{"id":246,"date":"2015-09-18T20:16:55","date_gmt":"2015-09-18T20:16:55","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=246"},"modified":"2015-11-02T18:33:53","modified_gmt":"2015-11-02T18:33:53","slug":"using-the-negative-rule-of-exponents","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/using-the-negative-rule-of-exponents\/","title":{"raw":"Using the Negative Rule of Exponents","rendered":"Using the Negative Rule of Exponents"},"content":{"raw":"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]\\frac{{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 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
[latex]\\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>\r\nIf we were to simplify the original expression using the quotient rule, we would have\r\n
[latex]\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\\hfill \\\\ & =& \\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
[latex]\\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}& \\text{and}& {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, 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
\r\n

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
[latex]{a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n
\r\n

Example 5: 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
    \r\n\t
  1. [latex]\\frac{{\\theta }^{3}}{{\\theta }^{10}}[\/latex]<\/li>\r\n\t
  2. [latex]\\frac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\r\n\t
  3. [latex]\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\n
      \r\n\t
    1. [latex]\\frac{{\\theta }^{3}}{{\\theta }^{10}}={\\theta }^{3 - 10}={\\theta }^{-7}=\\frac{1}{{\\theta }^{7}}[\/latex]<\/li>\r\n\t
    2. [latex]\\frac{{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
    3. [latex]\\frac{{\\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<\/div>\r\n
      \r\n

      Try It 5<\/h3>\r\nWrite each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.\r\n

      a. [latex]\\frac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]\r\nb. [latex]\\frac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]\r\nc. [latex]\\frac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      \r\n

      Example 6: 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
        \r\n\t
      1. [latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\r\n\t
      2. [latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\r\n\t
      3. [latex]\\frac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Solution<\/h3>\r\n
          \r\n\t
        1. [latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\r\n\t
        2. [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
        3. [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<\/div>\r\n
          \r\n

          Try It 6<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further. Write answers with positive exponents.\r\n
            \r\n\t
          1. [latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\r\n\t
          2. [latex]\\frac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\r\n<\/ol>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

            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]\\frac{{h}^{3}}{{h}^{5}}[\/latex]? When [latex]mnegative rule of exponents<\/em> to simplify the expression to its reciprocal.<\/p>\n

            Divide one exponential expression by another with a larger exponent. Use our example, [latex]\\frac{{h}^{3}}{{h}^{5}}[\/latex].<\/p>\n

            [latex]\\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

            If we were to simplify the original expression using the quotient rule, we would have<\/p>\n

            [latex]\\begin{array}{ccc}\\hfill \\frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\\hfill \\\\ & =& \\text{ }{h}^{-2}\\hfill \\end{array}[\/latex]<\/div>\n

            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

            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

            [latex]\\begin{array}{ccc}{a}^{-n}=\\frac{1}{{a}^{n}}& \\text{and}& {a}^{n}=\\frac{1}{{a}^{-n}}\\end{array}[\/latex]<\/div>\n

            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

            \n

            A General Note: The Negative Rule of Exponents<\/h3>\n

            For any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that<\/p>\n

            [latex]{a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\n<\/div>\n
            \n

            Example 5: Using the Negative Exponent Rule<\/h3>\n

            Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n

              \n
            1. [latex]\\frac{{\\theta }^{3}}{{\\theta }^{10}}[\/latex]<\/li>\n
            2. [latex]\\frac{{z}^{2}\\cdot z}{{z}^{4}}[\/latex]<\/li>\n
            3. [latex]\\frac{{\\left(-5{t}^{3}\\right)}^{4}}{{\\left(-5{t}^{3}\\right)}^{8}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
              \n

              Solution<\/h3>\n
                \n
              1. [latex]\\frac{{\\theta }^{3}}{{\\theta }^{10}}={\\theta }^{3 - 10}={\\theta }^{-7}=\\frac{1}{{\\theta }^{7}}[\/latex]<\/li>\n
              2. [latex]\\frac{{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
              3. [latex]\\frac{{\\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
                \n

                Try It 5<\/h3>\n

                Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n

                a. [latex]\\frac{{\\left(-3t\\right)}^{2}}{{\\left(-3t\\right)}^{8}}[\/latex]
                \nb. [latex]\\frac{{f}^{47}}{{f}^{49}\\cdot f}[\/latex]
                \nc. [latex]\\frac{2{k}^{4}}{5{k}^{7}}[\/latex]<\/p>\n

                Solution<\/a><\/p>\n<\/div>\n

                \n

                Example 6: Using the Product and Quotient Rules<\/h3>\n

                Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n

                  \n
                1. [latex]{b}^{2}\\cdot {b}^{-8}[\/latex]<\/li>\n
                2. [latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}[\/latex]<\/li>\n
                3. [latex]\\frac{-7z}{{\\left(-7z\\right)}^{5}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
                  \n

                  Solution<\/h3>\n
                    \n
                  1. [latex]{b}^{2}\\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\\frac{1}{{b}^{6}}[\/latex]<\/li>\n
                  2. [latex]{\\left(-x\\right)}^{5}\\cdot {\\left(-x\\right)}^{-5}={\\left(-x\\right)}^{5 - 5}={\\left(-x\\right)}^{0}=1[\/latex]<\/li>\n
                  3. [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
                    \n

                    Try It 6<\/h3>\n

                    Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.<\/p>\n

                      \n
                    1. [latex]{t}^{-11}\\cdot {t}^{6}[\/latex]<\/li>\n
                    2. [latex]\\frac{{25}^{12}}{{25}^{13}}[\/latex]<\/li>\n<\/ol>\n

                      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

                      \n\t\t\t

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