{"id":2448,"date":"2023-05-10T16:23:08","date_gmt":"2023-05-10T16:23:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2448"},"modified":"2024-10-18T20:55:16","modified_gmt":"2024-10-18T20:55:16","slug":"simple-and-compound-interest-learn-it-3","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/simple-and-compound-interest-learn-it-3\/","title":{"raw":"Simple and Compound Interest: Learn It 3","rendered":"Simple and Compound Interest: Learn It 3"},"content":{"raw":"<h2>Compound Interest<\/h2>\r\n<p>We want to simplify the process for calculating compounding because creating a table like the one on the previous page is time-consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if [latex]P_m[\/latex] represents the amount of money after [latex]m[\/latex] months, then we could write the recursive equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_0 = $1000[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_m= (1+0.0025)P_{m-1}[\/latex]<\/p>\r\n<p>You may recognize this as the recursive form of exponential growth.<\/p>\r\n<section class=\"textbox recall\">\r\n<p><strong>recursive growth<\/strong><\/p>\r\n<p>Recall the underlying process of recursive growth. From a starting amount, [latex]P_0[\/latex], each subsequent amount, [latex]P_m[\/latex], grows in proportion to itself, [latex]P_{m-1}[\/latex], at some rate [latex]r[\/latex].<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]P_m=P_{m-1}+r\\cdot P_{m-1}[\/latex]<\/p>\r\n<p>Factoring out the [latex]P_{m-1}[\/latex] from each term on the right-hand side<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]P_m=(1+r)\\cdot P_{m-1}[\/latex].<\/p>\r\n<\/section>\r\n<p>In the example below, we'll build an explicit equation for the growth.<\/p>\r\n<section class=\"textbox recall\">\r\n<p><strong>Multiplying terms containing exponents<\/strong><\/p>\r\n<p>In the example below, you'll need to use the rules for multiplying like bases containing exponents<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]a^{m}a^{n}=a^{m+n}[\/latex]<\/p>\r\n<p>That is, when multiplying like bases, we add the exponents.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Build an explicit equation for the growth of [latex]$1000[\/latex] deposited in a bank account offering [latex]3\\%[\/latex] interest, compounded monthly.<br \/>\r\n[reveal-answer q=\"530288\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"530288\"]\r\n\r\n<ul>\r\n\t<li>[latex]P_0 = $1000[\/latex]<\/li>\r\n\t<li>[latex]P_1 = 1.0025P_0 = 1.0025 (1000)[\/latex]<\/li>\r\n\t<li>[latex]P_2 = 1.0025P_1 = 1.0025 (1.0025 (1000)) = 1.0025^2(1000)[\/latex]<\/li>\r\n\t<li>[latex]P_3 = 1.0025P_2 = 1.0025 (1.0025^2(1000)) = 1.0025^3(1000)[\/latex]<\/li>\r\n\t<li>[latex]P_4 = 1.0025P_3 = 1.0025 (1.0025^3(1000)) = 1.0025^4(1000)[\/latex]<\/li>\r\n<\/ul>\r\n<p>Observing a pattern, we could conclude<\/p>\r\n<p style=\"text-align: center;\">[latex]P_m = (1.0025)^m($1000)[\/latex]<\/p>\r\n<p><br \/>\r\nNotice that the [latex]$1000[\/latex] in the equation was [latex]P_0[\/latex], the starting amount. We found [latex]1.0025[\/latex] by adding one to the growth rate divided by [latex]12[\/latex], since we were compounding [latex]12[\/latex] times per year.<\/p>\r\n<p>Generalizing our result, we could write<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{m}}={{P}_{0}}{{\\left(1+\\frac{r}{n}\\right)}^{m}}[\/latex]<\/p>\r\n<p>In this formula:<\/p>\r\n<ul>\r\n\t<li>[latex]m[\/latex] is the number of compounding periods (months in our example)<\/li>\r\n\t<li>[latex]r[\/latex] is the annual interest rate<\/li>\r\n\t<li>[latex]n[\/latex] is the number of compounds per year.<\/li>\r\n<\/ul>\r\n<p>View this video for a walkthrough of this example.<\/p>\r\n<p>[embed]https:\/\/youtu.be\/xuQTFmP9nNg[\/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\/Compound+Interest.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cCompound Interest\u201d here (opens in new window).<\/a><\/p>\r\n<p><em>Note: This video uses [latex]k[\/latex] as the variable instead of [latex]n[\/latex].<\/em><br \/>\r\n[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If [latex]t[\/latex] is the number of years, then [latex]m = nt[\/latex]. Making this change gives us the standard formula for compound interest.<\/p>\r\n<section class=\"textbox questionHelp\">How did we get [latex]m = nt[\/latex]?<br \/>\r\n<br \/>\r\nRecall that [latex]m[\/latex] represents the number of compounding periods that an investment remains in the account, and [latex]n[\/latex] represents the number of times per year that your interest is compounded. If your deposit earns interest compounded monthly, then [latex]n = 12[\/latex]. If you leave the deposit in for [latex]1[\/latex] year, then [latex]m = 12[\/latex]. But if you leave the deposit in for [latex]2[\/latex] years, then [latex]m = 2*12 = 24[\/latex]. Looking at that another way, [latex]m = t(\\text{years}) * n[\/latex].<\/section>\r\n<section class=\"textbox example\">\r\n<p>An investment of [latex]$1000[\/latex] earning interest of [latex]4\\%[\/latex] compounded quarterly ([latex]4[\/latex] times per year) is left in the account for [latex]3[\/latex] years.<\/p>\r\n<p>We have [latex]4[\/latex] compounding periods per year, so [latex]n = 4[\/latex].<\/p>\r\n<p>If we leave our money in for [latex]1[\/latex] year, the number of compounding periods is [latex]1*4: m=4[\/latex].<\/p>\r\n<p>If we leave our money in for [latex]3[\/latex] years, [latex]m = 3*4[\/latex], or [latex]12[\/latex].<\/p>\r\n<p>Knowing that investments are usually left to grow over years than over a number of compounding periods, we'll adjust the formula slightly and just write [latex]nt[\/latex]. This will make it easier to load the formula into a spreadsheet.<\/p>\r\n<\/section>\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>compound interest<\/h3>\r\n<p style=\"text-align: center;\">[latex]P_{t}=P_{0}\\left(1+\\frac{r}{n}\\right)^{nt}[\/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]P_0[\/latex] is the starting balance of the account (also called initial deposit, or principal)<\/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\r\n\r\n<ul>\r\n\t<li>If the compounding is done annually (once a year), [latex]n=1[\/latex].<\/li>\r\n\t<li>If the compounding is done quarterly, [latex]n=4[\/latex].<\/li>\r\n\t<li>If the compounding is done monthly, [latex]n=12[\/latex].<\/li>\r\n\t<li>If the compounding is done daily, [latex]n=365[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox questionHelp\"><strong>The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.\u00a0<\/strong><\/section>\r\n<p>In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after [latex]20[\/latex] years.<\/p>\r\n<section class=\"textbox proTip\">\r\n<p><strong>Don't forget to convert percent to a decimal<\/strong><\/p>\r\n<p>Usually, in order to perform calculations on a number expressed in percent form, you\u2019ll need to convert it to decimal form. The rate [latex]r[\/latex] in interest formulas must be converted from percent to decimal form before you use the formula.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit [latex]$3000[\/latex] in a CD paying [latex]6\\%[\/latex] interest, compounded monthly. How much will you have in the account after [latex]20[\/latex] years?[reveal-answer q=\"788137\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"788137\"]In this example,\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 25%;\">[latex]P_0 = $3000[\/latex]<\/td>\r\n<td>the initial 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>[latex]12[\/latex] months in [latex]1[\/latex] year<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]t = 20[\/latex]<\/td>\r\n<td>\u00a0since we\u2019re looking for how much we\u2019ll have after [latex]20[\/latex] years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>So [latex]{{P}_{20}}=3000{{\\left(1+\\frac{0.06}{12}\\right)}^{20\\times12}}=\\$9930.61[\/latex] (round your answer to the nearest penny)<\/p>\r\n<p>A video walkthrough of this example problem\u00a0is available below.<\/p>\r\n<p>https:\/\/youtu.be\/8NazxAjhpJw<\/p>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Compound+interest+CD+example.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cCompound interest CD example\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><\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6936[\/ohm2_question]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6937[\/ohm2_question]<\/section>\r\n<h3>Solving For Time<\/h3>\r\n<section class=\"textbox connectIt\">\r\n<p>Note: This section assumes you\u2019ve covered solving exponential equations using logarithms, either in prior classes or in the growth models module.<\/p>\r\n<\/section>\r\n<p>Often we are interested in how long it will take to accumulate money or how long we\u2019d need to extend a loan to bring payments down to a reasonable level.<\/p>\r\n<section class=\"textbox example\">If you invest [latex]$2000[\/latex] at [latex]6\\%[\/latex] compounded monthly, how long will it take the account to double in value?<br \/>\r\n[reveal-answer q=\"610603\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"610603\"]This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,\r\n\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]P_0 = $2000[\/latex]<\/td>\r\n<td>the initial 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>[latex]12[\/latex] months in [latex]1[\/latex] year<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>&nbsp;<\/p>\r\n<p>So our general equation is [latex]{{P}_{t}}=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{t\\times12}}[\/latex]. We also know that we want our ending amount to be double of [latex]$2000[\/latex], which is [latex]$4000[\/latex], so we\u2019re looking for [latex]t[\/latex] so that [latex]P_t = 4000[\/latex]. To solve this, we set our equation for [latex]P_t[\/latex] equal to [latex]4000[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r@{\\hfill}l}<br \/>\r\n4000=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{t\\times12}} &amp;&amp;&amp; \\text{Divide both sides by } 2000 \\\\<br \/>\r\n2={{\\left(1.005\\right)}^{12t}} &amp;&amp;&amp; \\text{To solve for the exponent, take the log of both sides} \\\\<br \/>\r\n\\log\\left(2\\right)=\\log\\left({{\\left(1.005\\right)}^{12t}}\\right) &amp;&amp;&amp; \\text{Use the exponent property of logs on the right side} \\\\<br \/>\r\n\\log\\left(2\\right)=12t\\log\\left(1.005\\right) &amp;&amp;&amp; \\text{Now we can divide both sides by } 12\\text{log}(1.005) \\\\<br \/>\r\n\\frac{\\log\\left(2\\right)}{12\\log\\left(1.005\\right)}=t &amp;&amp;&amp; \\text{Approximating this to a decimal} \\\\<br \/>\r\nt = 11.581 &amp;&amp;&amp;<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>It will take about [latex]11.581[\/latex] years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded [latex]\\text{log}(2)[\/latex] to [latex]0.301[\/latex] and [latex]\\text{log}(1.005)[\/latex] to [latex]0.00217[\/latex], then your final answer would have been about [latex]11.577[\/latex] years.<\/p>\r\n<p>View this video for a walkthrough of this example.<\/p>\r\n<p>[embed]https:\/\/youtu.be\/zHRTxtFiyxc[\/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\/Find+doubling+time+for+compound+interest.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cFind doubling time for compound interest\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":"<h2>Compound Interest<\/h2>\n<p>We want to simplify the process for calculating compounding because creating a table like the one on the previous page is time-consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if [latex]P_m[\/latex] represents the amount of money after [latex]m[\/latex] months, then we could write the recursive equation:<\/p>\n<p style=\"text-align: center;\">[latex]P_0 = $1000[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_m= (1+0.0025)P_{m-1}[\/latex]<\/p>\n<p>You may recognize this as the recursive form of exponential growth.<\/p>\n<section class=\"textbox recall\">\n<p><strong>recursive growth<\/strong><\/p>\n<p>Recall the underlying process of recursive growth. From a starting amount, [latex]P_0[\/latex], each subsequent amount, [latex]P_m[\/latex], grows in proportion to itself, [latex]P_{m-1}[\/latex], at some rate [latex]r[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">[latex]P_m=P_{m-1}+r\\cdot P_{m-1}[\/latex]<\/p>\n<p>Factoring out the [latex]P_{m-1}[\/latex] from each term on the right-hand side<\/p>\n<p style=\"padding-left: 30px;\">[latex]P_m=(1+r)\\cdot P_{m-1}[\/latex].<\/p>\n<\/section>\n<p>In the example below, we&#8217;ll build an explicit equation for the growth.<\/p>\n<section class=\"textbox recall\">\n<p><strong>Multiplying terms containing exponents<\/strong><\/p>\n<p>In the example below, you&#8217;ll need to use the rules for multiplying like bases containing exponents<\/p>\n<p style=\"padding-left: 30px;\">[latex]a^{m}a^{n}=a^{m+n}[\/latex]<\/p>\n<p>That is, when multiplying like bases, we add the exponents.<\/p>\n<\/section>\n<section class=\"textbox example\">Build an explicit equation for the growth of [latex]$1000[\/latex] deposited in a bank account offering [latex]3\\%[\/latex] interest, compounded monthly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q530288\">Show Solution<\/button><\/p>\n<div id=\"q530288\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>[latex]P_0 = $1000[\/latex]<\/li>\n<li>[latex]P_1 = 1.0025P_0 = 1.0025 (1000)[\/latex]<\/li>\n<li>[latex]P_2 = 1.0025P_1 = 1.0025 (1.0025 (1000)) = 1.0025^2(1000)[\/latex]<\/li>\n<li>[latex]P_3 = 1.0025P_2 = 1.0025 (1.0025^2(1000)) = 1.0025^3(1000)[\/latex]<\/li>\n<li>[latex]P_4 = 1.0025P_3 = 1.0025 (1.0025^3(1000)) = 1.0025^4(1000)[\/latex]<\/li>\n<\/ul>\n<p>Observing a pattern, we could conclude<\/p>\n<p style=\"text-align: center;\">[latex]P_m = (1.0025)^m($1000)[\/latex]<\/p>\n<p>\nNotice that the [latex]$1000[\/latex] in the equation was [latex]P_0[\/latex], the starting amount. We found [latex]1.0025[\/latex] by adding one to the growth rate divided by [latex]12[\/latex], since we were compounding [latex]12[\/latex] times per year.<\/p>\n<p>Generalizing our result, we could write<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{m}}={{P}_{0}}{{\\left(1+\\frac{r}{n}\\right)}^{m}}[\/latex]<\/p>\n<p>In this formula:<\/p>\n<ul>\n<li>[latex]m[\/latex] is the number of compounding periods (months in our example)<\/li>\n<li>[latex]r[\/latex] is the annual interest rate<\/li>\n<li>[latex]n[\/latex] is the number of compounds per year.<\/li>\n<\/ul>\n<p>View this video for a walkthrough of this example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Compound Interest\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xuQTFmP9nNg?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\/Compound+Interest.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cCompound Interest\u201d here (opens in new window).<\/a><\/p>\n<p><em>Note: This video uses [latex]k[\/latex] as the variable instead of [latex]n[\/latex].<\/em>\n<\/div>\n<\/div>\n<\/section>\n<p>While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If [latex]t[\/latex] is the number of years, then [latex]m = nt[\/latex]. Making this change gives us the standard formula for compound interest.<\/p>\n<section class=\"textbox questionHelp\">How did we get [latex]m = nt[\/latex]?<\/p>\n<p>Recall that [latex]m[\/latex] represents the number of compounding periods that an investment remains in the account, and [latex]n[\/latex] represents the number of times per year that your interest is compounded. If your deposit earns interest compounded monthly, then [latex]n = 12[\/latex]. If you leave the deposit in for [latex]1[\/latex] year, then [latex]m = 12[\/latex]. But if you leave the deposit in for [latex]2[\/latex] years, then [latex]m = 2*12 = 24[\/latex]. Looking at that another way, [latex]m = t(\\text{years}) * n[\/latex].<\/section>\n<section class=\"textbox example\">\n<p>An investment of [latex]$1000[\/latex] earning interest of [latex]4\\%[\/latex] compounded quarterly ([latex]4[\/latex] times per year) is left in the account for [latex]3[\/latex] years.<\/p>\n<p>We have [latex]4[\/latex] compounding periods per year, so [latex]n = 4[\/latex].<\/p>\n<p>If we leave our money in for [latex]1[\/latex] year, the number of compounding periods is [latex]1*4: m=4[\/latex].<\/p>\n<p>If we leave our money in for [latex]3[\/latex] years, [latex]m = 3*4[\/latex], or [latex]12[\/latex].<\/p>\n<p>Knowing that investments are usually left to grow over years than over a number of compounding periods, we&#8217;ll adjust the formula slightly and just write [latex]nt[\/latex]. This will make it easier to load the formula into a spreadsheet.<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>compound interest<\/h3>\n<p style=\"text-align: center;\">[latex]P_{t}=P_{0}\\left(1+\\frac{r}{n}\\right)^{nt}[\/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]P_0[\/latex] is the starting balance of the account (also called initial deposit, or principal)<\/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\n<ul>\n<li>If the compounding is done annually (once a year), [latex]n=1[\/latex].<\/li>\n<li>If the compounding is done quarterly, [latex]n=4[\/latex].<\/li>\n<li>If the compounding is done monthly, [latex]n=12[\/latex].<\/li>\n<li>If the compounding is done daily, [latex]n=365[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\"><strong>The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.\u00a0<\/strong><\/section>\n<p>In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after [latex]20[\/latex] years.<\/p>\n<section class=\"textbox proTip\">\n<p><strong>Don&#8217;t forget to convert percent to a decimal<\/strong><\/p>\n<p>Usually, in order to perform calculations on a number expressed in percent form, you\u2019ll need to convert it to decimal form. The rate [latex]r[\/latex] in interest formulas must be converted from percent to decimal form before you use the formula.<\/p>\n<\/section>\n<section class=\"textbox example\">A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit [latex]$3000[\/latex] in a CD paying [latex]6\\%[\/latex] interest, compounded monthly. 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=\"q788137\">Show Solution<\/button><\/p>\n<div id=\"q788137\" class=\"hidden-answer\" style=\"display: none\">In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"width: 25%;\">[latex]P_0 = $3000[\/latex]<\/td>\n<td>the initial 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>[latex]12[\/latex] months in [latex]1[\/latex] year<\/td>\n<\/tr>\n<tr>\n<td>[latex]t = 20[\/latex]<\/td>\n<td>\u00a0since we\u2019re looking for how much we\u2019ll have after [latex]20[\/latex] years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>So [latex]{{P}_{20}}=3000{{\\left(1+\\frac{0.06}{12}\\right)}^{20\\times12}}=\\$9930.61[\/latex] (round your answer to the nearest penny)<\/p>\n<p>A video walkthrough of this example problem\u00a0is available below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Compound interest CD example\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/8NazxAjhpJw?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\/Compound+interest+CD+example.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cCompound interest CD example\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><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6936\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6936&theme=lumen&iframe_resize_id=ohm6936&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6937\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6937&theme=lumen&iframe_resize_id=ohm6937&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Solving For Time<\/h3>\n<section class=\"textbox connectIt\">\n<p>Note: This section assumes you\u2019ve covered solving exponential equations using logarithms, either in prior classes or in the growth models module.<\/p>\n<\/section>\n<p>Often we are interested in how long it will take to accumulate money or how long we\u2019d need to extend a loan to bring payments down to a reasonable level.<\/p>\n<section class=\"textbox example\">If you invest [latex]$2000[\/latex] at [latex]6\\%[\/latex] compounded monthly, how long will it take the account to double in value?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q610603\">Show Solution<\/button><\/p>\n<div id=\"q610603\" class=\"hidden-answer\" style=\"display: none\">This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]P_0 = $2000[\/latex]<\/td>\n<td>the initial 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>[latex]12[\/latex] months in [latex]1[\/latex] year<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>So our general equation is [latex]{{P}_{t}}=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{t\\times12}}[\/latex]. We also know that we want our ending amount to be double of [latex]$2000[\/latex], which is [latex]$4000[\/latex], so we\u2019re looking for [latex]t[\/latex] so that [latex]P_t = 4000[\/latex]. To solve this, we set our equation for [latex]P_t[\/latex] equal to [latex]4000[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r@{\\hfill}l}<br \/>  4000=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{t\\times12}} &&& \\text{Divide both sides by } 2000 \\\\<br \/>  2={{\\left(1.005\\right)}^{12t}} &&& \\text{To solve for the exponent, take the log of both sides} \\\\<br \/>  \\log\\left(2\\right)=\\log\\left({{\\left(1.005\\right)}^{12t}}\\right) &&& \\text{Use the exponent property of logs on the right side} \\\\<br \/>  \\log\\left(2\\right)=12t\\log\\left(1.005\\right) &&& \\text{Now we can divide both sides by } 12\\text{log}(1.005) \\\\<br \/>  \\frac{\\log\\left(2\\right)}{12\\log\\left(1.005\\right)}=t &&& \\text{Approximating this to a decimal} \\\\<br \/>  t = 11.581 &&&<br \/>  \\end{array}[\/latex]<\/p>\n<p>It will take about [latex]11.581[\/latex] years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded [latex]\\text{log}(2)[\/latex] to [latex]0.301[\/latex] and [latex]\\text{log}(1.005)[\/latex] to [latex]0.00217[\/latex], then your final answer would have been about [latex]11.577[\/latex] years.<\/p>\n<p>View this video for a walkthrough of this example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Find doubling time for compound interest\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zHRTxtFiyxc?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\/Find+doubling+time+for+compound+interest.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cFind doubling time for compound interest\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":21,"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\/2448"}],"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":67,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2448\/revisions"}],"predecessor-version":[{"id":14605,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2448\/revisions\/14605"}],"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\/2448\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2448"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2448"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2448"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2448"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}