{"id":2499,"date":"2023-05-10T18:23:15","date_gmt":"2023-05-10T18:23:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2499"},"modified":"2025-08-27T00:38:28","modified_gmt":"2025-08-27T00:38:28","slug":"annuities-and-loans-learn-it-1","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/annuities-and-loans-learn-it-1\/","title":{"raw":"Annuities and Loans: Learn It 1","rendered":"Annuities and Loans: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Calculate annuity balance, interest earned, and payout<\/li>\r\n\t<li>Calculate loan payments, balance, and interest using the loan formula<\/li>\r\n\t<li>Compare loans in real-world applications<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Savings Annuity<\/h2>\r\n<p>For most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a <strong>savings annuity<\/strong>. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.<\/p>\r\n<center>\r\n[caption id=\"attachment_737\" align=\"aligncenter\" width=\"415\"]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\"><img class=\"wp-image-737\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\" alt=\"Glass jar labeled &quot;Retirement.&quot; Inside are crumpled $100 bills\" width=\"415\" height=\"363\" \/><\/a> Figure 1. Most retirement funds are savings annuities[\/caption]\r\n<\/center>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>savings annuity<\/h3>\r\n<p>A <strong>savings annuity<\/strong> allows an individual to save money and earn interest on a regular basis, typically over a long period of time.<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox proTip\">\r\n<p>Annuities are usually offered by insurance companies, banks, or other financial institutions, and require the individual to make regular payments, usually on a monthly or yearly basis, for a predetermined period of time.<\/p>\r\n<\/section>\r\n<p>An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{n}\\right){{P}_{m-1}}[\/latex]<\/p>\r\n<p>For a savings annuity, we simply need to add a deposit, [latex]d[\/latex], to the account with each compounding period:<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{n}\\right){{P}_{m-1}}+d[\/latex]<\/p>\r\n<p>Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working hypothetically.<\/p>\r\n<section class=\"textbox recall\">\r\n<p>For the following example, you'll need to recall the following skills:<\/p>\r\n<ul>\r\n\t<li>The distributive property: [latex]a\\left(b+c\\right)=ab+ac[\/latex]<\/li>\r\n\t<li>Factoring out a greatest common factor: [latex]m\\left(a+b\\right) + n\\left(a+b\\right)=\\left(a+b\\right)\\left(m+n\\right)[\/latex]<\/li>\r\n\t<li>How to multiply like bases with exponents: [latex]a^{m-1}\\cdot a=a^{m-1+1}=a^{m}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox example\">Suppose we will deposit [latex]$100[\/latex] each month into an account paying [latex]6\\%[\/latex] interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.<br \/>\r\n[reveal-answer q=\"747494\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"747494\"]In this example:\r\n\r\n<ul>\r\n\t<li>[latex]r = 0.06[\/latex] ([latex]6\\%[\/latex])<\/li>\r\n\t<li>[latex]n = 12[\/latex] ([latex]12[\/latex] compounds\/deposits per year)<\/li>\r\n\t<li>[latex]d = $100[\/latex] (our deposit per month)<\/li>\r\n<\/ul>\r\n<p>Writing out the recursive equation gives<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{0.06}{12}\\right){{P}_{m-1}}+100=\\left(1.005\\right){{P}_{m-1}}+100[\/latex]<\/p>\r\n<p>Assuming we start with an empty account, we can begin using this relationship:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_0=0[\/latex]<\/p>\r\n<center>[latex]P_1=(1.005)P_0+100=100[\/latex]<\/center>\r\n<p style=\"text-align: center;\">[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Continuing this pattern, after [latex]m[\/latex] deposits, we\u2019d have saved:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[\/latex]<\/p>\r\n<p style=\"text-align: left;\">In other words, after [latex]m[\/latex] months, the first deposit will have earned compound interest for [latex]m-1[\/latex] months. The second deposit will have earned interest for [latex]m-2[\/latex] months. The last month's deposit ([latex]L[\/latex]) would have earned only one month's worth of interest. The most recent deposit will have earned no interest yet.<\/p>\r\n<p style=\"text-align: left;\">This equation leaves a lot to be desired, though \u2013 it doesn\u2019t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by [latex]1.005[\/latex]:<\/p>\r\n<p style=\"text-align: left;\">[latex]1.005{{P}_{m}}=[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]1.005\\left(100{{\\left(1.005\\right)}^{m-1}}+100{{\\left(1.005\\right)}^{m-2}}+\\cdots+100(1.005)+100\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Distributing on the right side of the equation gives<\/p>\r\n<p style=\"text-align: center;\">[latex]1.005{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}+100{{\\left(1.005\\right)}^{m-1}}+\\cdots+100{{(1.005)}^{2}}+100(1.005)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we\u2019ll line this up with like terms from our original equation, and subtract each side<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\begin{matrix}1.005{{P}_{m}}&amp;=&amp;100{{\\left(1.005\\right)}^{m}}+&amp;100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&amp;100(1.005)&amp;{}\\\\{{P}_{m}}&amp;=&amp;{}&amp;100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&amp;100(1.005)&amp;+100\\\\\\end{matrix}\\\\&amp;\\\\\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Almost all the terms cancel on the right hand side when we subtract, leaving<\/p>\r\n<p style=\"text-align: center;\">[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}-100[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Factor [latex]P_m[\/latex] out of the terms on the left side.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}P_m(1.005-1)=100{{\\left(1.005\\right)}^{m}}-100\\\\(0.005)P_m=100{{\\left(1.005\\right)}^{m}}-100\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Solve for [latex]P_m[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;0.005{{P}_{m}}=100\\left({{\\left(1.005\\right)}^{m}}-1\\right)\\\\&amp;\\\\&amp;{{P}_{m}}=\\frac{100\\left({{\\left(1.005\\right)}^{m}}-1\\right)}{0.005}\\\\\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Replacing [latex]m[\/latex] months with [latex]12t[\/latex], where [latex]t[\/latex] is measured in years, gives<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{t}}=\\frac{100\\left({{\\left(1.005\\right)}^{12t}}-1\\right)}{0.005}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Recall [latex]0.005[\/latex] was [latex]\\frac{r}{n}[\/latex] and [latex]100[\/latex] was the deposit [latex]d[\/latex]. [latex]12[\/latex] was [latex]n[\/latex], the number of deposit each year.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>Generalizing this result, we get the savings annuity formula.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>annuity formula<\/h3>\r\n<p style=\"text-align: center;\">[latex]P_{t}=\\frac{d\\left(\\left(1+\\frac{r}{n}\\right)^{nt}-1\\right)}{\\left(\\frac{r}{n}\\right)}[\/latex]<\/p>\r\n<p>&nbsp;<\/p>\r\n<ul>\r\n\t<li>[latex]P_t[\/latex] is the balance in the account after [latex]t[\/latex] years.<\/li>\r\n\t<li>[latex]d[\/latex] is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\r\n\t<li>[latex]r[\/latex] is the annual interest rate in decimal form.<\/li>\r\n\t<li>[latex]n[\/latex] is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<p>If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.<\/p>\r\n<\/div>\r\n<\/section>\r\n<p>For example, if the compounding frequency isn\u2019t stated:<\/p>\r\n<ul>\r\n\t<li>If you make your deposits every month, use monthly compounding, [latex]n = 12[\/latex].<\/li>\r\n\t<li>If you make your deposits every year, use yearly compounding, [latex]n = 1[\/latex].<\/li>\r\n\t<li>If you make your deposits every quarter, use quarterly compounding, [latex]n = 4[\/latex].<\/li>\r\n\t<li>Etc.<\/li>\r\n<\/ul>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>When do you use this?<\/strong><\/p>\r\n<p>Annuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.<\/p>\r\n<p>Compound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.<\/p>\r\n<ul>\r\n\t<li>Compound interest: One deposit<\/li>\r\n\t<li>Annuity: Many deposits.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox recall\">\r\n<p>Using the order of operations correctly is essential when using complicated formulas like the annuity formula.<\/p>\r\n<p>Remember the acronym PEMDAS to guide you: start by simplifying expressions within parentheses, then resolve any exponents. Following that, address multiplication and division operations in the order they occur from left to right, and finally, tackle addition and subtraction, again moving from left to right.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit [latex]$100[\/latex] each month into an IRA earning [latex]6\\%[\/latex] interest, how much will you have in the account after [latex]20[\/latex] years?<br \/>\r\n[reveal-answer q=\"261481\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"261481\"]In this example,\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 25%;\">[latex]d = $100[\/latex]<\/td>\r\n<td>the monthly deposit<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]r = 0.06[\/latex]<\/td>\r\n<td>[latex]6\\%[\/latex] annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]n = 12[\/latex]<\/td>\r\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]t = 20[\/latex]<\/td>\r\n<td>\u00a0we want the amount after [latex]20[\/latex] years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>Putting this into the equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{{P}_{20}}=\\frac{100\\left({{\\left(1+\\frac{0.06}{12}\\right)}^{20(12)}}-1\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left({{\\left(1.005\\right)}^{240}}-1\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left(3.310-1\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{20}}=\\frac{100\\left(2.310\\right)}{\\left(0.005\\right)}=\\$46200 \\\\\\end{align}[\/latex]<\/p>\r\n<p>The account will grow to [latex]$46,200[\/latex] after [latex]20[\/latex] years.<\/p>\r\n<p>Notice that you deposited into the account a total of [latex]$24,000[\/latex] ([latex]$100[\/latex] a month for [latex]240[\/latex] months). The difference between what you end up with and how much you put in is the interest earned. In this case it is<\/p>\r\n<p style=\"text-align: center;\">[latex]$46,200 - $24,000 = $22,200[\/latex].<\/p>\r\n<p>This example is explained in detail here.<\/p>\r\n<p>[embed]https:\/\/youtu.be\/quLg4bRpxPA[\/embed]<\/p>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Saving+Annuities.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSaving Annuities\u201d here (opens in new window).<\/a><\/p>\r\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em><br \/>\r\n[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6938[\/ohm2_question]<\/section>\r\n<h3>Solving For The Deposit Amount<\/h3>\r\n<p>Financial planners typically recommend that you have a certain amount of savings upon retirement. \u00a0If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.<\/p>\r\n<section class=\"textbox example\">You want to have [latex]$200,000[\/latex] in your account when you retire in [latex]30[\/latex] years. Your retirement account earns [latex]8\\%[\/latex] interest. How much do you need to deposit each month to meet your retirement goal?<br \/>\r\n[reveal-answer q=\"897790\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"897790\"]In this example, we\u2019re looking for [latex]d[\/latex].\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]r = 0.08[\/latex]<\/td>\r\n<td>[latex]8\\%[\/latex] annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]n = 12[\/latex]<\/td>\r\n<td>since we\u2019re depositing monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]t = 30[\/latex]<\/td>\r\n<td>[latex]30[\/latex] years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]P_{30} = $200,000[\/latex]<\/td>\r\n<td>The amount we want to have in [latex]30[\/latex] years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>In this case, we\u2019re going to have to set up the equation, and solve for [latex]d[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;200,000=\\frac{d\\left({{\\left(1+\\frac{0.08}{12}\\right)}^{30(12)}}-1\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&amp;200,000=\\frac{d\\left({{\\left(1.00667\\right)}^{360}}-1\\right)}{\\left(0.00667\\right)}\\\\&amp;200,000=d(1491.57)\\\\&amp;d=\\frac{200,000}{1491.57}=\\$134.09 \\\\\\end{align}[\/latex]<\/p>\r\n<p>So you would need to deposit [latex]$134.09[\/latex] each month to have [latex]$200,000[\/latex] in [latex]30[\/latex] years if your account earns [latex]8\\%[\/latex] interest.<\/p>\r\n<p>View the solving of this problem\u00a0in the following video.<\/p>\r\n<p>[embed]https:\/\/youtu.be\/LB6pl7o0REc[\/embed]<\/p>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Savings+annuities+-+solving+for+the+deposit.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSavings annuities - solving for the deposit\u201d here (opens in new window).<\/a><\/p>\r\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em><br \/>\r\n[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h3>Solving For Time<\/h3>\r\n<p>We can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.<\/p>\r\n<section class=\"textbox recall\">\r\n<p>In the following example, you'll need to recall that you can solve for a variable contained in an exponent by taking the log of both sides of the equation.<\/p>\r\n<p>Ex. Solve for [latex]x[\/latex] in the following equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r@{\\hfill}l}a = b^{mx} &amp;&amp; \\text{we are solving for} x\\text{, in the exponent} \\\\ log(a) = log\\left(b^{mx}\\right) &amp;&amp; \\text{ take the log of both sides} \\\\ log(a)=mx\\ast log\\left(b\\right) &amp;&amp; \\text{use the exponent property} \\\\ \\frac{log(a)}{mb}=x\u00a0 &amp;&amp; \\text{divide away all non-}x \\text{ terms to isolate } x \\\\<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">If you invest [latex]$100[\/latex] each month into an account earning [latex]3\\%[\/latex] compounded monthly, how long will it take the account to grow to [latex]$10,000[\/latex]?<br \/>\r\n[reveal-answer q=\"181207\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"181207\"]This is a savings annuity problem since we are making regular deposits into the account.\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 25%;\">[latex]d = $100[\/latex]<\/td>\r\n<td>the monthly deposit<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]r = 0.03[\/latex]<\/td>\r\n<td>[latex]3\\%[\/latex] annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]n = 12[\/latex]<\/td>\r\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>We don\u2019t know [latex]t[\/latex], but we want[latex]P_t[\/latex] to be [latex]$10,000[\/latex].<\/p>\r\n<p>Putting this into the equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]10,000=\\frac{100\\left({{\\left(1+\\frac{0.03}{12}\\right)}^{t(12)}}-1\\right)}{\\left(\\frac{0.03}{12}\\right)}[\/latex]<\/p>\r\n<p>Simplifying the fractions a bit<\/p>\r\n<p style=\"text-align: center;\">[latex]10,000=\\frac{100\\left({{\\left(1.0025\\right)}^{12t}}-1\\right)}{0.0025}[\/latex]<\/p>\r\n<p>We want to isolate the exponential term, [latex]1.0025^{12t}[\/latex], so multiply both sides by [latex]0.0025[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} 25 = 100\\left({\\left(1.0025\\right)}^{12t} - 1\\right) &amp; \\text{Divide both sides by } 100 \\\\ 0.25 = {\\left(1.0025\\right)}^{12t} - 1 &amp; \\text{Add } 1 \\text{ to both sides} \\\\ 1.25 = {\\left(1.0025\\right)}^{12t} &amp; \\text{Now take the log of both sides} \\\\ \\log(1.25) = \\log\\left({\\left(1.0025\\right)}^{12t}\\right) &amp; \\text{Use the exponent property of logs} \\\\ \\log(1.25) = 12t\\log(1.0025) &amp; \\text{Divide by } 12\\log(1.0025) \\\\ \\frac{\\log(1.25)}{12\\log(1.0025)} = t &amp; \\text{Approximating to a decimal} \\\\ t = 7.447 \\\\ \\end{array} [\/latex]<\/p>\r\n<p>It will take about [latex]7.447[\/latex] years to grow the account to [latex]$10,000[\/latex].<\/p>\r\n<p>This example is demonstrated here:<\/p>\r\n<p>[embed]https:\/\/youtu.be\/F3QVyswCzRo[\/embed]<\/p>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Use+logs+to+find+the+time+it+takes+an+annuity+to+grow.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUse logs to find the time it takes an annuity to grow\u201d here (opens in new window).<\/a><\/p>\r\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em><br \/>\r\n[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate annuity balance, interest earned, and payout<\/li>\n<li>Calculate loan payments, balance, and interest using the loan formula<\/li>\n<li>Compare loans in real-world applications<\/li>\n<\/ul>\n<\/section>\n<h2>Savings Annuity<\/h2>\n<p>For most of us, we aren\u2019t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a <strong>savings annuity<\/strong>. Most retirement plans like 401k plans or IRA plans are examples of savings annuities.<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_737\" aria-describedby=\"caption-attachment-737\" style=\"width: 415px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-737\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02171735\/7027606047_cac49c3b79_z.jpg\" alt=\"Glass jar labeled &quot;Retirement.&quot; Inside are crumpled $100 bills\" width=\"415\" height=\"363\" \/><\/a><figcaption id=\"caption-attachment-737\" class=\"wp-caption-text\">Figure 1. Most retirement funds are savings annuities<\/figcaption><\/figure>\n<\/div>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>savings annuity<\/h3>\n<p>A <strong>savings annuity<\/strong> allows an individual to save money and earn interest on a regular basis, typically over a long period of time.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\">\n<p>Annuities are usually offered by insurance companies, banks, or other financial institutions, and require the individual to make regular payments, usually on a monthly or yearly basis, for a predetermined period of time.<\/p>\n<\/section>\n<p>An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{n}\\right){{P}_{m-1}}[\/latex]<\/p>\n<p>For a savings annuity, we simply need to add a deposit, [latex]d[\/latex], to the account with each compounding period:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{r}{n}\\right){{P}_{m-1}}+d[\/latex]<\/p>\n<p>Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working hypothetically.<\/p>\n<section class=\"textbox recall\">\n<p>For the following example, you&#8217;ll need to recall the following skills:<\/p>\n<ul>\n<li>The distributive property: [latex]a\\left(b+c\\right)=ab+ac[\/latex]<\/li>\n<li>Factoring out a greatest common factor: [latex]m\\left(a+b\\right) + n\\left(a+b\\right)=\\left(a+b\\right)\\left(m+n\\right)[\/latex]<\/li>\n<li>How to multiply like bases with exponents: [latex]a^{m-1}\\cdot a=a^{m-1+1}=a^{m}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">Suppose we will deposit [latex]$100[\/latex] each month into an account paying [latex]6\\%[\/latex] interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q747494\">Show Solution<\/button><\/p>\n<div id=\"q747494\" class=\"hidden-answer\" style=\"display: none\">In this example:<\/p>\n<ul>\n<li>[latex]r = 0.06[\/latex] ([latex]6\\%[\/latex])<\/li>\n<li>[latex]n = 12[\/latex] ([latex]12[\/latex] compounds\/deposits per year)<\/li>\n<li>[latex]d = $100[\/latex] (our deposit per month)<\/li>\n<\/ul>\n<p>Writing out the recursive equation gives<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}=\\left(1+\\frac{0.06}{12}\\right){{P}_{m-1}}+100=\\left(1.005\\right){{P}_{m-1}}+100[\/latex]<\/p>\n<p>Assuming we start with an empty account, we can begin using this relationship:<\/p>\n<p style=\"text-align: center;\">[latex]P_0=0[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]P_1=(1.005)P_0+100=100[\/latex]<\/div>\n<p style=\"text-align: center;\">[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[\/latex]<\/p>\n<p style=\"text-align: left;\">Continuing this pattern, after [latex]m[\/latex] deposits, we\u2019d have saved:<\/p>\n<p style=\"text-align: center;\">[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[\/latex]<\/p>\n<p style=\"text-align: left;\">In other words, after [latex]m[\/latex] months, the first deposit will have earned compound interest for [latex]m-1[\/latex] months. The second deposit will have earned interest for [latex]m-2[\/latex] months. The last month&#8217;s deposit ([latex]L[\/latex]) would have earned only one month&#8217;s worth of interest. The most recent deposit will have earned no interest yet.<\/p>\n<p style=\"text-align: left;\">This equation leaves a lot to be desired, though \u2013 it doesn\u2019t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by [latex]1.005[\/latex]:<\/p>\n<p style=\"text-align: left;\">[latex]1.005{{P}_{m}}=[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]1.005\\left(100{{\\left(1.005\\right)}^{m-1}}+100{{\\left(1.005\\right)}^{m-2}}+\\cdots+100(1.005)+100\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Distributing on the right side of the equation gives<\/p>\n<p style=\"text-align: center;\">[latex]1.005{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}+100{{\\left(1.005\\right)}^{m-1}}+\\cdots+100{{(1.005)}^{2}}+100(1.005)[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we\u2019ll line this up with like terms from our original equation, and subtract each side<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\begin{matrix}1.005{{P}_{m}}&=&100{{\\left(1.005\\right)}^{m}}+&100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&100(1.005)&{}\\\\{{P}_{m}}&=&{}&100{{\\left(1.005\\right)}^{m-1}}+\\cdots+&100(1.005)&+100\\\\\\end{matrix}\\\\&\\\\\\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Almost all the terms cancel on the right hand side when we subtract, leaving<\/p>\n<p style=\"text-align: center;\">[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\\left(1.005\\right)}^{m}}-100[\/latex]<\/p>\n<p style=\"text-align: left;\">Factor [latex]P_m[\/latex] out of the terms on the left side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}P_m(1.005-1)=100{{\\left(1.005\\right)}^{m}}-100\\\\(0.005)P_m=100{{\\left(1.005\\right)}^{m}}-100\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Solve for [latex]P_m[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&0.005{{P}_{m}}=100\\left({{\\left(1.005\\right)}^{m}}-1\\right)\\\\&\\\\&{{P}_{m}}=\\frac{100\\left({{\\left(1.005\\right)}^{m}}-1\\right)}{0.005}\\\\\\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Replacing [latex]m[\/latex] months with [latex]12t[\/latex], where [latex]t[\/latex] is measured in years, gives<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{t}}=\\frac{100\\left({{\\left(1.005\\right)}^{12t}}-1\\right)}{0.005}[\/latex]<\/p>\n<p style=\"text-align: left;\">Recall [latex]0.005[\/latex] was [latex]\\frac{r}{n}[\/latex] and [latex]100[\/latex] was the deposit [latex]d[\/latex]. [latex]12[\/latex] was [latex]n[\/latex], the number of deposit each year.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>Generalizing this result, we get the savings annuity formula.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>annuity formula<\/h3>\n<p style=\"text-align: center;\">[latex]P_{t}=\\frac{d\\left(\\left(1+\\frac{r}{n}\\right)^{nt}-1\\right)}{\\left(\\frac{r}{n}\\right)}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<ul>\n<li>[latex]P_t[\/latex] is the balance in the account after [latex]t[\/latex] years.<\/li>\n<li>[latex]d[\/latex] is the regular deposit (the amount you deposit each year, each month, etc.)<\/li>\n<li>[latex]r[\/latex] is the annual interest rate in decimal form.<\/li>\n<li>[latex]n[\/latex] is the number of compounding periods in one year.<\/li>\n<\/ul>\n<p>If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.<\/p>\n<\/div>\n<\/section>\n<p>For example, if the compounding frequency isn\u2019t stated:<\/p>\n<ul>\n<li>If you make your deposits every month, use monthly compounding, [latex]n = 12[\/latex].<\/li>\n<li>If you make your deposits every year, use yearly compounding, [latex]n = 1[\/latex].<\/li>\n<li>If you make your deposits every quarter, use quarterly compounding, [latex]n = 4[\/latex].<\/li>\n<li>Etc.<\/li>\n<\/ul>\n<section class=\"textbox questionHelp\">\n<p><strong>When do you use this?<\/strong><\/p>\n<p>Annuities assume that you put money in the account <strong>on a regular schedule<\/strong> (every month, year, quarter, etc.) and let it sit there earning interest.<\/p>\n<p>Compound interest assumes that you put money in the account <strong>once<\/strong> and let it sit there earning interest.<\/p>\n<ul>\n<li>Compound interest: One deposit<\/li>\n<li>Annuity: Many deposits.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\">\n<p>Using the order of operations correctly is essential when using complicated formulas like the annuity formula.<\/p>\n<p>Remember the acronym PEMDAS to guide you: start by simplifying expressions within parentheses, then resolve any exponents. Following that, address multiplication and division operations in the order they occur from left to right, and finally, tackle addition and subtraction, again moving from left to right.<\/p>\n<\/section>\n<section class=\"textbox example\">A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit [latex]$100[\/latex] each month into an IRA earning [latex]6\\%[\/latex] interest, how much will you have in the account after [latex]20[\/latex] years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q261481\">Show Solution<\/button><\/p>\n<div id=\"q261481\" class=\"hidden-answer\" style=\"display: none\">In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"width: 25%;\">[latex]d = $100[\/latex]<\/td>\n<td>the monthly deposit<\/td>\n<\/tr>\n<tr>\n<td>[latex]r = 0.06[\/latex]<\/td>\n<td>[latex]6\\%[\/latex] annual rate<\/td>\n<\/tr>\n<tr>\n<td>[latex]n = 12[\/latex]<\/td>\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td>[latex]t = 20[\/latex]<\/td>\n<td>\u00a0we want the amount after [latex]20[\/latex] years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>Putting this into the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{{P}_{20}}=\\frac{100\\left({{\\left(1+\\frac{0.06}{12}\\right)}^{20(12)}}-1\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&{{P}_{20}}=\\frac{100\\left({{\\left(1.005\\right)}^{240}}-1\\right)}{\\left(0.005\\right)}\\\\&{{P}_{20}}=\\frac{100\\left(3.310-1\\right)}{\\left(0.005\\right)}\\\\&{{P}_{20}}=\\frac{100\\left(2.310\\right)}{\\left(0.005\\right)}=\\$46200 \\\\\\end{align}[\/latex]<\/p>\n<p>The account will grow to [latex]$46,200[\/latex] after [latex]20[\/latex] years.<\/p>\n<p>Notice that you deposited into the account a total of [latex]$24,000[\/latex] ([latex]$100[\/latex] a month for [latex]240[\/latex] months). The difference between what you end up with and how much you put in is the interest earned. In this case it is<\/p>\n<p style=\"text-align: center;\">[latex]$46,200 - $24,000 = $22,200[\/latex].<\/p>\n<p>This example is explained in detail here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Saving Annuities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/quLg4bRpxPA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Saving+Annuities.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSaving Annuities\u201d here (opens in new window).<\/a><\/p>\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6938\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6938&theme=lumen&iframe_resize_id=ohm6938&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Solving For The Deposit Amount<\/h3>\n<p>Financial planners typically recommend that you have a certain amount of savings upon retirement. \u00a0If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.<\/p>\n<section class=\"textbox example\">You want to have [latex]$200,000[\/latex] in your account when you retire in [latex]30[\/latex] years. Your retirement account earns [latex]8\\%[\/latex] interest. How much do you need to deposit each month to meet your retirement goal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q897790\">Show Solution<\/button><\/p>\n<div id=\"q897790\" class=\"hidden-answer\" style=\"display: none\">In this example, we\u2019re looking for [latex]d[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]r = 0.08[\/latex]<\/td>\n<td>[latex]8\\%[\/latex] annual rate<\/td>\n<\/tr>\n<tr>\n<td>[latex]n = 12[\/latex]<\/td>\n<td>since we\u2019re depositing monthly<\/td>\n<\/tr>\n<tr>\n<td>[latex]t = 30[\/latex]<\/td>\n<td>[latex]30[\/latex] years<\/td>\n<\/tr>\n<tr>\n<td>[latex]P_{30} = $200,000[\/latex]<\/td>\n<td>The amount we want to have in [latex]30[\/latex] years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>In this case, we\u2019re going to have to set up the equation, and solve for [latex]d[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&200,000=\\frac{d\\left({{\\left(1+\\frac{0.08}{12}\\right)}^{30(12)}}-1\\right)}{\\left(\\frac{0.08}{12}\\right)}\\\\&200,000=\\frac{d\\left({{\\left(1.00667\\right)}^{360}}-1\\right)}{\\left(0.00667\\right)}\\\\&200,000=d(1491.57)\\\\&d=\\frac{200,000}{1491.57}=\\$134.09 \\\\\\end{align}[\/latex]<\/p>\n<p>So you would need to deposit [latex]$134.09[\/latex] each month to have [latex]$200,000[\/latex] in [latex]30[\/latex] years if your account earns [latex]8\\%[\/latex] interest.<\/p>\n<p>View the solving of this problem\u00a0in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Savings annuities - solving for the deposit\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/LB6pl7o0REc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Savings+annuities+-+solving+for+the+deposit.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cSavings annuities &#8211; solving for the deposit\u201d here (opens in new window).<\/a><\/p>\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em>\n<\/div>\n<\/div>\n<\/section>\n<h3>Solving For Time<\/h3>\n<p>We can solve the annuities formula for time, like we did the compounding interest formula, by using logarithms. In the next example we will work through how this is done.<\/p>\n<section class=\"textbox recall\">\n<p>In the following example, you&#8217;ll need to recall that you can solve for a variable contained in an exponent by taking the log of both sides of the equation.<\/p>\n<p>Ex. Solve for [latex]x[\/latex] in the following equation<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r@{\\hfill}l}a = b^{mx} && \\text{we are solving for} x\\text{, in the exponent} \\\\ log(a) = log\\left(b^{mx}\\right) && \\text{ take the log of both sides} \\\\ log(a)=mx\\ast log\\left(b\\right) && \\text{use the exponent property} \\\\ \\frac{log(a)}{mb}=x\u00a0 && \\text{divide away all non-}x \\text{ terms to isolate } x \\\\<br \/>  \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">If you invest [latex]$100[\/latex] each month into an account earning [latex]3\\%[\/latex] compounded monthly, how long will it take the account to grow to [latex]$10,000[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q181207\">Show Solution<\/button><\/p>\n<div id=\"q181207\" class=\"hidden-answer\" style=\"display: none\">This is a savings annuity problem since we are making regular deposits into the account.<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"width: 25%;\">[latex]d = $100[\/latex]<\/td>\n<td>the monthly deposit<\/td>\n<\/tr>\n<tr>\n<td>[latex]r = 0.03[\/latex]<\/td>\n<td>[latex]3\\%[\/latex] annual rate<\/td>\n<\/tr>\n<tr>\n<td>[latex]n = 12[\/latex]<\/td>\n<td>since we\u2019re doing monthly deposits, we\u2019ll compound monthly<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>We don\u2019t know [latex]t[\/latex], but we want[latex]P_t[\/latex] to be [latex]$10,000[\/latex].<\/p>\n<p>Putting this into the equation:<\/p>\n<p style=\"text-align: center;\">[latex]10,000=\\frac{100\\left({{\\left(1+\\frac{0.03}{12}\\right)}^{t(12)}}-1\\right)}{\\left(\\frac{0.03}{12}\\right)}[\/latex]<\/p>\n<p>Simplifying the fractions a bit<\/p>\n<p style=\"text-align: center;\">[latex]10,000=\\frac{100\\left({{\\left(1.0025\\right)}^{12t}}-1\\right)}{0.0025}[\/latex]<\/p>\n<p>We want to isolate the exponential term, [latex]1.0025^{12t}[\/latex], so multiply both sides by [latex]0.0025[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} 25 = 100\\left({\\left(1.0025\\right)}^{12t} - 1\\right) & \\text{Divide both sides by } 100 \\\\ 0.25 = {\\left(1.0025\\right)}^{12t} - 1 & \\text{Add } 1 \\text{ to both sides} \\\\ 1.25 = {\\left(1.0025\\right)}^{12t} & \\text{Now take the log of both sides} \\\\ \\log(1.25) = \\log\\left({\\left(1.0025\\right)}^{12t}\\right) & \\text{Use the exponent property of logs} \\\\ \\log(1.25) = 12t\\log(1.0025) & \\text{Divide by } 12\\log(1.0025) \\\\ \\frac{\\log(1.25)}{12\\log(1.0025)} = t & \\text{Approximating to a decimal} \\\\ t = 7.447 \\\\ \\end{array}[\/latex]<\/p>\n<p>It will take about [latex]7.447[\/latex] years to grow the account to [latex]$10,000[\/latex].<\/p>\n<p>This example is demonstrated here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Use logs to find the time it takes an annuity to grow\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/F3QVyswCzRo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Use+logs+to+find+the+time+it+takes+an+annuity+to+grow.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUse logs to find the time it takes an annuity to grow\u201d here (opens in new window).<\/a><\/p>\n<p><em>Note: This video uses [latex]k[\/latex] for [latex]n[\/latex] and [latex]N[\/latex] for [latex]t[\/latex].<\/em>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":26,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":89,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2499"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":63,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2499\/revisions"}],"predecessor-version":[{"id":15750,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2499\/revisions\/15750"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/89"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2499\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2499"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2499"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2499"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2499"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}