{"id":1330,"date":"2016-10-21T23:07:11","date_gmt":"2016-10-21T23:07:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1330"},"modified":"2025-10-13T20:06:41","modified_gmt":"2025-10-13T20:06:41","slug":"zero-and-negative-exponents","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/chapter\/zero-and-negative-exponents\/","title":{"raw":"Zero and Negative Exponents","rendered":"Zero and Negative Exponents"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Simplify expressions with exponents equal to zero.<\/li>\r\n \t<li>Simplify expressions with negative exponents.<\/li>\r\n \t<li>Simplify exponential expressions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\nReturn to the quotient rule. We made the condition that [latex]m&gt;n[\/latex] so that the difference [latex]m-n[\/latex] would never be zero or negative. What would happen if [latex]m=n[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to 1. To see how this is done, let us begin with an example.\r\n<p style=\"text-align: center\">[latex]\\dfrac{t^{8}}{t^{8}}=\\dfrac{\\cancel{t^{8}}}{\\cancel{t^{8}}}=1[\/latex]<\/p>\r\nIf we were to simplify the original expression using the quotient rule, we would have\r\n<div style=\"text-align: center\">[latex]\\dfrac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[\/latex]<\/div>\r\nIf we equate the two answers, the result is [latex]{t}^{0}=1[\/latex]. This is true for any nonzero real number, or any variable representing a real number.\r\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\r\nThe sole exception is the expression [latex]{0}^{0}[\/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Zero Exponent Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex], the zero exponent rule of exponents states that\r\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Zero Exponent Rule<\/h3>\r\nSimplify each expression using the zero exponent rule of exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"913171\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"913171\"]\r\nUse the zero exponent and other rules to simplify each expression.\r\n<ol>\r\n \t<li>[latex]\\begin{align}\\frac{c^{3}}{c^{3}} &amp; =c^{3-3} \\\\ &amp; =c^{0} \\\\ &amp; =1\\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} \\frac{-3{x}^{5}}{{x}^{5}}&amp; = -3\\cdot \\frac{{x}^{5}}{{x}^{5}} \\\\ &amp; = -3\\cdot {x}^{5 - 5} \\\\ &amp; = -3\\cdot {x}^{0} \\\\ &amp; = -3\\cdot 1 \\\\ &amp; = -3 \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}&amp; = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}} &amp;&amp; \\text{Use the product rule in the denominator}. \\\\ &amp; = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}} &amp;&amp; \\text{Simplify}. \\\\ &amp; = {\\left({j}^{2}k\\right)}^{4 - 4} &amp;&amp; \\text{Use the quotient rule}. \\\\ &amp; = {\\left({j}^{2}k\\right)}^{0} &amp;&amp; \\text{Simplify}. \\\\ &amp; = 1 \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}&amp; = 5{\\left(r{s}^{2}\\right)}^{2 - 2} &amp;&amp; \\text{Use the quotient rule}. \\\\ &amp; = 5{\\left(r{s}^{2}\\right)}^{0} &amp;&amp; \\text{Simplify}. \\\\ &amp; = 5\\cdot 1 &amp;&amp; \\text{Use the zero exponent rule}. \\\\ &amp; = 5 &amp;&amp; \\text{Simplify}. \\end{align}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify each expression using the zero exponent rule of exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{t}^{7}}{{t}^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(d{e}^{2}\\right)}^{11}}{2{\\left(d{e}^{2}\\right)}^{11}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{w}^{4}\\cdot {w}^{2}}{{w}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{t}^{3}\\cdot {t}^{4}}{{t}^{2}\\cdot {t}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"703483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703483\"]\r\n<ol>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=44120&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"450\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7833&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nIn this video we show more examples of how to simplify expressions with zero exponents.\r\n\r\nhttps:\/\/youtu.be\/rpoUg32utlc\r\n<h2>Using the Negative Rule of Exponents<\/h2>\r\nAnother useful result occurs if we relax the condition that [latex]m&gt;n[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\dfrac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]m&lt;n[\/latex]\u2014that is, where the difference [latex]m-n[\/latex] is negative\u2014we can use the <em>negative rule of exponents<\/em> to simplify the expression to its reciprocal.\r\n\r\nDivide one exponential expression by another with a larger exponent. Use our example, [latex]\\dfrac{{h}^{3}}{{h}^{5}}[\/latex].\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\frac{{h}^{3}}{{h}^{5}}&amp; = \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h} \\\\ &amp; = \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h} \\\\ &amp; = \\frac{1}{h\\cdot h} \\\\ &amp; = \\frac{1}{{h}^{2}} \\end{align}[\/latex]<\/div>\r\nIf we were to simplify the original expression using the quotient rule, we would have\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\frac{{h}^{3}}{{h}^{5}}&amp; = {h}^{3 - 5} \\\\ &amp; = {h}^{-2} \\end{align}[\/latex]<\/div>\r\nPutting the answers together, we have [latex]{h}^{-2}=\\dfrac{1}{{h}^{2}}[\/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number.\r\n\r\nA factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from numerator to denominator or vice versa.\r\n<div style=\"text-align: center\">[latex]{a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}[\/latex]<\/div>\r\nWe have shown that the exponential expression [latex]{a}^{n}[\/latex] is defined when [latex]n[\/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[\/latex] is defined for any integer [latex]n[\/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Negative Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that\r\n<div style=\"text-align: center\">[latex]{a}^{-n}=\\dfrac{1}{{a}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Negative Exponent Rule<\/h3>\r\nWrite each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\theta }^{3}}{{\\theta }^{10}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"746940\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746940\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\theta }^{3}}{{\\theta }^{10}}={\\theta }^{3 - 10}={\\theta }^{-7}=\\dfrac{1}{{\\theta }^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}=\\dfrac{{z}^{2+1}}{{z}^{4}}=\\dfrac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\dfrac{1}{z}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}={\\left(-5{t}^{3}\\right)}^{4 - 8}={\\left(-5{t}^{3}\\right)}^{-4}=\\dfrac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]\\dfrac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"85205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"85205\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{1}{{\\left(-3t\\right)}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{f}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{2}{5{k}^{3}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=51959&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109762&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109765&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nWatch this video to see more examples of simplifying expressions with negative exponents.\r\nhttps:\/\/youtu.be\/Gssi4dBtAEI\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product and Quotient Rules<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"803639\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"803639\"]\r\n<ol>\r\n \t<li>[latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}={\\left(-x\\right)}^{5 - 5}={\\left(-x\\right)}^{0}=1[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}=\\dfrac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"592096\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"592096\"]\r\n<ol>\r\n \t<li>[latex]{t}^{-5}=\\dfrac{1}{{t}^{5}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{25}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93393&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Finding the Power of a Product<\/h2>\r\nTo simplify the power of a product of two exponential expressions, we can use the <em>power of a product rule of exponents,<\/em> which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider [latex]{\\left(pq\\right)}^{3}[\/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.\r\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left(pq\\right)}^{3}&amp; = \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}} \\\\ &amp; = p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q \\\\ &amp; = \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}} \\\\ &amp; = {p}^{3}\\cdot {q}^{3} \\end{align}[\/latex]<\/div>\r\nIn other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power of a Product Rule of Exponents<\/h3>\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a product rule of exponents states that\r\n<div style=\"text-align: center\">[latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power of a Product Rule<\/h3>\r\nSimplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"189543\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"189543\"]\r\nUse the product and quotient rules and the new definitions to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{\\left(a{b}^{2}\\right)}^{3}={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}={a}^{3}{b}^{6}[\/latex]<\/li>\r\n \t<li>[latex]2{t}^{15}={\\left(2\\right)}^{15}\\cdot {\\left(t\\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}=-8\\cdot {w}^{3\\cdot 3}=-8{w}^{9}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}=\\dfrac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\dfrac{1}{2,401{z}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}={e}^{-14}{f}^{14}=\\dfrac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(5t\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"540817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"540817\"]\r\n<ol>\r\n \t<li>[latex]{g}^{10}{h}^{15}[\/latex]<\/li>\r\n \t<li>[latex]125{t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]-27{y}^{15}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{a}^{18}{b}^{21}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{r}^{12}}{{s}^{8}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14047&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14058&amp;theme=oea&amp;iframe_resize_id=mom30\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14059&amp;theme=oea&amp;iframe_resize_id=mom40\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to find hte power of a product.\r\n\r\nhttps:\/\/youtu.be\/p-2UkpJQWpo\r\n<h2>Finding the Power of a Quotient<\/h2>\r\nTo simplify the power of a quotient of two expressions, we can use the <em>power of a quotient rule,<\/em> which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let\u2019s look at the following example.\r\n<div style=\"text-align: center\">\r\n<div style=\"text-align: center\">[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\dfrac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\r\nLet\u2019s rewrite the original problem differently and look at the result.\r\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left({e}^{-2}{f}^{2}\\right)}^{7}&amp; = {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7} \\\\[1mm] &amp; = \\frac{{f}^{14}}{{e}^{14}} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\nIt appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.\r\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left({e}^{-2}{f}^{2}\\right)}^{7}&amp; = {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7} \\\\[1mm] &amp; = \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}} \\\\[1mm] &amp; = \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}} \\\\[1mm] &amp; = \\frac{{f}^{14}}{{e}^{14}} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power of a Quotient Rule of Exponents<\/h3>\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a quotient rule of exponents states that\r\n<div style=\"text-align: center\">[latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Power of a Quotient Rule<\/h3>\r\nSimplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"835767\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835767\"]\r\n<ol>\r\n \t<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}=\\dfrac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\dfrac{64}{{z}^{11\\cdot 3}}=\\dfrac{64}{{z}^{33}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}=\\dfrac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\dfrac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\dfrac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}=\\dfrac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\\dfrac{-1}{{t}^{2\\cdot 27}}=\\dfrac{-1}{{t}^{54}}=-\\dfrac{1}{{t}^{54}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\dfrac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\dfrac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\dfrac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\dfrac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\dfrac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\dfrac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\dfrac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\dfrac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\dfrac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.\r\n<ol>\r\n \t<li>[latex]{\\left(\\dfrac{{b}^{5}}{c}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{5}{{u}^{8}}\\right)}^{4}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{-1}{{w}^{3}}\\right)}^{35}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"815731\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815731\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{b}^{15}}{{c}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{625}{{u}^{32}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{-1}{{w}^{105}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{q}^{24}}{{p}^{32}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{c}^{20}{d}^{12}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14046&amp;theme=oea&amp;iframe_resize_id=mom50\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14051&amp;theme=oea&amp;iframe_resize_id=mom60\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43231&amp;theme=oea&amp;iframe_resize_id=mom60\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Simplifying Exponential Expressions<\/h2>\r\nRecall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Simplifying Exponential Expressions<\/h3>\r\nSimplify each expression and write the answer with positive exponents only.\r\n<ol>\r\n \t<li>[latex]{\\left(6{m}^{2}{n}^{-1}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"9384\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"9384\"]\r\n<ol>\r\n \t<li>[latex]\\begin{align} {\\left(6{m}^{2}{n}^{-1}\\right)}^{3}&amp; = {\\left(6\\right)}^{3}{\\left({m}^{2}\\right)}^{3}{\\left({n}^{-1}\\right)}^{3}&amp;&amp; \\text{The power of a product rule} \\\\ &amp; = {6}^{3}{m}^{2\\cdot 3}{n}^{-1\\cdot 3}&amp;&amp; \\text{The power rule} \\\\ &amp; = 216{m}^{6}{n}^{-3}&amp;&amp; \\text{Simplify}. \\\\ &amp; = \\frac{216{m}^{6}}{{n}^{3}}&amp;&amp; \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} {17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}&amp; =&amp; {17}^{5 - 4-3}&amp;&amp; \\text{The product rule} \\\\ &amp; = {17}^{-2}&amp;&amp; \\text{Simplify}. \\\\ &amp; = \\frac{1}{{17}^{2}}\\text{ or }\\frac{1}{289}&amp;&amp; \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} {\\left(\\frac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}&amp; = \\frac{{\\left({u}^{-1}v\\right)}^{2}}{{\\left({v}^{-1}\\right)}^{2}}&amp;&amp; \\text{The power of a quotient rule} \\\\ &amp; = \\frac{{u}^{-2}{v}^{2}}{{v}^{-2}}&amp;&amp; \\text{The power of a product rule} \\\\ &amp; = {u}^{-2}{v}^{2-\\left(-2\\right)}&amp;&amp; \\text{The quotient rule} \\\\ &amp; = {u}^{-2}{v}^{4}&amp;&amp; \\text{Simplify}. \\\\ &amp; = \\frac{{v}^{4}}{{u}^{2}}&amp;&amp; \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} \\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)&amp; =&amp; -2\\cdot 5\\cdot {a}^{3}\\cdot {a}^{-2}\\cdot {b}^{-1}\\cdot {b}^{2}&amp;&amp; \\text{Commutative and associative laws of multiplication} \\\\ &amp; = -10\\cdot {a}^{3 - 2}\\cdot {b}^{-1+2}&amp;&amp; \\text{The product rule} \\\\ &amp; = -10ab&amp;&amp; \\text{Simplify}. \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} {\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}&amp; = {\\left({x}^{2}\\sqrt{2}\\right)}^{4 - 4} &amp;&amp; \\text{The product rule} \\\\ &amp; = {\\left({x}^{2}\\sqrt{2}\\right)}^{0}&amp;&amp; \\text{Simplify}. \\\\ &amp; = 1&amp;&amp; \\text{The zero exponent rule} \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align} \\frac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}&amp; = \\frac{{\\left(3\\right)}^{5}\\cdot {\\left({w}^{2}\\right)}^{5}}{{\\left(6\\right)}^{2}\\cdot {\\left({w}^{-2}\\right)}^{2}}&amp;&amp; \\text{The power of a product rule} \\\\ &amp; = \\frac{{3}^{5}{w}^{2\\cdot 5}}{{6}^{2}{w}^{-2\\cdot 2}}&amp;&amp; \\text{The power rule} \\\\ &amp; = \\frac{243{w}^{10}}{36{w}^{-4}} &amp;&amp; \\text{Simplify}. \\\\ &amp; = \\frac{27{w}^{10-\\left(-4\\right)}}{4}&amp;&amp; \\text{The quotient rule and reduce fraction} \\\\ &amp; = \\frac{27{w}^{14}}{4}&amp;&amp; \\text{Simplify}. \\end{align}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify each expression and write the answer with positive exponents only.\r\n<ol>\r\n \t<li>[latex]{\\left(2u{v}^{-2}\\right)}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{8}\\cdot {x}^{-12}\\cdot x[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\\right)}^{2}[\/latex]<\/li>\r\n \t<li>[latex]\\left(9{r}^{-5}{s}^{3}\\right)\\left(3{r}^{6}{s}^{-4}\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\left(\\frac{4}{9}t{w}^{-2}\\right)}^{-3}{\\left(\\frac{4}{9}t{w}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{\\left(2{h}^{2}k\\right)}^{4}}{{\\left(7{h}^{-1}{k}^{2}\\right)}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"31244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"31244\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{v}^{6}}{8{u}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{1}{{x}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{e}^{4}}{{f}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{27r}{s}[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{16{h}^{10}}{49}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&amp;theme=oea&amp;iframe_resize_id=mom70\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&amp;theme=oea&amp;iframe_resize_id=mom80\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&amp;theme=oea&amp;iframe_resize_id=mom90\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&amp;theme=oea&amp;iframe_resize_id=mom95\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to find the power of a quotient.\r\nhttps:\/\/youtu.be\/BoBe31pRxFM","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Simplify expressions with exponents equal to zero.<\/li>\n<li>Simplify expressions with negative exponents.<\/li>\n<li>Simplify exponential expressions.<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\nReturn to the quotient rule. We made the condition that [latex]m>n[\/latex] so that the difference [latex]m-n[\/latex] would never be zero or negative. What would happen if [latex]m=n[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to 1. To see how this is done, let us begin with an example.<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{t^{8}}{t^{8}}=\\dfrac{\\cancel{t^{8}}}{\\cancel{t^{8}}}=1[\/latex]<\/p>\n<p>If we were to simplify the original expression using the quotient rule, we would have<\/p>\n<div style=\"text-align: center\">[latex]\\dfrac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}[\/latex]<\/div>\n<p>If we equate the two answers, the result is [latex]{t}^{0}=1[\/latex]. This is true for any nonzero real number, or any variable representing a real number.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\n<p>The sole exception is the expression [latex]{0}^{0}[\/latex]. This appears later in more advanced courses, but for now, we will consider the value to be undefined.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Zero Exponent Rule of Exponents<\/h3>\n<p>For any nonzero real number [latex]a[\/latex], the zero exponent rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]{a}^{0}=1[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Zero Exponent Rule<\/h3>\n<p>Simplify each expression using the zero exponent rule of exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{c}^{3}}{{c}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{-3{x}^{5}}{{x}^{5}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q913171\">Show Solution<\/span><\/p>\n<div id=\"q913171\" class=\"hidden-answer\" style=\"display: none\">\nUse the zero exponent and other rules to simplify each expression.<\/p>\n<ol>\n<li>[latex]\\begin{align}\\frac{c^{3}}{c^{3}} & =c^{3-3} \\\\ & =c^{0} \\\\ & =1\\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} \\frac{-3{x}^{5}}{{x}^{5}}& = -3\\cdot \\frac{{x}^{5}}{{x}^{5}} \\\\ & = -3\\cdot {x}^{5 - 5} \\\\ & = -3\\cdot {x}^{0} \\\\ & = -3\\cdot 1 \\\\ & = -3 \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}& = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}} && \\text{Use the product rule in the denominator}. \\\\ & = \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}} && \\text{Simplify}. \\\\ & = {\\left({j}^{2}k\\right)}^{4 - 4} && \\text{Use the quotient rule}. \\\\ & = {\\left({j}^{2}k\\right)}^{0} && \\text{Simplify}. \\\\ & = 1 \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}& = 5{\\left(r{s}^{2}\\right)}^{2 - 2} && \\text{Use the quotient rule}. \\\\ & = 5{\\left(r{s}^{2}\\right)}^{0} && \\text{Simplify}. \\\\ & = 5\\cdot 1 && \\text{Use the zero exponent rule}. \\\\ & = 5 && \\text{Simplify}. \\end{align}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify each expression using the zero exponent rule of exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{t}^{7}}{{t}^{7}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(d{e}^{2}\\right)}^{11}}{2{\\left(d{e}^{2}\\right)}^{11}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{w}^{4}\\cdot {w}^{2}}{{w}^{6}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{t}^{3}\\cdot {t}^{4}}{{t}^{2}\\cdot {t}^{5}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703483\">Show Solution<\/span><\/p>\n<div id=\"q703483\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{2}[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=44120&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"450\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7833&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>In this video we show more examples of how to simplify expressions with zero exponents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify Expressions Using the Quotient and Zero Exponent Rules\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/rpoUg32utlc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Using the Negative Rule of Exponents<\/h2>\n<p>Another useful result occurs if we relax the condition that [latex]m>n[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\dfrac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]m<n[\/latex]\u2014that is, where the difference [latex]m-n[\/latex] is negative\u2014we can use the <em>negative rule of exponents<\/em> to simplify the expression to its reciprocal.<\/p>\n<p>Divide one exponential expression by another with a larger exponent. Use our example, [latex]\\dfrac{{h}^{3}}{{h}^{5}}[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} \\frac{{h}^{3}}{{h}^{5}}& = \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h} \\\\ & = \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h} \\\\ & = \\frac{1}{h\\cdot h} \\\\ & = \\frac{1}{{h}^{2}} \\end{align}[\/latex]<\/div>\n<p>If we were to simplify the original expression using the quotient rule, we would have<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} \\frac{{h}^{3}}{{h}^{5}}& = {h}^{3 - 5} \\\\ & = {h}^{-2} \\end{align}[\/latex]<\/div>\n<p>Putting the answers together, we have [latex]{h}^{-2}=\\dfrac{1}{{h}^{2}}[\/latex]. This is true for any nonzero real number, or any variable representing a nonzero real number.<\/p>\n<p>A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar\u2014from numerator to denominator or vice versa.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}[\/latex]<\/div>\n<p>We have shown that the exponential expression [latex]{a}^{n}[\/latex] is defined when [latex]n[\/latex] is a natural number, 0, or the negative of a natural number. That means that [latex]{a}^{n}[\/latex] is defined for any integer [latex]n[\/latex]. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer [latex]n[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Negative Rule of Exponents<\/h3>\n<p>For any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]{a}^{-n}=\\dfrac{1}{{a}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Negative Exponent Rule<\/h3>\n<p>Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{\\theta }^{3}}{{\\theta }^{10}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q746940\">Show Solution<\/span><\/p>\n<div id=\"q746940\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{\\theta }^{3}}{{\\theta }^{10}}={\\theta }^{3 - 10}={\\theta }^{-7}=\\dfrac{1}{{\\theta }^{7}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}=\\dfrac{{z}^{2+1}}{{z}^{4}}=\\dfrac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\dfrac{1}{z}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}={\\left(-5{t}^{3}\\right)}^{4 - 8}={\\left(-5{t}^{3}\\right)}^{-4}=\\dfrac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]\\dfrac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]<\/li>\n<li>[latex]\\dfrac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q85205\">Show Solution<\/span><\/p>\n<div id=\"q85205\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{1}{{\\left(-3t\\right)}^{6}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{f}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{2}{5{k}^{3}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=51959&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109762&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109765&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>Watch this video to see more examples of simplifying expressions with negative exponents.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-2\" title=\"Simplify Expressions Using the Quotient and Negative Exponent Rules\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Gssi4dBtAEI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product and Quotient Rules<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\n<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\n<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q803639\">Show Solution<\/span><\/p>\n<div id=\"q803639\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\n<li>[latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}={\\left(-x\\right)}^{5 - 5}={\\left(-x\\right)}^{0}=1[\/latex]<\/li>\n<li>[latex]\\dfrac{-7z}{{\\left(-7z\\right)}^{5}}=\\dfrac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\n<li>[latex]\\dfrac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q592096\">Show Solution<\/span><\/p>\n<div id=\"q592096\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{t}^{-5}=\\dfrac{1}{{t}^{5}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{25}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93393&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Finding the Power of a Product<\/h2>\n<p>To simplify the power of a product of two exponential expressions, we can use the <em>power of a product rule of exponents,<\/em> which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider [latex]{\\left(pq\\right)}^{3}[\/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left(pq\\right)}^{3}& = \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}} \\\\ & = p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q \\\\ & = \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}} \\\\ & = {p}^{3}\\cdot {q}^{3} \\end{align}[\/latex]<\/div>\n<p>In other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Power of a Product Rule of Exponents<\/h3>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a product rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]\\large{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power of a Product Rule<\/h3>\n<p>Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q189543\">Show Solution<\/span><\/p>\n<div id=\"q189543\" class=\"hidden-answer\" style=\"display: none\">\nUse the product and quotient rules and the new definitions to simplify each expression.<\/p>\n<ol>\n<li>[latex]{\\left(a{b}^{2}\\right)}^{3}={\\left(a\\right)}^{3}\\cdot {\\left({b}^{2}\\right)}^{3}={a}^{1\\cdot 3}\\cdot {b}^{2\\cdot 3}={a}^{3}{b}^{6}[\/latex]<\/li>\n<li>[latex]2{t}^{15}={\\left(2\\right)}^{15}\\cdot {\\left(t\\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[\/latex]<\/li>\n<li>[latex]{\\left(-2{w}^{3}\\right)}^{3}={\\left(-2\\right)}^{3}\\cdot {\\left({w}^{3}\\right)}^{3}=-8\\cdot {w}^{3\\cdot 3}=-8{w}^{9}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{\\left(-7z\\right)}^{4}}=\\dfrac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\dfrac{1}{2,401{z}^{4}}[\/latex]<\/li>\n<li>[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}={\\left({e}^{-2}\\right)}^{7}\\cdot {\\left({f}^{2}\\right)}^{7}={e}^{-2\\cdot 7}\\cdot {f}^{2\\cdot 7}={e}^{-14}{f}^{14}=\\dfrac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]<\/li>\n<li>[latex]{\\left(5t\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]<\/li>\n<li>[latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q540817\">Show Solution<\/span><\/p>\n<div id=\"q540817\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{g}^{10}{h}^{15}[\/latex]<\/li>\n<li>[latex]125{t}^{3}[\/latex]<\/li>\n<li>[latex]-27{y}^{15}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{a}^{18}{b}^{21}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{r}^{12}}{{s}^{8}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14047&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14058&amp;theme=oea&amp;iframe_resize_id=mom30\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14059&amp;theme=oea&amp;iframe_resize_id=mom40\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show more examples of how to find hte power of a product.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Simplify Expressions Using Exponent Rules (Power of a Product)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/p-2UkpJQWpo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Finding the Power of a Quotient<\/h2>\n<p>To simplify the power of a quotient of two expressions, we can use the <em>power of a quotient rule,<\/em> which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let\u2019s look at the following example.<\/p>\n<div style=\"text-align: center\">\n<div style=\"text-align: center\">[latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\dfrac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\n<p>Let\u2019s rewrite the original problem differently and look at the result.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left({e}^{-2}{f}^{2}\\right)}^{7}& = {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7} \\\\[1mm] & = \\frac{{f}^{14}}{{e}^{14}} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<p>It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} {\\left({e}^{-2}{f}^{2}\\right)}^{7}& = {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7} \\\\[1mm] & = \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}} \\\\[1mm] & = \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}} \\\\[1mm] & = \\frac{{f}^{14}}{{e}^{14}} \\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Power of a Quotient Rule of Exponents<\/h3>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex], the power of a quotient rule of exponents states that<\/p>\n<div style=\"text-align: center\">[latex]\\large{\\left(\\dfrac{a}{b}\\right)}^{n}=\\dfrac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Power of a Quotient Rule<\/h3>\n<p>Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\n<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\n<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q835767\">Show Solution<\/span><\/p>\n<div id=\"q835767\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{\\left(\\dfrac{4}{{z}^{11}}\\right)}^{3}=\\dfrac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\dfrac{64}{{z}^{11\\cdot 3}}=\\dfrac{64}{{z}^{33}}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{p}{{q}^{3}}\\right)}^{6}=\\dfrac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\dfrac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\dfrac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{-1}{{t}^{2}}\\right)}^{27}=\\dfrac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\\dfrac{-1}{{t}^{2\\cdot 27}}=\\dfrac{-1}{{t}^{54}}=-\\dfrac{1}{{t}^{54}}[\/latex]<\/li>\n<li>[latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\dfrac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\dfrac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\dfrac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\dfrac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\n<li>[latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\dfrac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\dfrac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\dfrac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\dfrac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\dfrac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.<\/p>\n<ol>\n<li>[latex]{\\left(\\dfrac{{b}^{5}}{c}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{5}{{u}^{8}}\\right)}^{4}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{-1}{{w}^{3}}\\right)}^{35}[\/latex]<\/li>\n<li>[latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]<\/li>\n<li>[latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815731\">Show Solution<\/span><\/p>\n<div id=\"q815731\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{b}^{15}}{{c}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{625}{{u}^{32}}[\/latex]<\/li>\n<li>[latex]\\dfrac{-1}{{w}^{105}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{q}^{24}}{{p}^{32}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{c}^{20}{d}^{12}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14046&amp;theme=oea&amp;iframe_resize_id=mom50\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14051&amp;theme=oea&amp;iframe_resize_id=mom60\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43231&amp;theme=oea&amp;iframe_resize_id=mom60\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Simplifying Exponential Expressions<\/h2>\n<p>Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Simplifying Exponential Expressions<\/h3>\n<p>Simplify each expression and write the answer with positive exponents only.<\/p>\n<ol>\n<li>[latex]{\\left(6{m}^{2}{n}^{-1}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}[\/latex]<\/li>\n<li>[latex]{\\left(\\dfrac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}[\/latex]<\/li>\n<li>[latex]\\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)[\/latex]<\/li>\n<li>[latex]{\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9384\">Show Solution<\/span><\/p>\n<div id=\"q9384\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\begin{align} {\\left(6{m}^{2}{n}^{-1}\\right)}^{3}& = {\\left(6\\right)}^{3}{\\left({m}^{2}\\right)}^{3}{\\left({n}^{-1}\\right)}^{3}&& \\text{The power of a product rule} \\\\ & = {6}^{3}{m}^{2\\cdot 3}{n}^{-1\\cdot 3}&& \\text{The power rule} \\\\ & = 216{m}^{6}{n}^{-3}&& \\text{Simplify}. \\\\ & = \\frac{216{m}^{6}}{{n}^{3}}&& \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} {17}^{5}\\cdot {17}^{-4}\\cdot {17}^{-3}& =& {17}^{5 - 4-3}&& \\text{The product rule} \\\\ & = {17}^{-2}&& \\text{Simplify}. \\\\ & = \\frac{1}{{17}^{2}}\\text{ or }\\frac{1}{289}&& \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} {\\left(\\frac{{u}^{-1}v}{{v}^{-1}}\\right)}^{2}& = \\frac{{\\left({u}^{-1}v\\right)}^{2}}{{\\left({v}^{-1}\\right)}^{2}}&& \\text{The power of a quotient rule} \\\\ & = \\frac{{u}^{-2}{v}^{2}}{{v}^{-2}}&& \\text{The power of a product rule} \\\\ & = {u}^{-2}{v}^{2-\\left(-2\\right)}&& \\text{The quotient rule} \\\\ & = {u}^{-2}{v}^{4}&& \\text{Simplify}. \\\\ & = \\frac{{v}^{4}}{{u}^{2}}&& \\text{The negative exponent rule} \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} \\left(-2{a}^{3}{b}^{-1}\\right)\\left(5{a}^{-2}{b}^{2}\\right)& =& -2\\cdot 5\\cdot {a}^{3}\\cdot {a}^{-2}\\cdot {b}^{-1}\\cdot {b}^{2}&& \\text{Commutative and associative laws of multiplication} \\\\ & = -10\\cdot {a}^{3 - 2}\\cdot {b}^{-1+2}&& \\text{The product rule} \\\\ & = -10ab&& \\text{Simplify}. \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} {\\left({x}^{2}\\sqrt{2}\\right)}^{4}{\\left({x}^{2}\\sqrt{2}\\right)}^{-4}& = {\\left({x}^{2}\\sqrt{2}\\right)}^{4 - 4} && \\text{The product rule} \\\\ & = {\\left({x}^{2}\\sqrt{2}\\right)}^{0}&& \\text{Simplify}. \\\\ & = 1&& \\text{The zero exponent rule} \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align} \\frac{{\\left(3{w}^{2}\\right)}^{5}}{{\\left(6{w}^{-2}\\right)}^{2}}& = \\frac{{\\left(3\\right)}^{5}\\cdot {\\left({w}^{2}\\right)}^{5}}{{\\left(6\\right)}^{2}\\cdot {\\left({w}^{-2}\\right)}^{2}}&& \\text{The power of a product rule} \\\\ & = \\frac{{3}^{5}{w}^{2\\cdot 5}}{{6}^{2}{w}^{-2\\cdot 2}}&& \\text{The power rule} \\\\ & = \\frac{243{w}^{10}}{36{w}^{-4}} && \\text{Simplify}. \\\\ & = \\frac{27{w}^{10-\\left(-4\\right)}}{4}&& \\text{The quotient rule and reduce fraction} \\\\ & = \\frac{27{w}^{14}}{4}&& \\text{Simplify}. \\end{align}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify each expression and write the answer with positive exponents only.<\/p>\n<ol>\n<li>[latex]{\\left(2u{v}^{-2}\\right)}^{-3}[\/latex]<\/li>\n<li>[latex]{x}^{8}\\cdot {x}^{-12}\\cdot x[\/latex]<\/li>\n<li>[latex]{\\left(\\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\\right)}^{2}[\/latex]<\/li>\n<li>[latex]\\left(9{r}^{-5}{s}^{3}\\right)\\left(3{r}^{6}{s}^{-4}\\right)[\/latex]<\/li>\n<li>[latex]{\\left(\\frac{4}{9}t{w}^{-2}\\right)}^{-3}{\\left(\\frac{4}{9}t{w}^{-2}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]\\dfrac{{\\left(2{h}^{2}k\\right)}^{4}}{{\\left(7{h}^{-1}{k}^{2}\\right)}^{2}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q31244\">Show Solution<\/span><\/p>\n<div id=\"q31244\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{v}^{6}}{8{u}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{1}{{x}^{3}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{e}^{4}}{{f}^{4}}[\/latex]<\/li>\n<li>[latex]\\dfrac{27r}{s}[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]\\dfrac{16{h}^{10}}{49}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&amp;theme=oea&amp;iframe_resize_id=mom70\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&amp;theme=oea&amp;iframe_resize_id=mom80\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&amp;theme=oea&amp;iframe_resize_id=mom90\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&amp;theme=oea&amp;iframe_resize_id=mom95\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show more examples of how to find the power of a quotient.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-4\" title=\"Simplify Expressions Using Exponent Rules (Power of a Quotient)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BoBe31pRxFM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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-1330\">\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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <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>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Simplify Expressions With Zero Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/rpoUg32utlc\">https:\/\/youtu.be\/rpoUg32utlc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Simplify Expressions With Negative Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Gssi4dBtAEI\">https:\/\/youtu.be\/Gssi4dBtAEI<\/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>Power of a Product. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/p-2UkpJQWpo\">https:\/\/youtu.be\/p-2UkpJQWpo<\/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>Power of a Quotient. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BoBe31pRxFM\">https:\/\/youtu.be\/BoBe31pRxFM<\/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 44120, 43231. <strong>Authored by<\/strong>: Brenda Gardner. <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>Question ID 7833, 14060. <strong>Authored by<\/strong>: Tyler Wallace. <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>Question ID 109762, 109765. <strong>Authored 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>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 51959. <strong>Authored by<\/strong>: Roy Shahbazian. <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>Question ID 93393. <strong>Authored by<\/strong>: Michael Jenck. <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>Question ID 14047, 14058, 14059, 14046, 14051, 14056, 14057. <strong>Authored by<\/strong>: James Souza. <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>Question ID 43896. <strong>Authored by<\/strong>: Carla Kulinsky. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":21,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Simplify Expressions With Zero Exponents\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/rpoUg32utlc\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Simplify Expressions With Negative Exponents\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Gssi4dBtAEI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Power of a Product\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/p-2UkpJQWpo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Power of a Quotient\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/BoBe31pRxFM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 44120, 43231\",\"author\":\"Brenda Gardner\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 7833, 14060\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 109762, 109765\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 51959\",\"author\":\"Roy Shahbazian\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 93393\",\"author\":\"Michael Jenck\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 14047, 14058, 14059, 14046, 14051, 14056, 14057\",\"author\":\"James Souza\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 43896\",\"author\":\"Carla Kulinsky\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + 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