{"id":5617,"date":"2021-11-02T19:48:59","date_gmt":"2021-11-02T19:48:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/?post_type=chapter&#038;p=5617"},"modified":"2021-11-02T19:48:59","modified_gmt":"2021-11-02T19:48:59","slug":"exponential-functions-cut-parts","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/exponential-functions-cut-parts\/","title":{"raw":"Exponential Functions - cut parts","rendered":"Exponential Functions &#8211; cut parts"},"content":{"raw":"India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.[\/footnote] If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is \"exponential.\" To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.\r\n\r\n&nbsp;\r\n\r\n<span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXX<\/span>\r\n\r\n&nbsp;\r\n\r\nNow we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.5 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex] where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Real-World Exponential Model<\/h3>\r\nAt the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013 with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex] where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?\r\n\r\n[reveal-answer q=\"924755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924755\"]\r\n\r\nTo estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,\r\n<p style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\r\nThere will be about 1.549 billion people in India in the year 2031.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe population of China was about 1.39 billion in the year 2013 with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex] where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be in the year 2031? How does this compare to the population prediction we made for India in the previous example?\r\n\r\n[reveal-answer q=\"891037\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"891037\"]\r\n\r\nAbout 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129476&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXX<\/span>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nAs we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\u00a0shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.\r\n\r\nExamine the value of $1 invested at 100% interest for 1 year compounded at various frequencies.\r\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th>Frequency<\/th>\r\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n<th>Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Annually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n<td>$2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Semiannually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n<td>$2.25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Quarterly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n<td>$2.441406<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Monthly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n<td>$2.613035<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Daily<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n<td>$2.714567<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Hourly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n<td>$2.718127<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per minute<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n<td>$2.718279<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per second<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n<td>$2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThese values appear to be approaching a limit as <em>n<\/em>\u00a0increases without bound. In fact, as <em>n<\/em>\u00a0gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX<\/span>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing an Exponential Model When the Initial Value Is Known<\/h3>\r\nIn 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function <em>N<\/em>(<em>t<\/em>) representing the population <em>N<\/em>\u00a0of deer over time <em>t<\/em>.\r\n\r\n[reveal-answer q=\"910377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"910377\"]\r\n\r\nWe let our independent variable <em>t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, <em>a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find <em>b<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}N\\left(t\\right)\\hfill &amp; =80{b}^{t}\\hfill &amp; \\hfill \\\\ 180\\hfill &amp; =80{b}^{6}\\hfill &amp; \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\frac{9}{4}\\hfill &amp; ={b}^{6}\\hfill &amp; \\text{Divide and write in lowest terms}.\\hfill \\\\ b\\hfill &amp; ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill &amp; \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ b\\hfill &amp; \\approx 1.1447 &amp; \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Round the final answer to four places for the remainder of this section.<\/em>\r\n\r\nThe exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. Note that this exponential function models short-term growth. As the inputs get larger, the outputs will get increasingly larger resulting in the model not being useful in the long term due to extremely large output values.\r\n\r\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex] and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225444\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\" \/> Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t\u00a0years after 2006[\/caption]\r\n[\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population <em>N<\/em>\u00a0of wolves over time <em>t<\/em>.\r\n\r\n[reveal-answer q=\"222558\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"222558\"]\r\n\r\n[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX<\/span>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<h2>Equations of Exponential Functions<\/h2>\r\nIn the previous examples, we were given an exponential function which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, determine the constants <em>a<\/em>\u00a0and <em>b<\/em>, and evaluate the function.\r\n<div class=\"textbox\">\r\n<h3>How To: Given two data points, write an exponential model<\/h3>\r\n<ol>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li>If one of the data points has the form [latex]\\left(0,a\\right)[\/latex], then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{b}^{x}[\/latex], and solve for <em>b<\/em>.<\/li>\r\n \t<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. Solve the resulting system of two equations to find [latex]a[\/latex] and [latex]b[\/latex].<\/li>\r\n \t<li>Using the <em>a<\/em>\u00a0and <em>b<\/em>\u00a0found in the steps above, write the exponential function in the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing an Exponential Model When the Initial Value is Not Known<\/h3>\r\nFind an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].\r\n\r\n[reveal-answer q=\"904458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"904458\"]\r\n\r\nBecause we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and then solve the system for <em>a<\/em>\u00a0and <em>b<\/em>.\r\n<ul>\r\n \t<li>Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]<\/li>\r\n \t<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]<\/li>\r\n<\/ul>\r\nUse the first equation to solve for <em>a<\/em>\u00a0in terms of <em>b<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6=ab^{-2}\\\\\\frac{6}{b^{-2}}=a\\,\\,\\,\\,\\,\\,\\,\\,\\text{Divide.}\\\\a=6b^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to rewrite the denominator.}\\end{array}[\/latex]<\/p>\r\nSubstitute <em>a<\/em>\u00a0in the second equation and solve for <em>b<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=ab^{2}\\\\1=6b^{2}b^{2}=6b^{4}\\,\\,\\,\\,\\,\\text{Substitute }a.\\\\b=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to isolate }b.\\\\b\\approx0.6389\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Round 4 decimal places.}\\end{array}[\/latex]<\/p>\r\nUse the value of <em>b<\/em>\u00a0in the first equation to solve for the value of <em>a<\/em>:\r\n<p style=\"text-align: center;\">[latex]a=6b^{2}\\approx6\\left(0.6389\\right)^{2}\\approx2.4492[\/latex]<\/p>\r\nThus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].\r\n\r\nWe can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an <strong>exponential decay<\/strong> function.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225453\/CNX_Precalc_Figure_04_01_0032.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\" \/> The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the two points [latex]\\left(1,3\\right)[\/latex] and [latex]\\left(2,4.5\\right)[\/latex], find the equation of the exponential function that passes through these two points.\r\n\r\n[reveal-answer q=\"40110\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"40110\"]\r\n\r\n[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2942&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Do two points always determine a unique exponential function?<\/strong>\r\n\r\n<em>Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x,\u00a0<\/em><em>which in many real world cases involves time.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the graph of an exponential function, write its equation<\/h3>\r\n<ol>\r\n \t<li>First, identify two points on the graph. Choose the <em>y<\/em>-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.<\/li>\r\n \t<li>If one of the data points is the <em>y-<\/em>intercept [latex]\\left(0,a\\right)[\/latex] , then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and solve for <em>b<\/em>.<\/li>\r\n \t<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. Solve the resulting system of two equations to find <em>a<\/em>\u00a0and <em>b<\/em>.<\/li>\r\n \t<li>Write the exponential function, [latex]f\\left(x\\right)=a{b}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing an Exponential Function Given Its Graph<\/h3>\r\nFind an equation for the exponential function graphed below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225455\/CNX_Precalc_Figure_04_01_0042.jpg\" alt=\"Graph of an increasing exponential function with notable points at (0, 3) and (2, 12).\" width=\"731\" height=\"369\" \/>\r\n\r\n[reveal-answer q=\"440954\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"440954\"]\r\n\r\nWe can choose the <em>y<\/em>-intercept of the graph, [latex]\\left(0,3\\right)[\/latex], as our first point. This gives us the initial value [latex]a=3[\/latex]. Next, choose a point on the curve some distance away from [latex]\\left(0,3\\right)[\/latex] that has integer coordinates. One such point is [latex]\\left(2,12\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llllll}y=a{b}^{x}&amp; \\text{Write the general form of an exponential equation}. \\\\ y=3{b}^{x} &amp; \\text{Substitute the initial value 3 for }a. \\\\ 12=3{b}^{2} &amp; \\text{Substitute in 12 for }y\\text{ and 2 for }x. \\\\ 4={b}^{2} &amp; \\text{Divide by 3}. \\\\ b=\\pm 2 &amp; \\text{Take the square root}.\\end{array}[\/latex]<\/p>\r\nBecause we restrict ourselves to positive values of <em>b<\/em>, we will use <em>b\u00a0<\/em>= 2. Substitute <em>a<\/em>\u00a0and <em>b<\/em>\u00a0into standard form to yield the equation [latex]f\\left(x\\right)=3{\\left(2\\right)}^{x}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind an equation for the exponential function graphed below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225457\/CNX_Precalc_Figure_04_01_0052.jpg\" alt=\"Graph of an increasing function with a labeled point at (0, sqrt(2)).\" width=\"487\" height=\"294\" \/>\r\n\r\n[reveal-answer q=\"564720\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"564720\"][latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>","rendered":"<p>India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.<a class=\"footnote\" title=\"http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.\" id=\"return-footnote-5617-1\" href=\"#footnote-5617-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is &#8220;exponential.&#8221; To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXX<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.5 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex] where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Real-World Exponential Model<\/h3>\n<p>At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013 with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex] where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924755\">Show Solution<\/span><\/p>\n<div id=\"q924755\" class=\"hidden-answer\" style=\"display: none\">\n<p>To estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\n<p>There will be about 1.549 billion people in India in the year 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The population of China was about 1.39 billion in the year 2013 with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex] where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be in the year 2031? How does this compare to the population prediction we made for India in the previous example?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q891037\">Show Solution<\/span><\/p>\n<div id=\"q891037\" class=\"hidden-answer\" style=\"display: none\">\n<p>About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129476&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h3><\/h3>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXX<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\u00a0shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.<\/p>\n<p>Examine the value of $1 invested at 100% interest for 1 year compounded at various frequencies.<\/p>\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Frequency<\/th>\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>$2<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>$2.25<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>$2.441406<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>$2.613035<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>$2.714567<\/td>\n<\/tr>\n<tr>\n<td>Hourly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n<td>$2.718127<\/td>\n<\/tr>\n<tr>\n<td>Once per minute<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n<td>$2.718279<\/td>\n<\/tr>\n<tr>\n<td>Once per second<\/td>\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n<td>$2.718282<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>These values appear to be approaching a limit as <em>n<\/em>\u00a0increases without bound. In fact, as <em>n<\/em>\u00a0gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Exponential Model When the Initial Value Is Known<\/h3>\n<p>In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function <em>N<\/em>(<em>t<\/em>) representing the population <em>N<\/em>\u00a0of deer over time <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q910377\">Show Solution<\/span><\/p>\n<div id=\"q910377\" class=\"hidden-answer\" style=\"display: none\">\n<p>We let our independent variable <em>t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, <em>a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find <em>b<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}N\\left(t\\right)\\hfill & =80{b}^{t}\\hfill & \\hfill \\\\ 180\\hfill & =80{b}^{6}\\hfill & \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\frac{9}{4}\\hfill & ={b}^{6}\\hfill & \\text{Divide and write in lowest terms}.\\hfill \\\\ b\\hfill & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill & \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ b\\hfill & \\approx 1.1447 & \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Round the final answer to four places for the remainder of this section.<\/em><\/p>\n<p>The exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. Note that this exponential function models short-term growth. As the inputs get larger, the outputs will get increasingly larger resulting in the model not being useful in the long term due to extremely large output values.<\/p>\n<p>We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex] and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225444\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\" \/><\/p>\n<p class=\"wp-caption-text\">Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t\u00a0years after 2006<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population <em>N<\/em>\u00a0of wolves over time <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q222558\">Show Solution<\/span><\/p>\n<div id=\"q222558\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #ff0000;\">XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<h2>Equations of Exponential Functions<\/h2>\n<p>In the previous examples, we were given an exponential function which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, determine the constants <em>a<\/em>\u00a0and <em>b<\/em>, and evaluate the function.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given two data points, write an exponential model<\/h3>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li>If one of the data points has the form [latex]\\left(0,a\\right)[\/latex], then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{b}^{x}[\/latex], and solve for <em>b<\/em>.<\/li>\n<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. Solve the resulting system of two equations to find [latex]a[\/latex] and [latex]b[\/latex].<\/li>\n<li>Using the <em>a<\/em>\u00a0and <em>b<\/em>\u00a0found in the steps above, write the exponential function in the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Exponential Model When the Initial Value is Not Known<\/h3>\n<p>Find an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q904458\">Show Solution<\/span><\/p>\n<div id=\"q904458\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and then solve the system for <em>a<\/em>\u00a0and <em>b<\/em>.<\/p>\n<ul>\n<li>Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]<\/li>\n<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]<\/li>\n<\/ul>\n<p>Use the first equation to solve for <em>a<\/em>\u00a0in terms of <em>b<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6=ab^{-2}\\\\\\frac{6}{b^{-2}}=a\\,\\,\\,\\,\\,\\,\\,\\,\\text{Divide.}\\\\a=6b^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to rewrite the denominator.}\\end{array}[\/latex]<\/p>\n<p>Substitute <em>a<\/em>\u00a0in the second equation and solve for <em>b<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=ab^{2}\\\\1=6b^{2}b^{2}=6b^{4}\\,\\,\\,\\,\\,\\text{Substitute }a.\\\\b=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to isolate }b.\\\\b\\approx0.6389\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Round 4 decimal places.}\\end{array}[\/latex]<\/p>\n<p>Use the value of <em>b<\/em>\u00a0in the first equation to solve for the value of <em>a<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]a=6b^{2}\\approx6\\left(0.6389\\right)^{2}\\approx2.4492[\/latex]<\/p>\n<p>Thus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].<\/p>\n<p>We can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an <strong>exponential decay<\/strong> function.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225453\/CNX_Precalc_Figure_04_01_0032.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the two points [latex]\\left(1,3\\right)[\/latex] and [latex]\\left(2,4.5\\right)[\/latex], find the equation of the exponential function that passes through these two points.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q40110\">Show Solution<\/span><\/p>\n<div id=\"q40110\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2942&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Do two points always determine a unique exponential function?<\/strong><\/p>\n<p><em>Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x,\u00a0<\/em><em>which in many real world cases involves time.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the graph of an exponential function, write its equation<\/h3>\n<ol>\n<li>First, identify two points on the graph. Choose the <em>y<\/em>-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.<\/li>\n<li>If one of the data points is the <em>y-<\/em>intercept [latex]\\left(0,a\\right)[\/latex] , then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and solve for <em>b<\/em>.<\/li>\n<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. Solve the resulting system of two equations to find <em>a<\/em>\u00a0and <em>b<\/em>.<\/li>\n<li>Write the exponential function, [latex]f\\left(x\\right)=a{b}^{x}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Exponential Function Given Its Graph<\/h3>\n<p>Find an equation for the exponential function graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225455\/CNX_Precalc_Figure_04_01_0042.jpg\" alt=\"Graph of an increasing exponential function with notable points at (0, 3) and (2, 12).\" width=\"731\" height=\"369\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q440954\">Show Solution<\/span><\/p>\n<div id=\"q440954\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can choose the <em>y<\/em>-intercept of the graph, [latex]\\left(0,3\\right)[\/latex], as our first point. This gives us the initial value [latex]a=3[\/latex]. Next, choose a point on the curve some distance away from [latex]\\left(0,3\\right)[\/latex] that has integer coordinates. One such point is [latex]\\left(2,12\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llllll}y=a{b}^{x}& \\text{Write the general form of an exponential equation}. \\\\ y=3{b}^{x} & \\text{Substitute the initial value 3 for }a. \\\\ 12=3{b}^{2} & \\text{Substitute in 12 for }y\\text{ and 2 for }x. \\\\ 4={b}^{2} & \\text{Divide by 3}. \\\\ b=\\pm 2 & \\text{Take the square root}.\\end{array}[\/latex]<\/p>\n<p>Because we restrict ourselves to positive values of <em>b<\/em>, we will use <em>b\u00a0<\/em>= 2. Substitute <em>a<\/em>\u00a0and <em>b<\/em>\u00a0into standard form to yield the equation [latex]f\\left(x\\right)=3{\\left(2\\right)}^{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find an equation for the exponential function graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225457\/CNX_Precalc_Figure_04_01_0052.jpg\" alt=\"Graph of an increasing function with a labeled point at (0, sqrt(2)).\" width=\"487\" height=\"294\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q564720\">Show Solution<\/span><\/p>\n<div id=\"q564720\" class=\"hidden-answer\" style=\"display: none\">[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}[\/latex]. Answers may vary due to round-off error. The answer should be very close to [latex]1.4142{\\left(1.4142\\right)}^{x}[\/latex].<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-5617-1\">http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014. <a href=\"#return-footnote-5617-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":494375,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5617","chapter","type-chapter","status-publish","hentry"],"part":5539,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5617","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/494375"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5617\/revisions"}],"predecessor-version":[{"id":5618,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5617\/revisions\/5618"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5539"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5617\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5617"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5617"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5617"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5617"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}