
		</p></li>
	</ol>
</solution>
</example>
<p>
	In <xref ref="example-compound-interest" text="type-global"/>, you can see that the larger the value of <m>n</m>, the greater the value of <m>A</m>, keeping the other parameters fixed. More frequent compounding periods result in a higher account balance.
</p>
<exercise xml:id="exercise-compound-interest"><statement>
<p>
	Calculate the amount in Rashad’s account after <m>5</m> years if the interest is compounded daily. (See <xref ref="example-compound-interest" text="type-global"/>. There are <m>365</m> days in a year.)
</p>
</statement>
<answer><p><m>\$1221.39</m></p></answer>
</exercise>

</subsection>



<subsection><title>Solving Formulas</title>
<p>
	The techniques for solving exponential equations can also be used to solve formulas involving exponential expressions for one variable in terms of the others.
</p>
<example xml:id="example-exponential-formula">
	<p>
		Solve <m>2C = Cb^{kt}</m> for <m>t</m>. (Assume that <m>C</m> and <m>k \ne 0</m>.)
	</p>
<solution>
	<p>
		First, we divide both sides by <m>C</m> to isolate the power.
		<me>b^{kt} = 2</me>
		Next, we take the log base <m>10</m> of both sides.
		\begin{align*}
		\log b^{kt} \amp = \log 2 \\
		kt \log b \amp = \log 2 \amp\amp\text{Apply Log Property (3).}
		\end{align*}
		Finally, we divide both sides by <m>k \log b</m> to solve for <m>t</m>.
		<me>t = \frac{\log 2}{k \log b}</me>
	</p>
</solution>
</example>

<exercise xml:id="exercise-exponential-formula"><statement>
	<p>
		Solve <m>A = P(1 + r )^t</m> for <m>t</m>.
	</p>
</statement>
<answer><p><m>t=\dfrac{\log(A/P)}{\log(1+r)} </m></p></answer>
</exercise>


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

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