{"id":2335,"date":"2021-09-29T16:29:05","date_gmt":"2021-09-29T16:29:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-power-series-and-functions\/"},"modified":"2024-01-03T15:30:31","modified_gmt":"2024-01-03T15:30:31","slug":"skills-review-for-power-series-and-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-power-series-and-functions\/","title":{"raw":"Skills Review for Power Series and Functions","rendered":"Skills Review for Power Series and Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use summation notation<\/li>\r\n \t<li>Apply factorial notation<\/li>\r\n \t<li>Simplify expressions using the Product Property of Exponents<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Power Series and Functions section, we will look at power series and how they basically represent infinite polynomials. Here we will review how to expand sigma (summation) notation, apply factorial notation, and use the product rule for exponents.\r\n<h2>Expand Sigma (Summation) Notation<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Approximating Areas)<\/em><\/strong>\r\n\r\n<strong>Summation notation <\/strong>is used to represent long sums of values in a compact form. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>\u00a0to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms of the sum. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term of the sum. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term of the sum.\r\n\r\nIf we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{i}=2i[\/latex] for [latex]i=1[\/latex] through [latex]i=5[\/latex]. We can begin by substituting the terms for [latex]i[\/latex] and listing out the terms.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {a}_{1}=2\\left(1\\right)=2 \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/div>\r\nWe can find the sum by adding the terms:\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{5}2i=2+4+6+8+10=30[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Summation Notation<\/h3>\r\nThe sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{n}{a}_{i}[\/latex]<\/p>\r\nThis notation tells us to find the sum of [latex]{a}_{i}[\/latex] from [latex]i=1[\/latex] to [latex]i=n[\/latex].\r\n\r\n[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: EXpanding Summation Notation<\/h3>\r\nEvaluate [latex]\\displaystyle\\sum _{i=3}^{7}{i}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"14937\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14937\"]\r\n\r\nAccording to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{i}^{2}[\/latex] from [latex]i=3[\/latex] to [latex]i=7[\/latex]. We find the terms of the series by substituting [latex]i=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{i}^{2}[\/latex]. We add the terms to find the sum.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sum _{i=3}^{7}{i}^{2}&amp; ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill &amp; =9+16+25+36+49\\hfill \\\\ \\hfill &amp; =135\\hfill \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\displaystyle\\sum _{i=2}^{5}\\left(3i - 1\\right)[\/latex].\r\n\r\n[reveal-answer q=\"812548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812548\"]\r\n\r\n38\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]222190[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Apply Factorial Notation<\/h2>\r\n<strong><em>(also in Module 5, Skills Review for Alternating Series and Ratio and Root Tests)<\/em><\/strong>\r\n\r\nRecall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&amp;=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&amp;=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nAn example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nThe factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]\r\n<div class=\"textbox\">\r\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\r\n<strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\r\nThe special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&amp;amp;theme=oea&amp;amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex](n+3)![\/latex].\r\n[reveal-answer q=\"953701\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953701\"]\r\n\r\n[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]\r\n\r\nNotice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use the Product Rule for Exponents<\/h2>\r\n<strong><em>(also in Module 5, Skills Review for Alternating Series and Ratio and Root Tests)<\/em><\/strong>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Product Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product Rule<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"878162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"878162\"]\r\nUse the product rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\nAt first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\r\nNotice we get the same result by adding the three exponents in one step.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>RECALL<\/h3>\r\nFor any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that\r\n<center>[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/center> \u00a0\r\n\r\nFor an expression like\u00a0[latex](a^2)^3[\/latex], you have a base of\u00a0[latex]a[\/latex] raised to the power of [latex]2[\/latex], which is then raised to another power of [latex]3[\/latex]. Multiply the exponents\u00a0[latex]2[\/latex]\u00a0and\u00a0[latex]3[\/latex] to find the new exponent for [latex]a[\/latex]. This gives you\u00a0[latex]a^{2\\cdot3}[\/latex] or\u00a0[latex]a^6[\/latex]. Always remember: when an exponent is raised to another exponent, multiply the exponents to simplify the expression.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex].\r\n[reveal-answer q=\"953721\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953721\"]\r\n\r\n[latex](xy)^n=x^n \\cdot y^n[\/latex], so we apply the exponent [latex]3[\/latex] to both [latex]3a^2[\/latex] and [latex]b[\/latex] within the parentheses.\r\n\r\n<center>[latex](3a^2b)^3 = 3^3 \\cdot (a^2)^3 \\cdot b^3[\/latex]<\/center>\r\nNow we'll compute the powers individually.\r\n\r\n<center>[latex]\\begin{array}{l}\r\n3^3 = 27 \\\\\r\n(a^2)^3 = a^{2\\cdot3} = a^6 \\\\\r\nb^3 = b^3 \\text{ (no change since it's already in exponent form)}\r\n\\end{array}[\/latex]<\/center>\r\nPut together the results of the individual powers.\r\n\r\n<center>[latex](3a^2b)^3=27 \\cdot a^6 \\cdot b^3[\/latex]<\/center>\r\nNow we multiply the result by the second term [latex]2ab^4[\/latex]. Since we are dealing with multiplication, we can multiply coefficients (the numerical parts) and variables with the same base separately.\r\n\r\n<center>[latex](27\u22c5a^6 \\cdot b^3) \\cdot (2ab^4) = (27 \\cdot 2) \\cdot (a^6 \\cdot a) \\cdot (b^3 \\cdot b^4)[\/latex]<\/center>\r\nMultiply the coefficients and add the exponents for the variables with the same base.\r\n\r\n<center>[latex]\\begin{array}{l}\r\n27 \\cdot 2 = 54 \\\\\r\na^6 \\cdot a = a^{6+1} = a^7 \\\\\r\nb^3 \\cdot b^4 = b^{3+4} = b^7\r\n\\end{array}[\/latex]<\/center>\r\nFinally, combine all the simplified terms to get the final answer.\r\n\r\n<center>[latex]54 \\cdot a^7 \\cdot b^7[\/latex]<\/center>\r\nSo, the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex] simplifies to [latex]54a^7b^7[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify the expression [latex](2y^2)^3 \\cdot (4y^5).[\/latex]\r\n[reveal-answer q=\"953799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953799\"]\r\n\r\nFirst, apply the power to the first term, [latex](2y^2)^3[\/latex], by multiplying the exponent outside the parenthesis with each exponent inside the parenthesis:\r\n\r\n<center>[latex](2y^2)^3=2^3 \\cdot (y^2)^3=8y^6[\/latex]<\/center>\r\nNow, multiply this result by [latex]4y^5[\/latex]:\r\n\r\n<center>[latex]8y^6 \\cdot 4y^5 = (8 \\cdot 4) \\cdot y^{6+5} = 32y^{11}[\/latex]<\/center>\r\nTherefore, the expression simplifies to [latex]32y^{11}[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use summation notation<\/li>\n<li>Apply factorial notation<\/li>\n<li>Simplify expressions using the Product Property of Exponents<\/li>\n<\/ul>\n<\/div>\n<p>In the Power Series and Functions section, we will look at power series and how they basically represent infinite polynomials. Here we will review how to expand sigma (summation) notation, apply factorial notation, and use the product rule for exponents.<\/p>\n<h2>Expand Sigma (Summation) Notation<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Approximating Areas)<\/em><\/strong><\/p>\n<p><strong>Summation notation <\/strong>is used to represent long sums of values in a compact form. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>\u00a0to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms of the sum. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term of the sum. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term of the sum.<\/p>\n<p>If we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{i}=2i[\/latex] for [latex]i=1[\/latex] through [latex]i=5[\/latex]. We can begin by substituting the terms for [latex]i[\/latex] and listing out the terms.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {a}_{1}=2\\left(1\\right)=2 \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/div>\n<p>We can find the sum by adding the terms:<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{5}2i=2+4+6+8+10=30[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Summation Notation<\/h3>\n<p>The sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{i=1}^{n}{a}_{i}[\/latex]<\/p>\n<p>This notation tells us to find the sum of [latex]{a}_{i}[\/latex] from [latex]i=1[\/latex] to [latex]i=n[\/latex].<\/p>\n<p>[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: EXpanding Summation Notation<\/h3>\n<p>Evaluate [latex]\\displaystyle\\sum _{i=3}^{7}{i}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14937\">Show Solution<\/span><\/p>\n<div id=\"q14937\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{i}^{2}[\/latex] from [latex]i=3[\/latex] to [latex]i=7[\/latex]. We find the terms of the series by substituting [latex]i=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{i}^{2}[\/latex]. We add the terms to find the sum.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sum _{i=3}^{7}{i}^{2}& ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill & =9+16+25+36+49\\hfill \\\\ \\hfill & =135\\hfill \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\displaystyle\\sum _{i=2}^{5}\\left(3i - 1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812548\">Show Solution<\/span><\/p>\n<div id=\"q812548\" class=\"hidden-answer\" style=\"display: none\">\n<p>38<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm222190\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=222190&theme=oea&iframe_resize_id=ohm222190\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Apply Factorial Notation<\/h2>\n<p><strong><em>(also in Module 5, Skills Review for Alternating Series and Ratio and Root Tests)<\/em><\/strong><\/p>\n<p>Recall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>An example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>The factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\n<p><strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\n<p>The special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm61071\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&#38;theme=oea&#38;iframe_resize_id=ohm61071&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex](n+3)![\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953701\">Show Solution<\/span><\/p>\n<div id=\"q953701\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]<\/p>\n<p>Notice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use the Product Rule for Exponents<\/h2>\n<p><strong><em>(also in Module 5, Skills Review for Alternating Series and Ratio and Root Tests)<\/em><\/strong><\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Product Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product Rule<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\n<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q878162\">Show Solution<\/span><\/p>\n<div id=\"q878162\" class=\"hidden-answer\" style=\"display: none\">\nUse the product rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\n<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<p>At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\n<p>Notice we get the same result by adding the three exponents in one step.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1961\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&#38;theme=oea&#38;iframe_resize_id=ohm1961&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>RECALL<\/h3>\n<p>For any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\n<p> \u00a0<\/p>\n<p>For an expression like\u00a0[latex](a^2)^3[\/latex], you have a base of\u00a0[latex]a[\/latex] raised to the power of [latex]2[\/latex], which is then raised to another power of [latex]3[\/latex]. Multiply the exponents\u00a0[latex]2[\/latex]\u00a0and\u00a0[latex]3[\/latex] to find the new exponent for [latex]a[\/latex]. This gives you\u00a0[latex]a^{2\\cdot3}[\/latex] or\u00a0[latex]a^6[\/latex]. Always remember: when an exponent is raised to another exponent, multiply the exponents to simplify the expression.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953721\">Show Solution<\/span><\/p>\n<div id=\"q953721\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](xy)^n=x^n \\cdot y^n[\/latex], so we apply the exponent [latex]3[\/latex] to both [latex]3a^2[\/latex] and [latex]b[\/latex] within the parentheses.<\/p>\n<div style=\"text-align: center;\">[latex](3a^2b)^3 = 3^3 \\cdot (a^2)^3 \\cdot b^3[\/latex]<\/div>\n<p>Now we&#8217;ll compute the powers individually.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}  3^3 = 27 \\\\  (a^2)^3 = a^{2\\cdot3} = a^6 \\\\  b^3 = b^3 \\text{ (no change since it's already in exponent form)}  \\end{array}[\/latex]<\/div>\n<p>Put together the results of the individual powers.<\/p>\n<div style=\"text-align: center;\">[latex](3a^2b)^3=27 \\cdot a^6 \\cdot b^3[\/latex]<\/div>\n<p>Now we multiply the result by the second term [latex]2ab^4[\/latex]. Since we are dealing with multiplication, we can multiply coefficients (the numerical parts) and variables with the same base separately.<\/p>\n<div style=\"text-align: center;\">[latex](27\u22c5a^6 \\cdot b^3) \\cdot (2ab^4) = (27 \\cdot 2) \\cdot (a^6 \\cdot a) \\cdot (b^3 \\cdot b^4)[\/latex]<\/div>\n<p>Multiply the coefficients and add the exponents for the variables with the same base.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}  27 \\cdot 2 = 54 \\\\  a^6 \\cdot a = a^{6+1} = a^7 \\\\  b^3 \\cdot b^4 = b^{3+4} = b^7  \\end{array}[\/latex]<\/div>\n<p>Finally, combine all the simplified terms to get the final answer.<\/p>\n<div style=\"text-align: center;\">[latex]54 \\cdot a^7 \\cdot b^7[\/latex]<\/div>\n<p>So, the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex] simplifies to [latex]54a^7b^7[\/latex].\n<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify the expression [latex](2y^2)^3 \\cdot (4y^5).[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953799\">Show Solution<\/span><\/p>\n<div id=\"q953799\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, apply the power to the first term, [latex](2y^2)^3[\/latex], by multiplying the exponent outside the parenthesis with each exponent inside the parenthesis:<\/p>\n<div style=\"text-align: center;\">[latex](2y^2)^3=2^3 \\cdot (y^2)^3=8y^6[\/latex]<\/div>\n<p>Now, multiply this result by [latex]4y^5[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]8y^6 \\cdot 4y^5 = (8 \\cdot 4) \\cdot y^{6+5} = 32y^{11}[\/latex]<\/div>\n<p>Therefore, the expression simplifies to [latex]32y^{11}[\/latex].\n<\/p><\/div>\n<\/div>\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-2335\">\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>Modification and Revision. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/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>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen 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