chap4-summary.xml

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

<section xml:id="chap4-summary" xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Chapter Summary and Review</title>
<subsection><title>Key Concepts</title>
<p>
	<ol label="1">
		<li><p>If a quantity is multiplied by a constant factor, <m>b</m>, in each time period, we say that it undergoes <term>exponential growth</term><idx>exponential growth</idx> or <term>decay</term> . The constant <m>b</m> is called the <term>growth factor</term><idx>growth factor</idx> if <m>b\gt 1</m> and the <term>decay factor</term><idx>decay factor</idx> if <m>0 \lt b \lt 1</m>.</p></li>
		<li><p>Quantities that increase or decrease by a constant percent in each time period grow or decay exponentially.</p></li>
		<li><p>
			<assemblage><title>Exponential Growth and Decay</title><p>
				The function
				<me>P(t)=P_0 b^t</me>
				models exponential growth and decay.</p>
			<p><m>P_0 = P(0)</m> is the initial value of <m>P</m>;</p>
			<p><m>b</m> is the growth or decay factor.</p>
			<p><ol label="1">
				<li><p>If <m>b\gt 1</m>, then <m>P(t)</m> is increasing, and <m>b = 1 + r</m>, where <m>r</m> represents percent increase.</p></li>
				<li><p>If <m>0\lt b \lt 1</m>, then <m>P(t)</m> is decreasing, and <m>b= 1 - r</m>, where <m>r</m> represents percent decrease.</p></li>
			</ol></p>
			</assemblage>
		</p></li>
		<li><p>
			<assemblage><title>Interest Compounded Annually</title><p>
			The amount <m>A(t)</m> accumulated (principal plus interest) in an account bearing interest compounded annually is
			<me>A(t)=(1 + r)^t</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>
		</p></li>
		<li><p>
			In linear growth, a constant amount is <em>added</em> to the output for each unit increase in the input. In exponential growth, the output is <em>multiplied</em> by a constant factor for each unit increase in the input.
		</p></li>
		<li><p>
			An <term>exponential function</term><idx>exponential function</idx> has the form
			<me>f (x) = ab^x,~~ \text{ where }~~ b\gt 0 ~~\text{ and } b \ne 1, ~~~a\ne 0</me>
		</p></li>
		<li><p>
			<assemblage><title>Properties of Exponential Functions, <m>f(x)=ab^x, ~~a\gt 0 </m></title>
			<p><ol label="1">
				<li><p>Domain: all real numbers</p></li>
				<li><p>Range: all positive numbers</p></li>
				<li><p>If <m>b\gt 1</m>, the function is increasing and concave up; if <m>0\lt b \lt 1</m>, the function is decreasing and concave up.</p></li>
				<li><p>The <m>y</m>-intercept is <m>(a, 0)</m>. There is no <m>x</m>-intercept</p></li>
			</ol>
			</p></assemblage>
		</p></li>
		<li><p>
			The graphs of exponential functions can be transformed by shifts, stretches, and reflections.
		</p></li>
		<li><p>
			<assemblage><title>Reflections of Graphs</title>
			<p><ol label="1">
				<li><p>The graph of <m>y = -f (x)</m> is the reflection of the graph of <m>y = f (x)</m> about the <m>x</m>-axis.</p></li>
				<li><p>The graph of <m>y = f (-x)</m> is the reflection of the graph of <m>y = f (x)</m> about the <m>y</m>-axis.</p></li>
			</ol>
			</p></assemblage>
		</p></li>
		<li><p>
			Exponential functions <m>f (x) = ab^x</m> have different properties than power functions <m>f (x) = kx^p</m>.
		</p></li>
		<li><p>We can solve <term>exponential equations</term> by writing both sides with the same base and equating the exponents.</p></li>
		<li><p>We can use graphs to find approximate solutions to exponential equations</p></li>
		<li><p>We use logrithms to help us solve exponential equations.</p></li>
		<li><p>The <term>base <m>b</m> logarithm of <m>x</m></term>, 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></li>
		<li><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></li>
		<li><p>The operation of taking a base <m>b</m> logarithm is the inverse operation for raising the base <m>b</m> to a power.</p></li>
		<li><p>Base 10 logarithms are called <term>common logarithms</term>, and <m>\log x</m> means <m>\log_{10} x</m>.</p></li>
		<li><p>
			<assemblage><title>Steps for Solving Base 10 Exponential Equations</title>
			<p><ol label="1">
				<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>
		</p></li>
		<li><p>
			<assemblage><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 label="1">
				<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></li>
		<li><p>We can use the properties of logarithms to solve exponential equations with any base.</p></li>
		<li><p>
			<assemblage><title>Compounded 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)=\left(1 + \dfrac{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>
		</p></li>
		<li><p>We can use the ratio method to fit an exponential function through two points.<p></p>
		<assemblage><title>To find an exponential function <m>f (x) = ab^x</m> through two points:</title><p>
				<ol label="1">
					<li><p>Use the coordinates of the points to write two equations in <m>a</m> and <m>b</m>.</p></li>
					<li><p>Divide one equation by the other to eliminate <m>a</m>.</p></li>
					<li><p>Solve for <m>b</m>.</p></li>
					<li><p>Substitute <m>b</m> into either equation and solve for <m>a</m>.</p></li>
				</ol>
			</p></assemblage>
		</p></li>
		<li><p>Every increasing exponential has a fixed <term>doubling time</term>. Every decreasing exponential function has a fixed <term>half-life</term>.</p></li>
		<li><p>If <m>D</m> is the doubling time for a population, its growth law can be written as <m>P(t) = P_0 2^{t/D}</m>.</p></li>
		<li><p>If <m>H</m> is the half-life for a quantity, its decay law can be written as <m>Q(t) = Q_0 (0.5)^{t/H}</m>.</p></li>
		<li><p>
			<assemblage><title>Future Value of an Annuity</title><p>
				If you make <m>n</m> payments per year for <m>t</m> years into an annuity that pays interest rate <m>r</m> compounded <m>n</m> times per year, the future value, <m>FV</m>, of the annuity is
				<me>FV=\dfrac{P\left[\left(1+\dfrac{r}{n} \right)^{nt}-1 \right]}{\dfrac{r}{n}} </me>
				where each payment is <m>P</m> dollars.
			</p></assemblage>
		</p></li>
		<li><p>
			<assemblage><title>Present Value of an Annuity</title><p>
				If you wish to receive <m>n</m> payments per year for <m>t</m> years from a fund that earns interest rate <m>r</m> compounded <m>n</m> times per year, the present value, <m>PV</m>, of the annuity must be
				<me>PV=\dfrac{P\left[1- \left(1+\dfrac{r}{n} \right)^{-nt} \right]}{\dfrac{r}{n}} </me>
				where each payment is <m>P</m> dollars.
			</p></assemblage>
		</p></li>
	</ol>.
</p>
</subsection>
<!-- <subsubsection> -->
	<xi:include href="./chap4-rev-problems.xml" />   <!-- exercises  -->
<!-- </subsubsection> -->

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