{"id":380,"date":"2016-10-11T22:49:39","date_gmt":"2016-10-11T22:49:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=380"},"modified":"2019-05-30T16:54:41","modified_gmt":"2019-05-30T16:54:41","slug":"compound-interest","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/compound-interest\/","title":{"raw":"Compound Interest","rendered":"Compound Interest"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate one-time simple interest, and simple interest over time<\/li>\r\n \t<li>Determine APY given an interest scenario<\/li>\r\n \t<li>Calculate compound interest<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Compounding<\/h2>\r\nWith simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called <strong>compounding<\/strong>.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/23200323\/achievement-18134_1280.jpg\"><img class=\"aligncenter wp-image-552 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/23200323\/achievement-18134_1280-1024x759.jpg\" alt=\"a row of gold coin stacks. From left to right, they grown from one coin, to two, to four, ending with a stack of 32 coins\" width=\"1024\" height=\"759\" \/><\/a>\r\n\r\nSuppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?\r\n\r\nThe 3% interest is an annual percentage rate (APR) \u2013 the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\\frac{3%}{12}[\/latex]= 0.25% per month.\r\n\r\nIn the first month,\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> = $1000<\/li>\r\n \t<li><em>r<\/em> = 0.0025 (0.25%)<\/li>\r\n \t<li><em>I <\/em>= $1000 (0.0025) = $2.50<\/li>\r\n \t<li><em>A<\/em> = $1000 + $2.50 = $1002.50<\/li>\r\n<\/ul>\r\nIn the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.\r\n\r\n&nbsp;\r\n\r\nIn the second month,\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> = $1002.50<\/li>\r\n \t<li><em>I <\/em>= $1002.50 (0.0025) = $2.51 (rounded)<\/li>\r\n \t<li><em>A<\/em> = $1002.50 + $2.51 = $1005.01<\/li>\r\n<\/ul>\r\nNotice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that <strong>compounding<\/strong>\u00a0interest gives us.\r\n\r\nCalculating out a few more months gives the following:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Month<\/strong><\/td>\r\n<td><strong>Starting balance<\/strong><\/td>\r\n<td><strong>Interest earned<\/strong><\/td>\r\n<td><strong>Ending Balance<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>1000.00<\/td>\r\n<td>2.50<\/td>\r\n<td>1002.50<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>1002.50<\/td>\r\n<td>2.51<\/td>\r\n<td>1005.01<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>1005.01<\/td>\r\n<td>2.51<\/td>\r\n<td>1007.52<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4<\/td>\r\n<td>1007.52<\/td>\r\n<td>2.52<\/td>\r\n<td>1010.04<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5<\/td>\r\n<td>1010.04<\/td>\r\n<td>2.53<\/td>\r\n<td>1012.57<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6<\/td>\r\n<td>1012.57<\/td>\r\n<td>2.53<\/td>\r\n<td>1015.10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>7<\/td>\r\n<td>1015.10<\/td>\r\n<td>2.54<\/td>\r\n<td>1017.64<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>8<\/td>\r\n<td>1017.64<\/td>\r\n<td>2.54<\/td>\r\n<td>1020.18<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9<\/td>\r\n<td>1020.18<\/td>\r\n<td>2.55<\/td>\r\n<td>1022.73<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>10<\/td>\r\n<td>1022.73<\/td>\r\n<td>2.56<\/td>\r\n<td>1025.29<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>11<\/td>\r\n<td>1025.29<\/td>\r\n<td>2.56<\/td>\r\n<td>1027.85<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>12<\/td>\r\n<td>1027.85<\/td>\r\n<td>2.57<\/td>\r\n<td>1030.42<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe want to simplify the process for calculating compounding, because creating a table like the one above is time consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if <em>P<sub>m <\/sub><\/em>represents the amount of money after <em>m<\/em> months, then we could write the recursive equation:\r\n\r\n<em>P<sub>0<\/sub><\/em> = $1000\r\n\r\n<em>P<sub>m<\/sub><\/em> = (1+0.0025)<em>P<sub>m-1<\/sub><\/em>\r\n\r\nYou probably recognize this as the recursive form of exponential growth. If not, we go through the steps to build an explicit equation for the growth in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nBuild an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly.\r\n[reveal-answer q=\"530288\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530288\"]\r\n<ul>\r\n \t<li><em>P<sub>0<\/sub><\/em> = $1000<\/li>\r\n \t<li><em>P\u00ad<sub>1<\/sub><\/em> = 1.0025<em>P\u00ad<sub>0<\/sub><\/em> = 1.0025 (1000)<\/li>\r\n \t<li><em>P\u00ad<sub>2<\/sub><\/em> = 1.0025<em>P\u00ad<sub>1<\/sub><\/em> = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)<\/li>\r\n \t<li><em>P\u00ad<sub>3<\/sub><\/em> = 1.0025<em>P\u00ad<sub>2<\/sub><\/em> = 1.0025 (1.00252(1000)) = 1.00253(1000)<\/li>\r\n \t<li><em>P\u00ad<sub>4<\/sub><\/em> = 1.0025<em>P\u00ad<sub>3<\/sub><\/em> = 1.0025 (1.00253(1000)) = 1.00254(1000)<\/li>\r\n<\/ul>\r\nObserving a pattern, we could conclude\r\n<ul>\r\n \t<li><em>P<sub>m<\/sub><\/em> = (1.0025)<sup><em>m<\/em><\/sup>($1000)<\/li>\r\n<\/ul>\r\nNotice that the $1000 in the equation was <em>P<sub>0<\/sub><\/em>, the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year.\r\n\r\n&nbsp;\r\n\r\nGeneralizing our result, we could write\r\n\r\n[latex]{{P}_{m}}={{P}_{0}}{{\\left(1+\\frac{r}{k}\\right)}^{m}}[\/latex]\r\n\r\n&nbsp;\r\n\r\nIn this formula:\r\n<ul>\r\n \t<li><em>m<\/em> is the number of compounding periods (months in our example)<\/li>\r\n \t<li><em>r<\/em> is the annual interest rate<\/li>\r\n \t<li><em>k<\/em> is the number of compounds per year.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\nView this video for a walkthrough of the concept of compound interest.\r\n\r\nhttps:\/\/youtu.be\/xuQTFmP9nNg\r\n\r\n<\/div>\r\nWhile 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 <em>N<\/em> is the number of years, then <em>m = N k<\/em>. Making this change gives us the standard formula for compound interest.\r\n<div class=\"textbox\">\r\n<h3>Compound Interest<\/h3>\r\n[latex]P_{N}=P_{0}\\left(1+\\frac{r}{k}\\right)^{Nk}[\/latex]\r\n<ul>\r\n \t<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\r\n \t<li><em>P<sub>0 <\/sub><\/em> is the starting balance of the account (also called initial deposit, or principal)<\/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\r\n<ul>\r\n \t<li>If the compounding is done annually (once a year), <em>k<\/em> = 1.<\/li>\r\n \t<li>If the compounding is done quarterly, <em>k<\/em> = 4.<\/li>\r\n \t<li>If the compounding is done monthly, <em>k<\/em> = 12.<\/li>\r\n \t<li>If the compounding is done daily, <em>k<\/em> = 365.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\"><strong><em>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<\/em><\/strong><\/div>\r\nIn the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after 20 years.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA 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 $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?\r\n[reveal-answer q=\"788137\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788137\"]\r\n\r\nIn this example,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $3000<\/td>\r\n<td>the initial deposit<\/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>12 months in 1 year<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 20<\/td>\r\n<td>\u00a0since we\u2019re looking for how much we\u2019ll have after 20 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]{{P}_{20}}=3000{{\\left(1+\\frac{0.06}{12}\\right)}^{20\\times12}}=\\$9930.61[\/latex] (round your answer to the nearest penny)\r\n\r\n[\/hidden-answer]\r\n\r\nA video walkthrough of this example problem\u00a0is available below.\r\n\r\nhttps:\/\/youtu.be\/8NazxAjhpJw\r\n\r\n<\/div>\r\nLet us compare the amount of money earned from compounding against the amount you would earn from simple interest\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Years<\/td>\r\n<td>Simple Interest ($15 per month)<\/td>\r\n<td>6% compounded monthly = 0.5% each month.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5<\/td>\r\n<td>$3900<\/td>\r\n<td>$4046.55<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>10<\/td>\r\n<td>$4800<\/td>\r\n<td>$5458.19<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>15<\/td>\r\n<td>$5700<\/td>\r\n<td>$7362.28<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>20<\/td>\r\n<td>$6600<\/td>\r\n<td>$9930.61<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>25<\/td>\r\n<td>$7500<\/td>\r\n<td>$13394.91<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>30<\/td>\r\n<td>$8400<\/td>\r\n<td>$18067.73<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>35<\/td>\r\n<td>$9300<\/td>\r\n<td>$24370.65<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<img class=\"aligncenter wp-image-381\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11224717\/accountbalanceyears.png\" alt=\"Line graph. Vertical axis: Account Balance ($), in increments of 5000 from 5000 to 25000. Horizontal axis: years, in increments of five, from 0 to 25. A blue dotted line shows a gradual increase over time, from roughly $2500 at year 0 to roughly $10000 at year 35. A pink dotted line shows a more dramatic increase, from roughly $2500 at year 0 to $25000 at year 35.\" width=\"427\" height=\"345\" \/>\r\n\r\nAs you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6693&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Evaluating exponents on the calculator<\/h3>\r\nWhen we need to calculate something like [latex]5^3[\/latex] it is easy enough to just multiply [latex]5\\cdot{5}\\cdot{5}=125[\/latex].\u00a0 But when we need to calculate something like [latex]1.005^{240}[\/latex], it would be very tedious to calculate this by multiplying [latex]1.005[\/latex] by itself [latex]240[\/latex] times!\u00a0 So to make things easier, we can harness the power of our scientific calculators.\r\n\r\nMost scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:\r\n<p style=\"text-align: center;\">^ , \u00a0 [latex]y^x[\/latex] ,\u00a0\u00a0 or [latex]x^y[\/latex] .<\/p>\r\nTo evaluate [latex]1.005^{240}[\/latex] we'd type [latex]1.005[\/latex] ^ [latex]240[\/latex], or [latex]1.005 \\space{y^{x}}\\space 240[\/latex].\u00a0 Try it out - you should get something around 3.3102044758.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nYou know that you will need $40,000 for your child\u2019s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?\r\n[reveal-answer q=\"842460\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"842460\"]\r\n\r\nIn this example, we\u2019re looking for <em>P<sub>0<\/sub><\/em>.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>r<\/em> = 0.04<\/td>\r\n<td>4%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 4<\/td>\r\n<td>4 quarters in 1 year<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N <\/em>= 18<\/td>\r\n<td>Since we know the balance in 18 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>18<\/sub><\/em> = $40,000<\/td>\r\n<td>The amount we have in 18 years<\/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>P<sub>0<\/sub><\/em>.\r\n\r\n[latex]\\begin{align}&amp;40000={{P}_{0}}{{\\left(1+\\frac{0.04}{4}\\right)}^{18\\times4}}\\\\&amp;40000={{P}_{0}}(2.0471)\\\\&amp;{{P}_{0}}=\\frac{40000}{2.0471}=\\$19539.84 \\\\\\end{align}[\/latex]\r\n\r\nSo you would need to deposit $19,539.84 now to have $40,000 in 18 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6692&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Rounding<\/h3>\r\nIt is important to be very careful about rounding when calculating things with exponents. In general, you want to keep as many decimals during calculations as you can. Be sure to <strong>keep at least 3 significant digits<\/strong> (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a \u201cclose enough\u201d answer, but keeping more digits is always better.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nTo see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $1000<\/td>\r\n<td>the initial deposit<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>r<\/em> = 0.05<\/td>\r\n<td>5%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>k<\/em> = 12<\/td>\r\n<td>12 months in 1 year<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>N<\/em> = 30<\/td>\r\n<td>since we\u2019re looking for the amount after 30 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIf we first compute <em>r\/k<\/em>, we find 0.05\/12 = 0.00416666666667\r\n\r\nHere is the effect of rounding this to different values:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>&nbsp;\r\n\r\n<strong><em>r\/k<\/em> rounded to:<\/strong><\/td>\r\n<td><strong>Gives <em>P\u00ad30<\/em>\u00ad to be:<\/strong><\/td>\r\n<td><strong>Error<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.004<\/td>\r\n<td>$4208.59<\/td>\r\n<td>$259.15<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.0042<\/td>\r\n<td>$4521.45<\/td>\r\n<td>$53.71<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.00417<\/td>\r\n<td>$4473.09<\/td>\r\n<td>$5.35<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.004167<\/td>\r\n<td>$4468.28<\/td>\r\n<td>$0.54<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.0041667<\/td>\r\n<td>$4467.80<\/td>\r\n<td>$0.06<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>no rounding<\/td>\r\n<td>$4467.74<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIf you\u2019re working in a bank, of course you wouldn\u2019t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough - $5 off of $4500 isn\u2019t too bad. Certainly keeping that fourth decimal place wouldn\u2019t have hurt.\r\n\r\nView the following for a demonstration of this example.\r\n\r\nhttps:\/\/youtu.be\/VhhYtaMN6mo\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Using your calculator<\/h3>\r\nIn many cases, you can avoid rounding completely by how you enter things in your calculator. For example, in the example above, we needed to calculate [latex]{{P}_{30}}=1000{{\\left(1+\\frac{0.05}{12}\\right)}^{12\\times30}}[\/latex]\r\n\r\nWe can quickly calculate 12\u00d730 = 360, giving [latex]{{P}_{30}}=1000{{\\left(1+\\frac{0.05}{12}\\right)}^{360}}[\/latex].\r\n\r\nNow we can use the calculator.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Type this<\/strong><\/td>\r\n<td><strong>Calculator shows<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.05 \u00f7 12 = .<\/td>\r\n<td>0.00416666666667<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>+ 1 = .<\/td>\r\n<td>1.00416666666667<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>yx 360 = .<\/td>\r\n<td>4.46774431400613<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\u00d7 1000 = .<\/td>\r\n<td>4467.74431400613<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h3>Using your calculator continued<\/h3>\r\nThe previous steps were assuming you have a \u201cone operation at a time\u201d calculator; a more advanced calculator will often allow you to type in the entire expression to be evaluated. If you have a calculator like this, you will probably just need to enter:\r\n\r\n1000 \u00d7 \u00a0( 1 + 0.05 \u00f7 12 ) y<sup>x<\/sup> 360 =\r\n\r\n<\/div>\r\n<h2>Solving For Time<\/h2>\r\n<div class=\"textbox shaded\">Note: This section assumes you\u2019ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.<\/div>\r\nOften 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nIf you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?\r\n[reveal-answer q=\"610603\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"610603\"]\r\n\r\nThis is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>P<sub>0<\/sub><\/em> = $2000<\/td>\r\n<td>the initial deposit<\/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>12 months in 1 year<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo our general equation is [latex]{{P}_{N}}=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{N\\times12}}[\/latex]. We also know that we want our ending amount to be double of $2000, which is $4000, so we\u2019re looking for <em>N<\/em> so that <em>P<sub>N<\/sub><\/em> = 4000. To solve this, we set our equation for <em>P<sub>N<\/sub><\/em> equal to 4000.\r\n<p style=\"text-align: center;\">[latex]4000=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{N\\times12}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 2000<\/p>\r\n<p style=\"text-align: center;\">[latex]2={{\\left(1.005\\right)}^{12N}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 To solve for the exponent, take the log of both sides<\/p>\r\n<p style=\"text-align: center;\">[latex]\\log\\left(2\\right)=\\log\\left({{\\left(1.005\\right)}^{12N}}\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\r\n<p style=\"text-align: center;\">[latex]\\log\\left(2\\right)=12N\\log\\left(1.005\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now we can divide both sides by 12log(1.005)<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\log\\left(2\\right)}{12\\log\\left(1.005\\right)}=N[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Approximating this to a decimal<\/p>\r\n<em>N<\/em> = 11.581\r\n\r\nIt will take about 11.581 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 log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.\r\n\r\n[\/hidden-answer]\r\n\r\nGet additional guidance for this example in the following:\r\n\r\nhttps:\/\/youtu.be\/zHRTxtFiyxc\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate one-time simple interest, and simple interest over time<\/li>\n<li>Determine APY given an interest scenario<\/li>\n<li>Calculate compound interest<\/li>\n<\/ul>\n<\/div>\n<h2>Compounding<\/h2>\n<p>With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called <strong>compounding<\/strong>.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/23200323\/achievement-18134_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-552 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/11\/23200323\/achievement-18134_1280-1024x759.jpg\" alt=\"a row of gold coin stacks. From left to right, they grown from one coin, to two, to four, ending with a stack of 32 coins\" width=\"1024\" height=\"759\" \/><\/a><\/p>\n<p>Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?<\/p>\n<p>The 3% interest is an annual percentage rate (APR) \u2013 the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\\frac{3%}{12}[\/latex]= 0.25% per month.<\/p>\n<p>In the first month,<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> = $1000<\/li>\n<li><em>r<\/em> = 0.0025 (0.25%)<\/li>\n<li><em>I <\/em>= $1000 (0.0025) = $2.50<\/li>\n<li><em>A<\/em> = $1000 + $2.50 = $1002.50<\/li>\n<\/ul>\n<p>In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.<\/p>\n<p>&nbsp;<\/p>\n<p>In the second month,<\/p>\n<ul>\n<li><em>P<sub>0<\/sub><\/em> = $1002.50<\/li>\n<li><em>I <\/em>= $1002.50 (0.0025) = $2.51 (rounded)<\/li>\n<li><em>A<\/em> = $1002.50 + $2.51 = $1005.01<\/li>\n<\/ul>\n<p>Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that <strong>compounding<\/strong>\u00a0interest gives us.<\/p>\n<p>Calculating out a few more months gives the following:<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>Month<\/strong><\/td>\n<td><strong>Starting balance<\/strong><\/td>\n<td><strong>Interest earned<\/strong><\/td>\n<td><strong>Ending Balance<\/strong><\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>1000.00<\/td>\n<td>2.50<\/td>\n<td>1002.50<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>1002.50<\/td>\n<td>2.51<\/td>\n<td>1005.01<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>1005.01<\/td>\n<td>2.51<\/td>\n<td>1007.52<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>1007.52<\/td>\n<td>2.52<\/td>\n<td>1010.04<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>1010.04<\/td>\n<td>2.53<\/td>\n<td>1012.57<\/td>\n<\/tr>\n<tr>\n<td>6<\/td>\n<td>1012.57<\/td>\n<td>2.53<\/td>\n<td>1015.10<\/td>\n<\/tr>\n<tr>\n<td>7<\/td>\n<td>1015.10<\/td>\n<td>2.54<\/td>\n<td>1017.64<\/td>\n<\/tr>\n<tr>\n<td>8<\/td>\n<td>1017.64<\/td>\n<td>2.54<\/td>\n<td>1020.18<\/td>\n<\/tr>\n<tr>\n<td>9<\/td>\n<td>1020.18<\/td>\n<td>2.55<\/td>\n<td>1022.73<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>1022.73<\/td>\n<td>2.56<\/td>\n<td>1025.29<\/td>\n<\/tr>\n<tr>\n<td>11<\/td>\n<td>1025.29<\/td>\n<td>2.56<\/td>\n<td>1027.85<\/td>\n<\/tr>\n<tr>\n<td>12<\/td>\n<td>1027.85<\/td>\n<td>2.57<\/td>\n<td>1030.42<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We want to simplify the process for calculating compounding, because creating a table like the one above is time consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if <em>P<sub>m <\/sub><\/em>represents the amount of money after <em>m<\/em> months, then we could write the recursive equation:<\/p>\n<p><em>P<sub>0<\/sub><\/em> = $1000<\/p>\n<p><em>P<sub>m<\/sub><\/em> = (1+0.0025)<em>P<sub>m-1<\/sub><\/em><\/p>\n<p>You probably recognize this as the recursive form of exponential growth. If not, we go through the steps to build an explicit equation for the growth in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Build an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530288\">Show Solution<\/span><\/p>\n<div id=\"q530288\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li><em>P<sub>0<\/sub><\/em> = $1000<\/li>\n<li><em>P\u00ad<sub>1<\/sub><\/em> = 1.0025<em>P\u00ad<sub>0<\/sub><\/em> = 1.0025 (1000)<\/li>\n<li><em>P\u00ad<sub>2<\/sub><\/em> = 1.0025<em>P\u00ad<sub>1<\/sub><\/em> = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)<\/li>\n<li><em>P\u00ad<sub>3<\/sub><\/em> = 1.0025<em>P\u00ad<sub>2<\/sub><\/em> = 1.0025 (1.00252(1000)) = 1.00253(1000)<\/li>\n<li><em>P\u00ad<sub>4<\/sub><\/em> = 1.0025<em>P\u00ad<sub>3<\/sub><\/em> = 1.0025 (1.00253(1000)) = 1.00254(1000)<\/li>\n<\/ul>\n<p>Observing a pattern, we could conclude<\/p>\n<ul>\n<li><em>P<sub>m<\/sub><\/em> = (1.0025)<sup><em>m<\/em><\/sup>($1000)<\/li>\n<\/ul>\n<p>Notice that the $1000 in the equation was <em>P<sub>0<\/sub><\/em>, the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year.<\/p>\n<p>&nbsp;<\/p>\n<p>Generalizing our result, we could write<\/p>\n<p>[latex]{{P}_{m}}={{P}_{0}}{{\\left(1+\\frac{r}{k}\\right)}^{m}}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>In this formula:<\/p>\n<ul>\n<li><em>m<\/em> is the number of compounding periods (months in our example)<\/li>\n<li><em>r<\/em> is the annual interest rate<\/li>\n<li><em>k<\/em> is the number of compounds per year.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>View this video for a walkthrough of the concept of compound interest.<\/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<\/div>\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 <em>N<\/em> is the number of years, then <em>m = N k<\/em>. Making this change gives us the standard formula for compound interest.<\/p>\n<div class=\"textbox\">\n<h3>Compound Interest<\/h3>\n<p>[latex]P_{N}=P_{0}\\left(1+\\frac{r}{k}\\right)^{Nk}[\/latex]<\/p>\n<ul>\n<li><em>P<sub>N<\/sub><\/em> is the balance in the account after <em>N<\/em> years.<\/li>\n<li><em>P<sub>0 <\/sub><\/em> is the starting balance of the account (also called initial deposit, or principal)<\/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\n<ul>\n<li>If the compounding is done annually (once a year), <em>k<\/em> = 1.<\/li>\n<li>If the compounding is done quarterly, <em>k<\/em> = 4.<\/li>\n<li>If the compounding is done monthly, <em>k<\/em> = 12.<\/li>\n<li>If the compounding is done daily, <em>k<\/em> = 365.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\"><strong><em>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<\/em><\/strong><\/div>\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 20 years.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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 $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788137\">Show Solution<\/span><\/p>\n<div id=\"q788137\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example,<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $3000<\/td>\n<td>the initial deposit<\/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>12 months in 1 year<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 20<\/td>\n<td>\u00a0since we\u2019re looking for how much we\u2019ll have after 20 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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<\/div>\n<\/div>\n<p>A video walkthrough of this example problem\u00a0is available below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<\/div>\n<p>Let us compare the amount of money earned from compounding against the amount you would earn from simple interest<\/p>\n<table>\n<tbody>\n<tr>\n<td>Years<\/td>\n<td>Simple Interest ($15 per month)<\/td>\n<td>6% compounded monthly = 0.5% each month.<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>$3900<\/td>\n<td>$4046.55<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>$4800<\/td>\n<td>$5458.19<\/td>\n<\/tr>\n<tr>\n<td>15<\/td>\n<td>$5700<\/td>\n<td>$7362.28<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>$6600<\/td>\n<td>$9930.61<\/td>\n<\/tr>\n<tr>\n<td>25<\/td>\n<td>$7500<\/td>\n<td>$13394.91<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>$8400<\/td>\n<td>$18067.73<\/td>\n<\/tr>\n<tr>\n<td>35<\/td>\n<td>$9300<\/td>\n<td>$24370.65<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-381\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11224717\/accountbalanceyears.png\" alt=\"Line graph. Vertical axis: Account Balance ($), in increments of 5000 from 5000 to 25000. Horizontal axis: years, in increments of five, from 0 to 25. A blue dotted line shows a gradual increase over time, from roughly $2500 at year 0 to roughly $10000 at year 35. A pink dotted line shows a more dramatic increase, from roughly $2500 at year 0 to $25000 at year 35.\" width=\"427\" height=\"345\" \/><\/p>\n<p>As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6693&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Evaluating exponents on the calculator<\/h3>\n<p>When we need to calculate something like [latex]5^3[\/latex] it is easy enough to just multiply [latex]5\\cdot{5}\\cdot{5}=125[\/latex].\u00a0 But when we need to calculate something like [latex]1.005^{240}[\/latex], it would be very tedious to calculate this by multiplying [latex]1.005[\/latex] by itself [latex]240[\/latex] times!\u00a0 So to make things easier, we can harness the power of our scientific calculators.<\/p>\n<p>Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:<\/p>\n<p style=\"text-align: center;\">^ , \u00a0 [latex]y^x[\/latex] ,\u00a0\u00a0 or [latex]x^y[\/latex] .<\/p>\n<p>To evaluate [latex]1.005^{240}[\/latex] we&#8217;d type [latex]1.005[\/latex] ^ [latex]240[\/latex], or [latex]1.005 \\space{y^{x}}\\space 240[\/latex].\u00a0 Try it out &#8211; you should get something around 3.3102044758.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>You know that you will need $40,000 for your child\u2019s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q842460\">Show Solution<\/span><\/p>\n<div id=\"q842460\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this example, we\u2019re looking for <em>P<sub>0<\/sub><\/em>.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>r<\/em> = 0.04<\/td>\n<td>4%<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 4<\/td>\n<td>4 quarters in 1 year<\/td>\n<\/tr>\n<tr>\n<td><em>N <\/em>= 18<\/td>\n<td>Since we know the balance in 18 years<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>18<\/sub><\/em> = $40,000<\/td>\n<td>The amount we have in 18 years<\/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>P<sub>0<\/sub><\/em>.<\/p>\n<p>[latex]\\begin{align}&40000={{P}_{0}}{{\\left(1+\\frac{0.04}{4}\\right)}^{18\\times4}}\\\\&40000={{P}_{0}}(2.0471)\\\\&{{P}_{0}}=\\frac{40000}{2.0471}=\\$19539.84 \\\\\\end{align}[\/latex]<\/p>\n<p>So you would need to deposit $19,539.84 now to have $40,000 in 18 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6692&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Rounding<\/h3>\n<p>It is important to be very careful about rounding when calculating things with exponents. In general, you want to keep as many decimals during calculations as you can. Be sure to <strong>keep at least 3 significant digits<\/strong> (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a \u201cclose enough\u201d answer, but keeping more digits is always better.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>P<sub>0<\/sub><\/em> = $1000<\/td>\n<td>the initial deposit<\/td>\n<\/tr>\n<tr>\n<td><em>r<\/em> = 0.05<\/td>\n<td>5%<\/td>\n<\/tr>\n<tr>\n<td><em>k<\/em> = 12<\/td>\n<td>12 months in 1 year<\/td>\n<\/tr>\n<tr>\n<td><em>N<\/em> = 30<\/td>\n<td>since we\u2019re looking for the amount after 30 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>If we first compute <em>r\/k<\/em>, we find 0.05\/12 = 0.00416666666667<\/p>\n<p>Here is the effect of rounding this to different values:<\/p>\n<table>\n<tbody>\n<tr>\n<td>&nbsp;<\/p>\n<p><strong><em>r\/k<\/em> rounded to:<\/strong><\/td>\n<td><strong>Gives <em>P\u00ad30<\/em>\u00ad to be:<\/strong><\/td>\n<td><strong>Error<\/strong><\/td>\n<\/tr>\n<tr>\n<td>0.004<\/td>\n<td>$4208.59<\/td>\n<td>$259.15<\/td>\n<\/tr>\n<tr>\n<td>0.0042<\/td>\n<td>$4521.45<\/td>\n<td>$53.71<\/td>\n<\/tr>\n<tr>\n<td>0.00417<\/td>\n<td>$4473.09<\/td>\n<td>$5.35<\/td>\n<\/tr>\n<tr>\n<td>0.004167<\/td>\n<td>$4468.28<\/td>\n<td>$0.54<\/td>\n<\/tr>\n<tr>\n<td>0.0041667<\/td>\n<td>$4467.80<\/td>\n<td>$0.06<\/td>\n<\/tr>\n<tr>\n<td>no rounding<\/td>\n<td>$4467.74<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>If you\u2019re working in a bank, of course you wouldn\u2019t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough &#8211; $5 off of $4500 isn\u2019t too bad. Certainly keeping that fourth decimal place wouldn\u2019t have hurt.<\/p>\n<p>View the following for a demonstration of this example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Compound interest - the importance of rounding\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VhhYtaMN6mo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Using your calculator<\/h3>\n<p>In many cases, you can avoid rounding completely by how you enter things in your calculator. For example, in the example above, we needed to calculate [latex]{{P}_{30}}=1000{{\\left(1+\\frac{0.05}{12}\\right)}^{12\\times30}}[\/latex]<\/p>\n<p>We can quickly calculate 12\u00d730 = 360, giving [latex]{{P}_{30}}=1000{{\\left(1+\\frac{0.05}{12}\\right)}^{360}}[\/latex].<\/p>\n<p>Now we can use the calculator.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>Type this<\/strong><\/td>\n<td><strong>Calculator shows<\/strong><\/td>\n<\/tr>\n<tr>\n<td>0.05 \u00f7 12 = .<\/td>\n<td>0.00416666666667<\/td>\n<\/tr>\n<tr>\n<td>+ 1 = .<\/td>\n<td>1.00416666666667<\/td>\n<\/tr>\n<tr>\n<td>yx 360 = .<\/td>\n<td>4.46774431400613<\/td>\n<\/tr>\n<tr>\n<td>\u00d7 1000 = .<\/td>\n<td>4467.74431400613<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3>Using your calculator continued<\/h3>\n<p>The previous steps were assuming you have a \u201cone operation at a time\u201d calculator; a more advanced calculator will often allow you to type in the entire expression to be evaluated. If you have a calculator like this, you will probably just need to enter:<\/p>\n<p>1000 \u00d7 \u00a0( 1 + 0.05 \u00f7 12 ) y<sup>x<\/sup> 360 =<\/p>\n<\/div>\n<h2>Solving For Time<\/h2>\n<div class=\"textbox shaded\">Note: This section assumes you\u2019ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.<\/div>\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<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q610603\">Show Solution<\/span><\/p>\n<div id=\"q610603\" class=\"hidden-answer\" style=\"display: none\">\n<p>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><em>P<sub>0<\/sub><\/em> = $2000<\/td>\n<td>the initial deposit<\/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>12 months in 1 year<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So our general equation is [latex]{{P}_{N}}=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{N\\times12}}[\/latex]. We also know that we want our ending amount to be double of $2000, which is $4000, so we\u2019re looking for <em>N<\/em> so that <em>P<sub>N<\/sub><\/em> = 4000. To solve this, we set our equation for <em>P<sub>N<\/sub><\/em> equal to 4000.<\/p>\n<p style=\"text-align: center;\">[latex]4000=2000{{\\left(1+\\frac{0.06}{12}\\right)}^{N\\times12}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 2000<\/p>\n<p style=\"text-align: center;\">[latex]2={{\\left(1.005\\right)}^{12N}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 To solve for the exponent, take the log of both sides<\/p>\n<p style=\"text-align: center;\">[latex]\\log\\left(2\\right)=\\log\\left({{\\left(1.005\\right)}^{12N}}\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\n<p style=\"text-align: center;\">[latex]\\log\\left(2\\right)=12N\\log\\left(1.005\\right)[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now we can divide both sides by 12log(1.005)<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\log\\left(2\\right)}{12\\log\\left(1.005\\right)}=N[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Approximating this to a decimal<\/p>\n<p><em>N<\/em> = 11.581<\/p>\n<p>It will take about 11.581 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 log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.<\/p>\n<\/div>\n<\/div>\n<p>Get additional guidance for this example in the following:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" 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<\/div>\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-380\">\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>Question ID 6692. <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><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>Math in Society. <strong>Authored by<\/strong>: Open Textbook Store, Transition Math Project, and the Open Course Library. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <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>achievement-bar-business-chart-18134. <strong>Authored by<\/strong>: PublicDomainPictures. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/achievement-bar-business-chart-18134\/\">https:\/\/pixabay.com\/en\/achievement-bar-business-chart-18134\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Compound Interest. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xuQTFmP9nNg\">https:\/\/youtu.be\/xuQTFmP9nNg<\/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>Compound interest CD example. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/8NazxAjhpJw\">https:\/\/youtu.be\/8NazxAjhpJw<\/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>Compound interest - the importance of rounding. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VhhYtaMN6mo\">https:\/\/youtu.be\/VhhYtaMN6mo<\/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 6693. <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 <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":20,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Open Textbook Store, Transition Math Project, and the Open Course Library\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"achievement-bar-business-chart-18134\",\"author\":\"PublicDomainPictures\",\"organization\":\"\",\"url\":\"https:\/\/pixabay.com\/en\/achievement-bar-business-chart-18134\/\",\"project\":\"\",\"license\":\"cc0\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Compound Interest\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xuQTFmP9nNg\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Compound interest CD example\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/8NazxAjhpJw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Compound interest - 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