{"id":388,"date":"2016-10-12T16:29:23","date_gmt":"2016-10-12T16:29:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=388"},"modified":"2021-02-05T23:58:41","modified_gmt":"2021-02-05T23:58:41","slug":"loans","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/loans\/","title":{"raw":"Loans","rendered":"Loans"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Conventional Loans<\/h2>\r\nIn the last section, you learned about payout annuities. In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920.png\"><img class=\"aligncenter size-large wp-image-747\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920-1024x681.png\" alt=\"Hand holding green pen, which has just written &quot;Approved!&quot; in a circle \" width=\"1024\" height=\"681\" \/><\/a>\r\n<div class=\"textbox examples\">\r\n<h3>new topic, same formula!<\/h3>\r\nMathematical formulas sometimes overlap, applying to more than one application. All the exercises and examples in this section use the same formula and techniques that you've already seen.\r\n\r\n<\/div>\r\nOne great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you\u2019re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.\r\n<div class=\"textbox\">\r\n<h2>Loans Formula<\/h2>\r\n[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\r\n \t<li><em>d <\/em> is your loan payment (your monthly payment, annual payment, etc)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\r\n \t<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\r\n \t<li><em>N<\/em> is the length of the loan, in years.<\/li>\r\n<\/ul>\r\n<\/div>\r\nLike before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.\r\n<div class=\"textbox\">\r\n<h2>When do you use this?<\/h2>\r\nThe loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.\r\n<ul>\r\n \t<li>Compound interest: One deposit<\/li>\r\n \t<li>Annuity: Many deposits<\/li>\r\n \t<li>Payout Annuity: Many withdrawals<\/li>\r\n \t<li>Loans: Many payments<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive a car can you afford? In other words, what loan amount can you pay off with $200 per month?\r\n[reveal-answer q=\"129373\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"129373\"]\r\n\r\nIn this example,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $200<\/td>\r\n<td>the monthly loan payment<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.03<\/td>\r\n<td>3% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 5<\/td>\r\n<td>since we\u2019re making monthly payments for 5 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe\u2019re looking for <em>P<sub>0<\/sub><\/em>, the starting amount of the loan.\r\n\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{200\\left(1-{{\\left(1+\\frac{0.03}{12}\\right)}^{-5(12)}}\\right)}{\\left(\\frac{0.03}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{200\\left(1-{{\\left(1.0025\\right)}^{-60}}\\right)}{\\left(0.0025\\right)}\\\\&amp;{{P}_{0}}=\\frac{200\\left(1-0.861\\right)}{\\left(0.0025\\right)}=\\$11,120 \\\\\\end{align}[\/latex]\r\n\r\nYou can afford a $11,120 loan.\r\n\r\nYou will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you\u2019re paying $12,000-$11,120 = $880 interest total.\r\n\r\n[\/hidden-answer]\r\n\r\nDetails of this example are examined in this video.\r\n\r\nhttps:\/\/youtu.be\/5NiNcdYytvY\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Solving for [latex]d[\/latex]<\/h3>\r\nIn the example above, you computed [latex]P_{0}[\/latex], the initial loan amount. In the example below, you are given the loan amount and must solve for the amount of the monthly payment, [latex]d[\/latex]. Use the same technique that you used in the previous sections.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6685[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?\r\n[reveal-answer q=\"538293\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538293\"]\r\n\r\nIn this example, we\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re paying monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $140,000<\/td>\r\n<td>\u00a0the starting loan amount<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;140,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;140,000=\\frac{d\\left(1-{{\\left(1.005\\right)}^{-360}}\\right)}{\\left(0.005\\right)}\\\\&amp;140,000=d(166.792)\\\\&amp;d=\\frac{140,000}{166.792}=\\$839.37 \\\\\\end{align}[\/latex]\r\n\r\nYou will make payments of $839.37 per month for 30 years.\r\n\r\nYou\u2019re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 - $140,000 = $162,173.20 in interest over the life of the loan.\r\n\r\n[\/hidden-answer]\r\n\r\nView more about this example here.\r\n\r\nhttps:\/\/youtu.be\/BYCECTyUc68\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6684[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nJanine bought $3,000 of new furniture on credit. Because her credit score isn\u2019t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?\r\n[reveal-answer q=\"642704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"642704\"]\r\n<div>\r\n\r\n<em>d<\/em> = \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 unknown\r\n\r\n<em>r<\/em> = 0.16 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16% annual rate\r\n\r\n<em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments\r\n\r\n<em>N<\/em> = 2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 years to repay\r\n\r\n<em>P0<\/em> = 3,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $3,000 loan\r\n\r\n[latex]\\begin{array}{c}3000=\\frac{{d}\\left(1-\\left(1+\\frac{0.06}{12}\\right)^{-2*12}\\right)}{\\frac{0.16}{12}}\\\\\\\\3000=20.42d\\end{array}[\/latex]\r\n\r\nSolve for d to get monthly payments of $146.89\r\n\r\nTwo years to repay means $146.89(24) = $3525.36 in total payments. \u00a0This means Janine will pay $3525.36 - $3000 = $525.36 in interest.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculating the Balance<\/h2>\r\nWith loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\"><img class=\"aligncenter size-full wp-image-751\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\" alt=\"Pair of glasses resting on a Mortgage Loan Statement\" width=\"640\" height=\"478\" \/><\/a>\r\n\r\nTo determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don\u2019t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will <em>not<\/em> have paid off $12,000 of the loan balance.\r\n\r\nTo determine the remaining loan balance, we can think \u201chow much loan will these loan payments be able to pay off in the remaining time on the loan?\u201d\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?\r\n[reveal-answer q=\"146377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"146377\"]\r\n\r\nTo determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we\u2019re looking for P<sub>0<\/sub> when\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $1,000<\/td>\r\n<td>the monthly loan payment<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.06<\/td>\r\n<td>6% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 10<\/td>\r\n<td>\u00a0since we\u2019re making monthly payments for 10 more years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-10(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1.005\\right)}^{-120}}\\right)}{\\left(0.005\\right)}\\\\&amp;{{P}_{0}}=\\frac{1000\\left(1-0.5496\\right)}{\\left(0.005\\right)}=\\$90,073.45 \\\\\\end{align}[\/latex]\r\n\r\nThe loan balance with 10 years remaining on the loan will be $90,073.45.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is explained in the following video:\r\n\r\nhttps:\/\/youtu.be\/fXLzeyCfAwE\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nOftentimes answering remaining balance questions requires two steps:\r\n<ol>\r\n \t<li>Calculating the monthly payments on the loan<\/li>\r\n \t<li>Calculating the remaining loan balance based on the <em>remaining time<\/em> on the loan<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?\r\n[reveal-answer q=\"423248\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"423248\"]\r\n\r\nFirst we will calculate their monthly payments.\r\n\r\nWe\u2019re looking for <em>d<\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.04<\/td>\r\n<td>4% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since they\u2019re paying monthly<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>30 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $180,000<\/td>\r\n<td>the starting loan amount<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe set up the equation and solve for <em>d<\/em>.\r\n\r\n[latex]\\begin{align}&amp;180,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&amp;180,000=\\frac{d\\left(1-{{\\left(1.00333\\right)}^{-360}}\\right)}{\\left(0.00333\\right)}\\\\&amp;180,000=d(209.562)\\\\&amp;d=\\frac{180,000}{209.562}=\\$858.93 \\\\\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\nNow that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>d<\/em> = $858.93<\/td>\r\n<td>the monthly loan payment we calculated above<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r <\/em>= 0.04<\/td>\r\n<td>4% annual rate<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>since they\u2019re doing monthly payments<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 25<\/td>\r\n<td>since they\u2019d be making monthly payments for 25 more years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]\\begin{align}&amp;{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-25(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&amp;{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1.00333\\right)}^{-300}}\\right)}{\\left(0.00333\\right)}\\\\&amp;{{P}_{0}}=\\frac{858.93\\left(1-0.369\\right)}{\\left(0.00333\\right)}=\\$162,758.21 \\\\\\end{align}[\/latex]\r\n\r\nThe loan balance after 5 years, with 25 years remaining on the loan, will be $162,758.21.\r\n\r\nOver that 5 years, the couple has paid off $180,000 - $162,758.21 = $17,241.79 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 - $17,241.79 = $340,294.01 of what they have paid so far has been interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving for Time<\/h2>\r\nRecall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nJoel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?\r\n[reveal-answer q=\"427851\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"427851\"]\r\n<div>\r\n\r\n<em>d<\/em> = $30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The monthly payments\r\n\r\n<em>r<\/em> = 0.12 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12% annual rate\r\n\r\n<em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments\r\n\r\n<em>P0<\/em> = 1,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $1,000 loan\r\n\r\nWe are solving for <em>N<\/em>, the time to pay off the loan\r\n\r\n[latex]1000=\\frac{30\\left(1-\\left(1+\\frac{0.12}{12}\\right)^{-N*12}\\right)}{\\frac{0.12}{12}}[\/latex]\r\n<div>\r\n\r\nSolving for <em>N<\/em> gives 3.396. It will take about 3.4 years to pay off the purchase.\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>FYI<\/h2>\r\nHome loans are typically paid off through an amortization process, <strong>amortization<\/strong> refers to paying off a debt (often from a loan or mortgage) over time through regular payments.\u00a0An <b>amortization schedule<\/b> is a <b>table<\/b> detailing each periodic payment on an\u00a0<b>amortizing<\/b> loan as generated by an <b>amortization calculator<\/b>.\r\n\r\nIf you want to know more, click on the link below to view the website\u00a0\u201cHow is an Amortization Schedule Calculated?\u201d by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.\r\n<ul>\r\n \t<li><a href=\"http:\/\/www.myamortizationchart.com\/articles\/how-is-an-amortization-schedule-calculated\/\" target=\"_blank\" rel=\"noopener\">How is an Amortization Schedule Calculated?<\/a><\/li>\r\n<\/ul>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the loan formula to calculate loan payments, loan balance, or interest accrued on a loan<\/li>\n<\/ul>\n<\/div>\n<h2>Conventional Loans<\/h2>\n<p>In the last section, you learned about payout annuities. In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-747\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/02205943\/approved-1049259_1920-1024x681.png\" alt=\"Hand holding green pen, which has just written &quot;Approved!&quot; in a circle\" width=\"1024\" height=\"681\" \/><\/a><\/p>\n<div class=\"textbox examples\">\n<h3>new topic, same formula!<\/h3>\n<p>Mathematical formulas sometimes overlap, applying to more than one application. All the exercises and examples in this section use the same formula and techniques that you&#8217;ve already seen.<\/p>\n<\/div>\n<p>One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you\u2019re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.<\/p>\n<div class=\"textbox\">\n<h2>Loans Formula<\/h2>\n<p>[latex]P_{0}=\\frac{d\\left(1-\\left(1+\\frac{r}{k}\\right)^{-Nk}\\right)}{\\left(\\frac{r}{k}\\right)}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> is the balance in the account at the beginning (the principal, or amount of the loan).<\/li>\n<li><em>d <\/em> is your loan payment (your monthly payment, annual payment, etc)<\/li>\n<li><em>r<\/em> is the annual interest rate in decimal form.<\/li>\n<li><em>k<\/em> is the number of compounding periods in one year.<\/li>\n<li><em>N<\/em> is the length of the loan, in years.<\/li>\n<\/ul>\n<\/div>\n<p>Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.<\/p>\n<div class=\"textbox\">\n<h2>When do you use this?<\/h2>\n<p>The loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.<\/p>\n<ul>\n<li>Compound interest: One deposit<\/li>\n<li>Annuity: Many deposits<\/li>\n<li>Payout Annuity: Many withdrawals<\/li>\n<li>Loans: Many payments<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive a car can you afford? In other words, what loan amount can you pay off with $200 per month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129373\">Show Solution<\/span><\/p>\n<div id=\"q129373\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $200<\/td>\n<td>the monthly loan payment<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.03<\/td>\n<td>3% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 5<\/td>\n<td>since we\u2019re making monthly payments for 5 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We\u2019re looking for <em>P<sub>0<\/sub><\/em>, the starting amount of the loan.<\/p>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{200\\left(1-{{\\left(1+\\frac{0.03}{12}\\right)}^{-5(12)}}\\right)}{\\left(\\frac{0.03}{12}\\right)}\\\\&{{P}_{0}}=\\frac{200\\left(1-{{\\left(1.0025\\right)}^{-60}}\\right)}{\\left(0.0025\\right)}\\\\&{{P}_{0}}=\\frac{200\\left(1-0.861\\right)}{\\left(0.0025\\right)}=\\$11,120 \\\\\\end{align}[\/latex]<\/p>\n<p>You can afford a $11,120 loan.<\/p>\n<p>You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you\u2019re paying $12,000-$11,120 = $880 interest total.<\/p>\n<\/div>\n<\/div>\n<p>Details of this example are examined in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Car loan\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5NiNcdYytvY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Solving for [latex]d[\/latex]<\/h3>\n<p>In the example above, you computed [latex]P_{0}[\/latex], the initial loan amount. In the example below, you are given the loan amount and must solve for the amount of the monthly payment, [latex]d[\/latex]. Use the same technique that you used in the previous sections.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6685\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6685&theme=oea&iframe_resize_id=ohm6685&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538293\">Show Solution<\/span><\/p>\n<div id=\"q538293\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example, we\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re paying monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $140,000<\/td>\n<td>\u00a0the starting loan amount<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In this case, we\u2019re going to have to set up the equation, and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&140,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&140,000=\\frac{d\\left(1-{{\\left(1.005\\right)}^{-360}}\\right)}{\\left(0.005\\right)}\\\\&140,000=d(166.792)\\\\&d=\\frac{140,000}{166.792}=\\$839.37 \\\\\\end{align}[\/latex]<\/p>\n<p>You will make payments of $839.37 per month for 30 years.<\/p>\n<p>You\u2019re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 &#8211; $140,000 = $162,173.20 in interest over the life of the loan.<\/p>\n<\/div>\n<\/div>\n<p>View more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Calculating payment on a home loan\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BYCECTyUc68?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6684&theme=oea&iframe_resize_id=ohm6684&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Janine bought $3,000 of new furniture on credit. Because her credit score isn\u2019t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q642704\">Show Solution<\/span><\/p>\n<div id=\"q642704\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 unknown<\/p>\n<p><em>r<\/em> = 0.16 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16% annual rate<\/p>\n<p><em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments<\/p>\n<p><em>N<\/em> = 2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 years to repay<\/p>\n<p><em>P0<\/em> = 3,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $3,000 loan<\/p>\n<p>[latex]\\begin{array}{c}3000=\\frac{{d}\\left(1-\\left(1+\\frac{0.06}{12}\\right)^{-2*12}\\right)}{\\frac{0.16}{12}}\\\\\\\\3000=20.42d\\end{array}[\/latex]<\/p>\n<p>Solve for d to get monthly payments of $146.89<\/p>\n<p>Two years to repay means $146.89(24) = $3525.36 in total payments. \u00a0This means Janine will pay $3525.36 &#8211; $3000 = $525.36 in interest.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Calculating the Balance<\/h2>\n<p>With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-751\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/07163930\/5445898743_bef11a7c87_z.jpg\" alt=\"Pair of glasses resting on a Mortgage Loan Statement\" width=\"640\" height=\"478\" \/><\/a><\/p>\n<p>To determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don\u2019t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will <em>not<\/em> have paid off $12,000 of the loan balance.<\/p>\n<p>To determine the remaining loan balance, we can think \u201chow much loan will these loan payments be able to pay off in the remaining time on the loan?\u201d<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q146377\">Show Solution<\/span><\/p>\n<div id=\"q146377\" class=\"hidden-answer\" style=\"display: none\">\n<p>To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we\u2019re looking for P<sub>0<\/sub> when<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $1,000<\/td>\n<td>the monthly loan payment<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.06<\/td>\n<td>6% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since we\u2019re doing monthly payments, we\u2019ll compound monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 10<\/td>\n<td>\u00a0since we\u2019re making monthly payments for 10 more years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1+\\frac{0.06}{12}\\right)}^{-10(12)}}\\right)}{\\left(\\frac{0.06}{12}\\right)}\\\\&{{P}_{0}}=\\frac{1000\\left(1-{{\\left(1.005\\right)}^{-120}}\\right)}{\\left(0.005\\right)}\\\\&{{P}_{0}}=\\frac{1000\\left(1-0.5496\\right)}{\\left(0.005\\right)}=\\$90,073.45 \\\\\\end{align}[\/latex]<\/p>\n<p>The loan balance with 10 years remaining on the loan will be $90,073.45.<\/p>\n<\/div>\n<\/div>\n<p>This example is explained in the following video:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Calculate remaining balance on loan from payment\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fXLzeyCfAwE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Oftentimes answering remaining balance questions requires two steps:<\/p>\n<ol>\n<li>Calculating the monthly payments on the loan<\/li>\n<li>Calculating the remaining loan balance based on the <em>remaining time<\/em> on the loan<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q423248\">Show Solution<\/span><\/p>\n<div id=\"q423248\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we will calculate their monthly payments.<\/p>\n<p>We\u2019re looking for <em>d<\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.04<\/td>\n<td>4% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since they\u2019re paying monthly<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>30 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $180,000<\/td>\n<td>the starting loan amount<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We set up the equation and solve for <em>d<\/em>.<\/p>\n<p>[latex]\\begin{align}&180,000=\\frac{d\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-30(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&180,000=\\frac{d\\left(1-{{\\left(1.00333\\right)}^{-360}}\\right)}{\\left(0.00333\\right)}\\\\&180,000=d(209.562)\\\\&d=\\frac{180,000}{209.562}=\\$858.93 \\\\\\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>d<\/em> = $858.93<\/td>\n<td>the monthly loan payment we calculated above<\/td>\n<\/tr>\n<tr>\n<td><em>r <\/em>= 0.04<\/td>\n<td>4% annual rate<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>since they\u2019re doing monthly payments<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 25<\/td>\n<td>since they\u2019d be making monthly payments for 25 more years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]\\begin{align}&{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1+\\frac{0.04}{12}\\right)}^{-25(12)}}\\right)}{\\left(\\frac{0.04}{12}\\right)}\\\\&{{P}_{0}}=\\frac{858.93\\left(1-{{\\left(1.00333\\right)}^{-300}}\\right)}{\\left(0.00333\\right)}\\\\&{{P}_{0}}=\\frac{858.93\\left(1-0.369\\right)}{\\left(0.00333\\right)}=\\$162,758.21 \\\\\\end{align}[\/latex]<\/p>\n<p>The loan balance after 5 years, with 25 years remaining on the loan, will be $162,758.21.<\/p>\n<p>Over that 5 years, the couple has paid off $180,000 &#8211; $162,758.21 = $17,241.79 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 &#8211; $17,241.79 = $340,294.01 of what they have paid so far has been interest.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving for Time<\/h2>\n<p>Recall that we have used logarithms to solve for time, since it is an exponent in interest calculations. We can apply the same idea to finding how long it will take to pay off a loan.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Joel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q427851\">Show Solution<\/span><\/p>\n<div id=\"q427851\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p><em>d<\/em> = $30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The monthly payments<\/p>\n<p><em>r<\/em> = 0.12 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12% annual rate<\/p>\n<p><em>k<\/em> = 12\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 since we\u2019re making monthly payments<\/p>\n<p><em>P0<\/em> = 1,000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we\u2019re starting with a $1,000 loan<\/p>\n<p>We are solving for <em>N<\/em>, the time to pay off the loan<\/p>\n<p>[latex]1000=\\frac{30\\left(1-\\left(1+\\frac{0.12}{12}\\right)^{-N*12}\\right)}{\\frac{0.12}{12}}[\/latex]<\/p>\n<div>\n<p>Solving for <em>N<\/em> gives 3.396. It will take about 3.4 years to pay off the purchase.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>FYI<\/h2>\n<p>Home loans are typically paid off through an amortization process, <strong>amortization<\/strong> refers to paying off a debt (often from a loan or mortgage) over time through regular payments.\u00a0An <b>amortization schedule<\/b> is a <b>table<\/b> detailing each periodic payment on an\u00a0<b>amortizing<\/b> loan as generated by an <b>amortization calculator<\/b>.<\/p>\n<p>If you want to know more, click on the link below to view the website\u00a0\u201cHow is an Amortization Schedule Calculated?\u201d by MyAmortizationChart.com. This website provides a brief overlook of Amortization Schedules.<\/p>\n<ul>\n<li><a href=\"http:\/\/www.myamortizationchart.com\/articles\/how-is-an-amortization-schedule-calculated\/\" target=\"_blank\" rel=\"noopener\">How is an Amortization Schedule Calculated?<\/a><\/li>\n<\/ul>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-388\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Loans. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>Project<\/strong>: Math in Society. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>approved-finance-business-loan-1049259. <strong>Authored by<\/strong>: InspiredImages. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/approved-finance-business-loan-1049259\/\">https:\/\/pixabay.com\/en\/approved-finance-business-loan-1049259\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Car loan. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/5NiNcdYytvY\">https:\/\/youtu.be\/5NiNcdYytvY<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculating payment on a home loan. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BYCECTyUc68\">https:\/\/youtu.be\/BYCECTyUc68<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6684, 6685. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t 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