{"id":557,"date":"2019-04-29T14:42:31","date_gmt":"2019-04-29T18:42:31","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=557"},"modified":"2020-02-13T14:58:30","modified_gmt":"2020-02-13T19:58:30","slug":"6-2-negative-exponents","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/6-2-negative-exponents\/","title":{"raw":"6.2 Negative Exponents","rendered":"6.2 Negative Exponents"},"content":{"raw":"[latexpage]\r\n\r\nConsider the following chart that shows the expansion of \\(a\\) for several exponents:\r\n\r\n\\(\\begin{array}{lllllllll}\r\na^4\\phantom{-}&=&a&\\times &a&\\times &a&\\times &a \\\\\r\na^3&=&a&\\times &a&\\times &a&& \\\\\r\na^2&=&a&\\times &a&&&& \\\\\r\na^1&=&a&&&&&& \\\\\r\na^0&=&1&&&&&&\r\n\\end{array}\\)\r\n\r\n\\(\\begin{array}{lll}\r\na^{-1}&=&\\dfrac{1}{a} \\\\ \\\\\r\na^{-2}&=&\\dfrac{1}{(a\\times a)} \\\\ \\\\\r\na^{-3}&=&\\dfrac{1}{(a\\times a\\times a)} \\\\ \\\\\r\na^{-4}&=&\\dfrac{1}{(a\\times a\\times a\\times a)}\r\n\\end{array}\\)\r\n\r\nIf zero and negative exponents are expanded to base 2, the result is the following:\r\n\r\n\\(\\begin{array}{llcll}\r\n2^0&=&1&& \\\\ \\\\\r\n2^{-1}&=&\\dfrac{1}{2}&& \\\\ \\\\\r\n2^{-2}&=&\\dfrac{1}{2\\times 2}&\\text{or}&\\dfrac{1}{4} \\\\ \\\\\r\n2^{-3}&=&\\dfrac{1}{2\\times 2\\times 2}&\\text{or}&\\dfrac{1}{8} \\\\ \\\\\r\n2^{-4}&=&\\dfrac{1}{2\\times 2\\times 2\\times 2}&\\text{or}&\\dfrac{1}{16}\r\n\\end{array}\\)\r\n\r\nThe most unusual of these is the exponent 0. Any base that is not equal to zero to the zeroth exponent is always 1. The simplest explanation of this is by example.\r\n
\r\n

Example 6.2.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSimplify \\(\\dfrac{x^3}{x^3}\\).\r\n\r\nUsing the quotient rule of exponents, we know that this simplifies to \\(x^{3-3}\\), which equals \\(x^0.\\) And we know\r\n

\\(\\dfrac{2^3}{2^3} = \\dfrac{8}{8}=1\\),<\/p>\r\n

\\(\\dfrac{3^3}{3^3} = \\dfrac{27}{27}=1\\),<\/p>\r\n

\\(\\dfrac{4^3}{4^3}=\\dfrac{64}{64}=1\\),<\/p>\r\n

\\(\\dfrac{5^3}{5^3}=\\dfrac{125}{125}=1\\),<\/p>\r\nand so on. A base raised to an exponent divided by that same base raised to that same exponent will always equal 1 unless the base is 0. This leads us to the zero power rule of exponents:\r\n

\\(\\text{Zero Power Rule of Exponents: }x^0 = 1\\hspace{0.25in} (x \\ne 0)\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThis zero rule of exponents can make difficult problems elementary simply because whatever the 0 exponent is attached to reduces to 1. Consider the following examples:\r\n

\r\n

Example 6.2.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSimplify the following expressions.\r\n
    \r\n \t
  1. \\(5x^0y^2\\)\r\nSince \\(x^0 = 1,\\) this simplifies to \\(5y^2.\\)<\/li>\r\n \t
  2. \\((5x^0y^2)^0\\)\r\nSince the zero exponent is on the outside of the parentheses, everything contained inside the parentheses is cancelled out to 1.<\/li>\r\n \t
  3. \\([(15x^3y^2)(25x^2y^2)]^0\\)\r\nSince the zero exponent is on the outside of the brackets, everything contained inside the brackets cancels out to 1.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nWhen encountering these types of problems, always remain aware of what the zero power is attached to, since only what it is attached to cancels to 1.\r\n\r\nWhen dealing with negative exponents, the simplest solution is to reciprocate the power. For instance:\r\n
    \r\n

    Example 6.2.3<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nSimplify the following expressions.\r\n
      \r\n \t
    1. \\(3x^{-2}y^2\\)\r\nSince the only negative exponent is \\(x^{-2}\\), this simplifies to \\(\\dfrac{3y^2}{x^2}.\\)<\/li>\r\n \t
    2. \\(4x^2y^{-3}\\)\r\nSince the only negative exponent is \\(y^{-3}\\), this simplifies to \\(\\dfrac{4x^2}{y^3}.\\)<\/li>\r\n \t
    3. \\((4x^2y^{-3})^{-1}\\)\r\nUsing the power of a power rule of exponents, we get \\(4^{-1}x^{-2}y^3\\).\r\nSimplifying the negative exponents of \\(4^{-1}x^{-2}\\), we get \\(\\dfrac{y^3}{4x^2}.\\)<\/li>\r\n \t
    4. \\((2m^{-1}n^{-3})(2m^{-1}n^{-3})^4\\)\r\nFirst using the power of a power rule on \\((2m^{-1}n^{-3})^4\\) yields \\(2^4m^{-4}n^{-12}\\).\r\nNow we multiply \\(2m^{-1}n^{-3}\\) by \\(2^4m^{-4}n^{-12}\\), yielding \\(2^5m^{-5}n^{-15}\\).\r\nWe can write this without any negative exponents as \\(\\dfrac{2^5}{m^5n^{15}}.\\)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

      <\/b>Four Rules of Negative Exponents<\/h1>\r\n

      \\(x^{-n}=\\dfrac{1}{x^n}\\hspace{0.25in} (x \\ne 0)\\)<\/p>\r\n

      \\(\\dfrac{1}{x^{-n}}=x^n\\hspace{0.25in} (x \\ne 0)\\)<\/p>\r\n

      \\(\\left(\\dfrac{x}{y}\\right)^{-n}=\\left(\\dfrac{y}{x}\\right)^n \\hspace{0.25in} (x,y \\ne 0)\\)<\/p>\r\n

      \\(\\left(\\dfrac{x^ay^b}{z^c}\\right)^{-n}=\\dfrac{z^{cn}}{x^{an}y^{bn}}\\hspace{0.25in} (x,y,z \\ne 0)\\)<\/p>\r\n\r\n

      Questions<\/h1>\r\nSimplify. Your answer should contain only positive exponents<\/strong>.\r\n
        \r\n \t
      1. \\((2x^4y^{-2})(2xy^3)^4\\)<\/li>\r\n \t
      2. \\((2a^{-2}b^{-3})(2a^0b^4)^4\\)<\/li>\r\n \t
      3. \\((2x^2y^2)^4x^{-4}\\)<\/li>\r\n \t
      4. \\([(m^0n^3)(2m^{-3}n^{-3})]^0\\)<\/li>\r\n \t
      5. \\((2x^{-3}y^2)\\div (3x^{-3}y^3\\cdot 3x^0)\\)<\/li>\r\n \t
      6. \\(3y^3\\div [(3yx^3)(2x^4y^{-3})]\\)<\/li>\r\n \t
      7. \\(2y\\div (x^0y^2)^4\\)<\/li>\r\n \t
      8. \\((a^4)^4\\div 2b\\)<\/li>\r\n \t
      9. \\((2a^2b^3)^4\\div a^{-1}\\)<\/li>\r\n \t
      10. \\((2y^{-4})^{-2}\\div x^2\\)<\/li>\r\n \t
      11. \\((2mn)^4\\div m^0n^{-2}\\)<\/li>\r\n \t
      12. \\(2x^{-3}\\div (x^4y^{-3})^{-1}\\)<\/li>\r\n \t
      13. \\([(2u^{-2}v^3)(2uv^4)^{-1}]\\div 2u^{-4}v^0\\)<\/li>\r\n \t
      14. \\([(2yx^2)(x^{-2})]\\div (2x^0y^4)^{-1}\\)<\/li>\r\n \t
      15. \\(b^{-1}\\div [(2a^4b^0)^0(2a^{-3}b^2)]\\)<\/li>\r\n \t
      16. \\(2yzx^2\\div [(2x^4y^4z^{-2})(zy^2)^4]\\)<\/li>\r\n \t
      17. \\([(cb^3)^2(2a^{-3}b^2)]\\div (a^3b^{-2}c^3)^3\\)<\/li>\r\n \t
      18. \\(2q^4(m^2p^2q)\\div (2m\\cdot 4p^2)^3\\)<\/li>\r\n \t
      19. \\((yx^{-4}z^2)^{-1}\\div z^3\\cdot x^2y^3z^{-1}\\)<\/li>\r\n \t
      20. \\(2mpn^{-3}\\div [2n^2p^0(m^0n^{-4}p^2)^3]\\)<\/li>\r\n<\/ol>\r\nAnswer Key 6.2<\/a>","rendered":"

        Consider the following chart that shows the expansion of \"a\" for several exponents:<\/p>\n

        \"\begin{array}{lllllllll}<\/p>\n

        \"\begin{array}{lll}<\/p>\n

        If zero and negative exponents are expanded to base 2, the result is the following:<\/p>\n

        \"\begin{array}{llcll}<\/p>\n

        The most unusual of these is the exponent 0. Any base that is not equal to zero to the zeroth exponent is always 1. The simplest explanation of this is by example.<\/p>\n

        \n
        \n

        Example 6.2.1<\/p>\n<\/header>\n

        \n

        Simplify \"\dfrac{x^3}{x^3}\".<\/p>\n

        Using the quotient rule of exponents, we know that this simplifies to \"x^{3-3}\", which equals \"x^0.\" And we know<\/p>\n

        \"\dfrac{2^3}{2^3} = \dfrac{8}{8}=1\",<\/p>\n

        \"\dfrac{3^3}{3^3} = \dfrac{27}{27}=1\",<\/p>\n

        \"\dfrac{4^3}{4^3}=\dfrac{64}{64}=1\",<\/p>\n

        \"\dfrac{5^3}{5^3}=\dfrac{125}{125}=1\",<\/p>\n

        and so on. A base raised to an exponent divided by that same base raised to that same exponent will always equal 1 unless the base is 0. This leads us to the zero power rule of exponents:<\/p>\n

        \"\text{Zero Power Rule of Exponents: }x^0 = 1\hspace{0.25in} (x \ne 0)\"<\/p>\n<\/div>\n<\/div>\n

        This zero rule of exponents can make difficult problems elementary simply because whatever the 0 exponent is attached to reduces to 1. Consider the following examples:<\/p>\n

        \n
        \n

        Example 6.2.2<\/p>\n<\/header>\n

        \n

        Simplify the following expressions.<\/p>\n

          \n
        1. \"5x^0y^2\"
          \nSince \"x^0 = 1,\" this simplifies to \"5y^2.\"<\/li>\n
        2. \"(5x^0y^2)^0\"
          \nSince the zero exponent is on the outside of the parentheses, everything contained inside the parentheses is cancelled out to 1.<\/li>\n
        3. \"[(15x^3y^2)(25x^2y^2)]^0\"
          \nSince the zero exponent is on the outside of the brackets, everything contained inside the brackets cancels out to 1.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

          When encountering these types of problems, always remain aware of what the zero power is attached to, since only what it is attached to cancels to 1.<\/p>\n

          When dealing with negative exponents, the simplest solution is to reciprocate the power. For instance:<\/p>\n

          \n
          \n

          Example 6.2.3<\/p>\n<\/header>\n

          \n

          Simplify the following expressions.<\/p>\n

            \n
          1. \"3x^{-2}y^2\"
            \nSince the only negative exponent is \"x^{-2}\", this simplifies to \"\dfrac{3y^2}{x^2}.\"<\/li>\n
          2. \"4x^2y^{-3}\"
            \nSince the only negative exponent is \"y^{-3}\", this simplifies to \"\dfrac{4x^2}{y^3}.\"<\/li>\n
          3. \"(4x^2y^{-3})^{-1}\"
            \nUsing the power of a power rule of exponents, we get \"4^{-1}x^{-2}y^3\".
            \nSimplifying the negative exponents of \"4^{-1}x^{-2}\", we get \"\dfrac{y^3}{4x^2}.\"<\/li>\n
          4. \"(2m^{-1}n^{-3})(2m^{-1}n^{-3})^4\"
            \nFirst using the power of a power rule on \"(2m^{-1}n^{-3})^4\" yields \"2^4m^{-4}n^{-12}\".
            \nNow we multiply \"2m^{-1}n^{-3}\" by \"2^4m^{-4}n^{-12}\", yielding \"2^5m^{-5}n^{-15}\".
            \nWe can write this without any negative exponents as \"\dfrac{2^5}{m^5n^{15}}.\"<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

            <\/b>Four Rules of Negative Exponents<\/h1>\n

            \"x^{-n}=\dfrac{1}{x^n}\hspace{0.25in} (x \ne 0)\"<\/p>\n

            \"\dfrac{1}{x^{-n}}=x^n\hspace{0.25in} (x \ne 0)\"<\/p>\n

            \"\left(\dfrac{x}{y}\right)^{-n}=\left(\dfrac{y}{x}\right)^n \hspace{0.25in} (x,y \ne 0)\"<\/p>\n

            \"\left(\dfrac{x^ay^b}{z^c}\right)^{-n}=\dfrac{z^{cn}}{x^{an}y^{bn}}\hspace{0.25in} (x,y,z \ne 0)\"<\/p>\n

            Questions<\/h1>\n

            Simplify. Your answer should contain only positive exponents<\/strong>.<\/p>\n

              \n
            1. \"(2x^4y^{-2})(2xy^3)^4\"<\/li>\n
            2. \"(2a^{-2}b^{-3})(2a^0b^4)^4\"<\/li>\n
            3. \"(2x^2y^2)^4x^{-4}\"<\/li>\n
            4. \"[(m^0n^3)(2m^{-3}n^{-3})]^0\"<\/li>\n
            5. \"(2x^{-3}y^2)\div (3x^{-3}y^3\cdot 3x^0)\"<\/li>\n
            6. \"3y^3\div [(3yx^3)(2x^4y^{-3})]\"<\/li>\n
            7. \"2y\div (x^0y^2)^4\"<\/li>\n
            8. \"(a^4)^4\div 2b\"<\/li>\n
            9. \"(2a^2b^3)^4\div a^{-1}\"<\/li>\n
            10. \"(2y^{-4})^{-2}\div x^2\"<\/li>\n
            11. \"(2mn)^4\div m^0n^{-2}\"<\/li>\n
            12. \"2x^{-3}\div (x^4y^{-3})^{-1}\"<\/li>\n
            13. \"[(2u^{-2}v^3)(2uv^4)^{-1}]\div 2u^{-4}v^0\"<\/li>\n
            14. \"[(2yx^2)(x^{-2})]\div (2x^0y^4)^{-1}\"<\/li>\n
            15. \"b^{-1}\div [(2a^4b^0)^0(2a^{-3}b^2)]\"<\/li>\n
            16. \"2yzx^2\div [(2x^4y^4z^{-2})(zy^2)^4]\"<\/li>\n
            17. \"[(cb^3)^2(2a^{-3}b^2)]\div (a^3b^{-2}c^3)^3\"<\/li>\n
            18. \"2q^4(m^2p^2q)\div (2m\cdot 4p^2)^3\"<\/li>\n
            19. \"(yx^{-4}z^2)^{-1}\div z^3\cdot x^2y^3z^{-1}\"<\/li>\n
            20. \"2mpn^{-3}\div [2n^2p^0(m^0n^{-4}p^2)^3]\"<\/li>\n<\/ol>\n

              Answer Key 6.2<\/a><\/p>\n","protected":false},"author":540,"menu_order":15,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-557","chapter","type-chapter","status-publish","hentry"],"part":376,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/557","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":12,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/557\/revisions"}],"predecessor-version":[{"id":3820,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/557\/revisions\/3820"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/376"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/557\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=557"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=557"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=557"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=557"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}