{"id":248,"date":"2015-09-18T20:18:06","date_gmt":"2015-09-18T20:18:06","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=248"},"modified":"2015-11-02T19:49:11","modified_gmt":"2015-11-02T19:49:11","slug":"finding-the-power-of-a-product-and-a-quotient","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/finding-the-power-of-a-product-and-a-quotient\/","title":{"raw":"Finding the Power of a Product and a Quotient","rendered":"Finding the Power of a Product and a Quotient"},"content":{"raw":"

Finding the Power of a Product<\/h2>\r\nTo simplify the power of a product of two exponential expressions, we can use the 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
[latex]\\begin{array}{ccc}\\hfill {\\left(pq\\right)}^{3}& =& \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}}\\hfill \\\\ & =& p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q\\hfill \\\\ & =& \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}}\\hfill \\\\ & =& {p}^{3}\\cdot {q}^{3}\\hfill \\end{array}[\/latex]<\/div>\r\nIn other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].\r\n
\r\n

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
[latex]{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\r\n<\/div>\r\n
\r\n

Example 7: 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
    \r\n\t
  1. [latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\r\n\t
  2. [latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n\t
  3. [latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\r\n\t
  4. [latex]\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\r\n\t
  5. [latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nUse the product and quotient rules and the new definitions to simplify each expression.\r\n
      \r\n\t
    1. [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
    2. [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
    3. [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
    4. [latex]\\frac{1}{{\\left(-7z\\right)}^{4}}=\\frac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\frac{1}{2,401{z}^{4}}[\/latex]<\/li>\r\n\t
    5. [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}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      Try It 7<\/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

      a. [latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]\r\nb. [latex]{\\left(5t\\right)}^{3}[\/latex]\r\nc. [latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]\r\nd. [latex]\\frac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]\r\ne. [latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      Finding the Power of a Quotient<\/h2>\r\nTo simplify the power of a quotient of two expressions, we can use the 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
      [latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\r\nLet\u2019s rewrite the original problem differently and look at the result.\r\n
      [latex]\\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/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
      [latex]\\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}}\\hfill \\\\ & =& \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\r\n
      \r\n

      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
      [latex]{\\left(\\frac{a}{b}\\right)}^{n}=\\frac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n
      \r\n

      Example 8: 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
        \r\n\t
      1. [latex]{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\r\n\t
      2. [latex]{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\r\n\t
      3. [latex]{\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\r\n\t
      4. [latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\r\n\t
      5. [latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Solution<\/h3>\r\n
          \r\n\t
        1. [latex]{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}=\\frac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\frac{64}{{z}^{11\\cdot 3}}=\\frac{64}{{z}^{33}}[\/latex]<\/li>\r\n\t
        2. [latex]{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}=\\frac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\frac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\frac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\r\n\t
        3. [latex]{\\\\left(\\frac{-1}{{t}^{2}}\\\\right)}^{27}=\\frac{{\\\\left(-1\\\\right)}^{27}}{{\\\\left({t}^{2}\\\\right)}^{27}}=\\frac{-1}{{t}^{2\\cdot 27}}=\\frac{-1}{{t}^{54}}=-\\frac{1}{{t}^{54}}[\/latex]<\/li>\r\n\t
        4. [latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\frac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\frac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\frac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\frac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\r\n\t
        5. [latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\frac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\frac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\frac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\frac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\frac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
          \r\n

          Try It 8<\/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

          a. [latex]{\\left(\\frac{{b}^{5}}{c}\\right)}^{3}[\/latex]\r\nb. [latex]{\\left(\\frac{5}{{u}^{8}}\\right)}^{4}[\/latex]\r\nc. [latex]{\\left(\\frac{-1}{{w}^{3}}\\right)}^{35}[\/latex]\r\nd. [latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]\r\ne. [latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

          Finding the Power of a Product<\/h2>\n

          To simplify the power of a product of two exponential expressions, we can use the 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

          [latex]\\begin{array}{ccc}\\hfill {\\left(pq\\right)}^{3}& =& \\stackrel{3\\text{ factors}}{{\\left(pq\\right)\\cdot \\left(pq\\right)\\cdot \\left(pq\\right)}}\\hfill \\\\ & =& p\\cdot q\\cdot p\\cdot q\\cdot p\\cdot q\\hfill \\\\ & =& \\stackrel{3\\text{ factors}}{{p\\cdot p\\cdot p}}\\cdot \\stackrel{3\\text{ factors}}{{q\\cdot q\\cdot q}}\\hfill \\\\ & =& {p}^{3}\\cdot {q}^{3}\\hfill \\end{array}[\/latex]<\/div>\n

          In other words, [latex]{\\left(pq\\right)}^{3}={p}^{3}\\cdot {q}^{3}[\/latex].<\/p>\n

          \n

          A General Note: The Power of a Product Rule of Exponents<\/h3>\n

          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

          [latex]{\\left(ab\\right)}^{n}={a}^{n}{b}^{n}[\/latex]<\/div>\n<\/div>\n
          \n

          Example 7: Using the Power of a Product Rule<\/h3>\n

          Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.<\/p>\n

            \n
          1. [latex]{\\left(a{b}^{2}\\right)}^{3}[\/latex]<\/li>\n
          2. [latex]{\\left(2t\\right)}^{15}[\/latex]<\/li>\n
          3. [latex]{\\left(-2{w}^{3}\\right)}^{3}[\/latex]<\/li>\n
          4. [latex]\\frac{1}{{\\left(-7z\\right)}^{4}}[\/latex]<\/li>\n
          5. [latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
            \n

            Solution<\/h3>\n

            Use the product and quotient rules and the new definitions to simplify each expression.<\/p>\n

              \n
            1. [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
            2. [latex]2{t}^{15}={\\left(2\\right)}^{15}\\cdot {\\left(t\\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[\/latex]<\/li>\n
            3. [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
            4. [latex]\\frac{1}{{\\left(-7z\\right)}^{4}}=\\frac{1}{{\\left(-7\\right)}^{4}\\cdot {\\left(z\\right)}^{4}}=\\frac{1}{2,401{z}^{4}}[\/latex]<\/li>\n
            5. [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}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
              \n

              Try It 7<\/h3>\n

              Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.<\/p>\n

              a. [latex]{\\left({g}^{2}{h}^{3}\\right)}^{5}[\/latex]
              \nb. [latex]{\\left(5t\\right)}^{3}[\/latex]
              \nc. [latex]{\\left(-3{y}^{5}\\right)}^{3}[\/latex]
              \nd. [latex]\\frac{1}{{\\left({a}^{6}{b}^{7}\\right)}^{3}}[\/latex]
              \ne. [latex]{\\left({r}^{3}{s}^{-2}\\right)}^{4}[\/latex]<\/p>\n

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

              Finding the Power of a Quotient<\/h2>\n

              To simplify the power of a quotient of two expressions, we can use the 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

              [latex]{\\left({e}^{-2}{f}^{2}\\right)}^{7}=\\frac{{f}^{14}}{{e}^{14}}[\/latex]<\/div>\n

              Let\u2019s rewrite the original problem differently and look at the result.<\/p>\n

              [latex]\\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\n

              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

              [latex]\\begin{array}{ccc}\\hfill {\\left({e}^{-2}{f}^{2}\\right)}^{7}& =& {\\left(\\frac{{f}^{2}}{{e}^{2}}\\right)}^{7}\\hfill \\\\ & =& \\frac{{\\left({f}^{2}\\right)}^{7}}{{\\left({e}^{2}\\right)}^{7}}\\hfill \\\\ & =& \\frac{{f}^{2\\cdot 7}}{{e}^{2\\cdot 7}}\\hfill \\\\ & =& \\frac{{f}^{14}}{{e}^{14}}\\hfill \\end{array}[\/latex]<\/div>\n
              \n

              A General Note: The Power of a Quotient Rule of Exponents<\/h3>\n

              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

              [latex]{\\left(\\frac{a}{b}\\right)}^{n}=\\frac{{a}^{n}}{{b}^{n}}[\/latex]<\/div>\n<\/div>\n
              \n

              Example 8: Using the Power of a Quotient Rule<\/h3>\n

              Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.<\/p>\n

                \n
              1. [latex]{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}[\/latex]<\/li>\n
              2. [latex]{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}[\/latex]<\/li>\n
              3. [latex]{\\left(\\frac{-1}{{t}^{2}}\\right)}^{27}[\/latex]<\/li>\n
              4. [latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}[\/latex]<\/li>\n
              5. [latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
                \n

                Solution<\/h3>\n
                  \n
                1. [latex]{\\left(\\frac{4}{{z}^{11}}\\right)}^{3}=\\frac{{\\left(4\\right)}^{3}}{{\\left({z}^{11}\\right)}^{3}}=\\frac{64}{{z}^{11\\cdot 3}}=\\frac{64}{{z}^{33}}[\/latex]<\/li>\n
                2. [latex]{\\left(\\frac{p}{{q}^{3}}\\right)}^{6}=\\frac{{\\left(p\\right)}^{6}}{{\\left({q}^{3}\\right)}^{6}}=\\frac{{p}^{1\\cdot 6}}{{q}^{3\\cdot 6}}=\\frac{{p}^{6}}{{q}^{18}}[\/latex]<\/li>\n
                3. [latex]{\\\\left(\\frac{-1}{{t}^{2}}\\\\right)}^{27}=\\frac{{\\\\left(-1\\\\right)}^{27}}{{\\\\left({t}^{2}\\\\right)}^{27}}=\\frac{-1}{{t}^{2\\cdot 27}}=\\frac{-1}{{t}^{54}}=-\\frac{1}{{t}^{54}}[\/latex]<\/li>\n
                4. [latex]{\\left({j}^{3}{k}^{-2}\\right)}^{4}={\\left(\\frac{{j}^{3}}{{k}^{2}}\\right)}^{4}=\\frac{{\\left({j}^{3}\\right)}^{4}}{{\\left({k}^{2}\\right)}^{4}}=\\frac{{j}^{3\\cdot 4}}{{k}^{2\\cdot 4}}=\\frac{{j}^{12}}{{k}^{8}}[\/latex]<\/li>\n
                5. [latex]{\\left({m}^{-2}{n}^{-2}\\right)}^{3}={\\left(\\frac{1}{{m}^{2}{n}^{2}}\\right)}^{3}=\\frac{{\\left(1\\right)}^{3}}{{\\left({m}^{2}{n}^{2}\\right)}^{3}}=\\frac{1}{{\\left({m}^{2}\\right)}^{3}{\\left({n}^{2}\\right)}^{3}}=\\frac{1}{{m}^{2\\cdot 3}\\cdot {n}^{2\\cdot 3}}=\\frac{1}{{m}^{6}{n}^{6}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n
                  \n

                  Try It 8<\/h3>\n

                  Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.<\/p>\n

                  a. [latex]{\\left(\\frac{{b}^{5}}{c}\\right)}^{3}[\/latex]
                  \nb. [latex]{\\left(\\frac{5}{{u}^{8}}\\right)}^{4}[\/latex]
                  \nc. [latex]{\\left(\\frac{-1}{{w}^{3}}\\right)}^{35}[\/latex]
                  \nd. [latex]{\\left({p}^{-4}{q}^{3}\\right)}^{8}[\/latex]
                  \ne. [latex]{\\left({c}^{-5}{d}^{-3}\\right)}^{4}[\/latex]<\/p>\n

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

                  \n\t\t\t

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