{"id":45,"date":"2023-06-21T13:22:27","date_gmt":"2023-06-21T13:22:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/zero-and-negative-exponents\/"},"modified":"2023-07-04T04:54:09","modified_gmt":"2023-07-04T04:54:09","slug":"zero-and-negative-exponents","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/zero-and-negative-exponents\/","title":{"raw":"\u25aa   Zero and Negative Exponents","rendered":"\u25aa   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\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&nbsp;\r\n<div class=\"textbox examples\">\r\n<h3>using order of operations with fractions<\/h3>\r\nWhen simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression\u00a0[latex]\\dfrac{5 a^m z^2}{a^mz}[\/latex]\u00a0using exponent rules, you may find it helpful to break the fraction up into a product of fractions, then simplify.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{5 a^m z^2}{a^mz}\\quad=\\quad 5\\cdot\\dfrac{a^m}{a^m}\\cdot\\dfrac{z^2}{z} \\quad=\\quad 5 \\cdot a^{m-m}\\cdot z^{2-1}\\quad=\\quad 5\\cdot a^0 \\cdot z^1 \\quad=\\quad 5z[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\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[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=44120&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7833&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\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}} \\text{ and } {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\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=51959&amp;theme=oea&amp;iframe_resize_id=mom5[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109762&amp;theme=oea&amp;iframe_resize_id=mom10[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109765&amp;theme=oea&amp;iframe_resize_id=mom15[\/embed]\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\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93393&amp;theme=oea&amp;iframe_resize_id=mom20[\/embed]\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\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14047&amp;theme=oea&amp;iframe_resize_id=mom25[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14058&amp;theme=oea&amp;iframe_resize_id=mom30[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14059&amp;theme=oea&amp;iframe_resize_id=mom40[\/embed]\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\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14046&amp;theme=oea&amp;iframe_resize_id=mom50[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14051&amp;theme=oea&amp;iframe_resize_id=mom60[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43231&amp;theme=oea&amp;iframe_resize_id=mom60[\/embed]\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\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&amp;theme=oea&amp;iframe_resize_id=mom70[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&amp;theme=oea&amp;iframe_resize_id=mom80[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&amp;theme=oea&amp;iframe_resize_id=mom90[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&amp;theme=oea&amp;iframe_resize_id=mom95[\/embed]\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>Return 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<p>&nbsp;<\/p>\n<div class=\"textbox examples\">\n<h3>using order of operations with fractions<\/h3>\n<p>When simplifying expressions with exponents, it is sometimes helpful to rely on the rule for multiplying fractions to separate the factors before doing work on them. For example, to simplify the expression\u00a0[latex]\\dfrac{5 a^m z^2}{a^mz}[\/latex]\u00a0using exponent rules, you may find it helpful to break the fraction up into a product of fractions, then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{5 a^m z^2}{a^mz}\\quad=\\quad 5\\cdot\\dfrac{a^m}{a^m}\\cdot\\dfrac{z^2}{z} \\quad=\\quad 5 \\cdot a^{m-m}\\cdot z^{2-1}\\quad=\\quad 5\\cdot a^0 \\cdot z^1 \\quad=\\quad 5z[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\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=\"ohm44120\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=44120&#38;theme=oea&#38;iframe_resize_id=ohm44120&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm7833\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7833&#38;theme=oea&#38;iframe_resize_id=ohm7833&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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}} \\text{ and } {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=\"ohm51959\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=51959&#38;theme=oea&#38;iframe_resize_id=ohm51959&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm109762\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109762&#38;theme=oea&#38;iframe_resize_id=ohm109762&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm109765\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109765&#38;theme=oea&#38;iframe_resize_id=ohm109765&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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=\"ohm93393\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93393&#38;theme=oea&#38;iframe_resize_id=ohm93393&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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=\"ohm14047\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14047&#38;theme=oea&#38;iframe_resize_id=ohm14047&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm14058\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14058&#38;theme=oea&#38;iframe_resize_id=ohm14058&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm14059\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14059&#38;theme=oea&#38;iframe_resize_id=ohm14059&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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=\"ohm14046\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14046&#38;theme=oea&#38;iframe_resize_id=ohm14046&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm14051\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14051&#38;theme=oea&#38;iframe_resize_id=ohm14051&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm43231\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43231&#38;theme=oea&#38;iframe_resize_id=ohm43231&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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=\"ohm14056\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14056&#38;theme=oea&#38;iframe_resize_id=ohm14056&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm14057\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14057&#38;theme=oea&#38;iframe_resize_id=ohm14057&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm14060\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14060&#38;theme=oea&#38;iframe_resize_id=ohm14060&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm43896\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43896&#38;theme=oea&#38;iframe_resize_id=ohm43896&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/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-45\">\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":395986,"menu_order":14,"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 + GPL\"}]","CANDELA_OUTCOMES_GUID":"1cff3a71-5ed2-4bf4-badf-d6eceaf22a0a","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-45","chapter","type-chapter","status-publish","hentry"],"part":31,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/45","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/45\/revisions"}],"predecessor-version":[{"id":836,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/45\/revisions\/836"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/31"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/45\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=45"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=45"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=45"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=45"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}