{"id":790,"date":"2016-06-01T20:49:09","date_gmt":"2016-06-01T20:49:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=790"},"modified":"2019-08-06T18:31:14","modified_gmt":"2019-08-06T18:31:14","slug":"read-negative-and-zero-exponent-rules-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-negative-and-zero-exponent-rules-2\/","title":{"raw":"Negative and Zero Exponent Rules","rendered":"Negative and Zero Exponent Rules"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Simplify exponential expressions containing negative exponents and exponents of 0 and 1<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Define and Use the Zero Exponent Rule<\/h2>\r\nReturn to the quotient rule. We worked with expressions for which\u00a0\u00a0[latex]a&gt;b[\/latex] so that the difference [latex]a-b[\/latex] would never be zero or negative.\r\n<div class=\"textbox shaded\">\r\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\r\nFor any non-zero number <i>x<\/i> and any integers <i>a<\/i> and <i>b<\/i>: [latex] \\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]\r\n\r\n<\/div>\r\nWhat would happen if [latex]a=b[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to\u00a0[latex]1[\/latex]. To see how this is done, let us begin with an example.\r\n<p style=\"text-align: center;\">[latex]\\frac{t^{8}}{t^{8}}=\\frac{\\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]\\frac{{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 shaded\">\r\n<h3>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<\/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=\"375469\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"375469\"]\r\n\r\nUse the zero exponent and other rules to simplify each expression.\r\n\r\n1.\r\n\r\n[latex]\\Large\\begin{array}\\text{ }\\frac{c^{3}}{c^{3}} \\hfill&amp; =c^{3-3} \\\\ \\hfill&amp; =c^{0} \\\\ \\hfill&amp; =1\\end{array}[\/latex]\r\n\r\n2.\r\n\r\n[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{-3{x}^{5}}{{x}^{5}}&amp; =&amp; -3\\cdot \\frac{{x}^{5}}{{x}^{5}}\\hfill \\\\ &amp; =&amp; -3\\cdot {x}^{5 - 5}\\hfill \\\\ &amp; =&amp; -3\\cdot {x}^{0}\\hfill \\\\ &amp; =&amp; -3\\cdot 1\\hfill \\\\ &amp; =&amp; -3\\hfill \\end{array}[\/latex]\r\n\r\n3.\r\n\r\n[latex]\\large\\begin{array}{cccc}\\hfill \\frac{{\\left({j}^{2}k\\right)}^{4}}{\\left({j}^{2}k\\right)\\cdot {\\left({j}^{2}k\\right)}^{3}}&amp; =&amp; \\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{1+3}}\\hfill &amp;\\text{Use the product rule in the denominator}.\\hfill \\\\ &amp; =&amp;\\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ &amp; =&amp;{\\left({j}^{2}k\\right)}^{4 - 4}\\hfill &amp;\\text{Use the quotient rule}.\\hfill \\\\ &amp; =&amp; {\\left({j}^{2}k\\right)}^{0}\\hfill&amp;\\text{Simplify}.\\hfill \\\\ &amp; =&amp; 1&amp; \\end{array}[\/latex]\r\n\r\n4.\r\n\r\n[latex]\\large\\begin{array}{cccc}\\hfill \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}&amp; =&amp;5{\\left(r{s}^{2}\\right)}^{2 - 2}\\hfill &amp;\\text{Use the quotient rule}.\\hfill \\\\ &amp; =&amp; 5{\\left(r{s}^{2}\\right)}^{0}\\hfill &amp;\\text{Simplify}.\\hfill \\\\ &amp; =&amp; 5\\cdot 1\\hfill &amp;\\text{Use the zero exponent rule}.\\hfill \\\\ &amp; =&amp; 5\\hfill &amp;\\text{Simplify}.\\hfill \\end{array}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000;\">In the following video, you will see more examples of simplifying expressions whose exponents may be zero.<\/span>\r\n\r\nhttps:\/\/youtu.be\/rpoUg32utlc\r\n<h2>Define and Use the Negative Exponent Rule<\/h2>\r\nAnother useful result occurs if we relax the condition that [latex]a&gt;b[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\frac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]a&lt;b[\/latex]\u2014that is, where the difference [latex]a-b[\/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]\\frac{{h}^{3}}{{h}^{5}}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}&amp; =&amp; \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{1}{h\\cdot h}\\hfill \\\\ &amp; =&amp; \\frac{1}{{h}^{2}}\\hfill \\end{array}[\/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]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}&amp; =&amp; {h}^{3 - 5}\\hfill \\\\ &amp; =&amp; \\text{ }{h}^{-2}\\hfill \\end{array}[\/latex]<\/div>\r\nPutting the answers together, we have [latex]{h}^{-2}=\\frac{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]\\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}&amp; \\text{and}&amp; {a}^{n}=\\frac{1}{{a}^{-n}}\\end{array}[\/latex]<\/div>\r\nWe have shown that the exponential expression [latex]{a}^{n}[\/latex] is defined when [latex]n[\/latex] is a natural number,\u00a0[latex]0[\/latex], 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 shaded\">\r\n<h3>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}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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{{(2b) }^{3}}{{(2b) }^{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=\"552758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"552758\"]\r\n<ol>\r\n \t<li>[latex]\\dfrac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\\frac{1}{{(2b)}^{7}}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}=\\frac{{z}^{2+1}}{{z}^{4}}=\\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\frac{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}=\\frac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000;\">In the following video, you will see examples of simplifying expressions with negative exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/Gssi4dBtAEI\r\n<h2>Combine Exponent Rules to Simplify Expressions<\/h2>\r\nNow we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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]\\frac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"163692\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"163692\"]\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]\\frac{-7z}{{\\left(-7z\\right)}^{5}}=\\frac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #000000;\">The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/span>\r\n\r\nhttps:\/\/youtu.be\/EkvN5tcp4Cc","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Simplify exponential expressions containing negative exponents and exponents of 0 and 1<\/li>\n<\/ul>\n<\/div>\n<h2>Define and Use the Zero Exponent Rule<\/h2>\n<p>Return to the quotient rule. We worked with expressions for which\u00a0\u00a0[latex]a>b[\/latex] so that the difference [latex]a-b[\/latex] would never be zero or negative.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Quotient (Division) Rule for Exponents<\/h3>\n<p>For any non-zero number <i>x<\/i> and any integers <i>a<\/i> and <i>b<\/i>: [latex]\\displaystyle \\frac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}[\/latex]<\/p>\n<\/div>\n<p>What would happen if [latex]a=b[\/latex]? In this case, we would use the <em>zero exponent rule of exponents<\/em> to simplify the expression to\u00a0[latex]1[\/latex]. To see how this is done, let us begin with an example.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{t^{8}}{t^{8}}=\\frac{\\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]\\frac{{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 shaded\">\n<h3>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<\/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=\"q375469\">Show Solution<\/span><\/p>\n<div id=\"q375469\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the zero exponent and other rules to simplify each expression.<\/p>\n<p>1.<\/p>\n<p>[latex]\\Large\\begin{array}\\text{ }\\frac{c^{3}}{c^{3}} \\hfill& =c^{3-3} \\\\ \\hfill& =c^{0} \\\\ \\hfill& =1\\end{array}[\/latex]<\/p>\n<p>2.<\/p>\n<p>[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{-3{x}^{5}}{{x}^{5}}& =& -3\\cdot \\frac{{x}^{5}}{{x}^{5}}\\hfill \\\\ & =& -3\\cdot {x}^{5 - 5}\\hfill \\\\ & =& -3\\cdot {x}^{0}\\hfill \\\\ & =& -3\\cdot 1\\hfill \\\\ & =& -3\\hfill \\end{array}[\/latex]<\/p>\n<p>3.<\/p>\n<p>[latex]\\large\\begin{array}{cccc}\\hfill \\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}}\\hfill &\\text{Use the product rule in the denominator}.\\hfill \\\\ & =&\\frac{{\\left({j}^{2}k\\right)}^{4}}{{\\left({j}^{2}k\\right)}^{4}}\\hfill & \\text{Simplify}.\\hfill \\\\ & =&{\\left({j}^{2}k\\right)}^{4 - 4}\\hfill &\\text{Use the quotient rule}.\\hfill \\\\ & =& {\\left({j}^{2}k\\right)}^{0}\\hfill&\\text{Simplify}.\\hfill \\\\ & =& 1& \\end{array}[\/latex]<\/p>\n<p>4.<\/p>\n<p>[latex]\\large\\begin{array}{cccc}\\hfill \\frac{5{\\left(r{s}^{2}\\right)}^{2}}{{\\left(r{s}^{2}\\right)}^{2}}& =&5{\\left(r{s}^{2}\\right)}^{2 - 2}\\hfill &\\text{Use the quotient rule}.\\hfill \\\\ & =& 5{\\left(r{s}^{2}\\right)}^{0}\\hfill &\\text{Simplify}.\\hfill \\\\ & =& 5\\cdot 1\\hfill &\\text{Use the zero exponent rule}.\\hfill \\\\ & =& 5\\hfill &\\text{Simplify}.\\hfill \\end{array}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000;\">In the following video, you will see more examples of simplifying expressions whose exponents may be zero.<\/span><\/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>Define and Use the Negative Exponent Rule<\/h2>\n<p>Another useful result occurs if we relax the condition that [latex]a>b[\/latex] in the quotient rule even further. For example, can we simplify [latex]\\frac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]a<b[\/latex]\u2014that is, where the difference [latex]a-b[\/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]\\frac{{h}^{3}}{{h}^{5}}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& \\frac{h\\cdot h\\cdot h}{h\\cdot h\\cdot h\\cdot h\\cdot h}\\hfill \\\\ & =& \\frac{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}}{\\cancel{h}\\cdot \\cancel{h}\\cdot \\cancel{h}\\cdot h\\cdot h}\\hfill \\\\ & =& \\frac{1}{h\\cdot h}\\hfill \\\\ & =& \\frac{1}{{h}^{2}}\\hfill \\end{array}[\/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]\\Large\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\\hfill \\\\ & =& \\text{ }{h}^{-2}\\hfill \\end{array}[\/latex]<\/div>\n<p>Putting the answers together, we have [latex]{h}^{-2}=\\frac{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]\\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}& \\text{and}& {a}^{n}=\\frac{1}{{a}^{-n}}\\end{array}[\/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,\u00a0[latex]0[\/latex], 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 shaded\">\n<h3>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}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/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{{(2b) }^{3}}{{(2b) }^{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=\"q552758\">Show Solution<\/span><\/p>\n<div id=\"q552758\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\dfrac{{(2b) }^{3}}{{(2b)}^{10}}={(2b)}^{3 - 10}={(2b) }^{-7}=\\frac{1}{{(2b)}^{7}}[\/latex]<\/li>\n<li>[latex]\\dfrac{{z}^{2}\\cdot z}{{z}^{4}}=\\frac{{z}^{2+1}}{{z}^{4}}=\\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\\frac{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}=\\frac{1}{{\\left(-5{t}^{3}\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000;\">In the following video, you will see examples of simplifying expressions with negative exponents.<\/span><\/p>\n<p><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<h2>Combine Exponent Rules to Simplify Expressions<\/h2>\n<p>Now we will combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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]\\frac{-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=\"q163692\">Show Solution<\/span><\/p>\n<div id=\"q163692\" 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]\\frac{-7z}{{\\left(-7z\\right)}^{5}}=\\frac{{\\left(-7z\\right)}^{1}}{{\\left(-7z\\right)}^{5}}={\\left(-7z\\right)}^{1 - 5}={\\left(-7z\\right)}^{-4}=\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #000000;\">The following video shows more examples of how to combine the use of the product and quotient rules to simplify expressions whose terms may have negative or zero exponents.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/EkvN5tcp4Cc?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-790\">\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><li>Simplify Expressions Using the Quotient and Zero Exponent Rules. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Simplify Expressions Using the Quotient and Negative Exponent Rules Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>Simplify Expressions Using Exponent Rules (Product, Quotient, Zero Exponent). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/EkvN5tcp4Cc\">https:\/\/youtu.be\/EkvN5tcp4Cc<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Jay Abrams et, al.. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/\">https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/<\/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>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Jay Abrams et, al.\",\"organization\":\"Lumen 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