{"id":728,"date":"2019-04-29T17:00:56","date_gmt":"2019-04-29T21:00:56","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=728"},"modified":"2019-12-05T13:41:00","modified_gmt":"2019-12-05T18:41:00","slug":"10-2-solving-exponential-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/10-2-solving-exponential-equations\/","title":{"raw":"10.2 Solving Exponential Equations","rendered":"10.2 Solving Exponential Equations"},"content":{"raw":"[latexpage]\r\n\r\nExponential equations are often reduced by using radicals\u2014similar to using exponents to solve for radical equations. There is one caveat, though: while odd index roots can be solved for either negative or positive values, even-powered roots can only be taken for even values, but have both positive and negative solutions. This is shown below:\r\n\r\n\\[\\begin{array}{l}\r\n\\text{For odd values of }n,\\text{ then }a^n=b\\text{ and }a=\\sqrt[n]{b} \\\\\r\n\\text{For even values of }n,\\text{ then }a^n=b\\text{ and }a=\\pm \\sqrt[n]{b}\r\n\\end{array}\\]\r\n
\r\n

Example 10.2.1<\/p>\r\n\r\n<\/header>\r\n

Solve for \\(x\\) in the equation \\(x^5 = 32\\).<\/div>\r\n
\r\n\r\nThe solution for this requires that you take the fifth root of both sides.\r\n

\\(\\begin{array}{ccc}\r\n(x^5)^{\\frac{1}{5}}&=&(32)^{\\frac{1}{5}} \\\\\r\nx&=&2\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nWhen taking a positive root, there will be two solutions. For example:\r\n

\r\n

Example 10.2.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for \\(x\\) in the equation \\(x^4 = 16\\).\r\n\r\nThe solution for this requires that the fourth root of both sides is taken.\r\n

\\(\\begin{array}{rcl}\r\n(x^4)^{\\frac{1}{4}}&=&(16)^{\\frac{1}{4}} \\\\ \\\\\r\nx&=&\\pm 2 \\\\ \\\\\r\n\\end{array}\\)<\/p>\r\nThe answer is \\(\\pm 2\\) because \\((2)^4=16\\) and \\((-2)^4=16\\).\r\n\r\n<\/div>\r\n<\/div>\r\nWhen encountering more complicated problems that require radical solutions,\u00a0 work the problem so that there is a single power to reduce as the starting point of the solution. This strategy makes for an easier solution.\r\n

\r\n

Example 10.2.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for \\(x\\) in the equation \\(2(2x + 4)^2 = 72\\).\r\n\r\nThe first step should be to isolate \\((2x+4)^2\\), which is done by dividing both sides by 2. This results in \\((2x + 4)^2 = 36\\).\r\n\r\nOnce isolated,\u00a0 take the square root of both sides of this equation:\r\n\r\n\\[\\begin{array}{rrcrrrr}\r\n[(2x&+&4)^2]^{\\frac{1}{2}}&=&36^{\\frac{1}{2}}&& \\\\\r\n2x&+&4&=&\\pm 6&& \\\\\r\n&&2x&=&-4 &\\pm &6 \\\\\r\n&&x&=&-2 &\\pm& 3 \\\\ \\\\\r\n&&x&=&-5, &1&\r\n\\end{array}\\]\r\n\r\nChecking these solutions in the original equation indicates that both work.\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 10.2.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for \\(x\\) in the equation \\((x + 4)^3 + 6 = -119\\).\r\n\r\nFirst, isolate \\((x + 4)^3\\) by subtracting 6 from both sides. This results in \\((x + 4)^3 = -125\\).\r\n\r\nNow,\u00a0 take the cube root of both sides, which leaves:\r\n\r\n\\[\\begin{array}{rrrrl}\r\n[(x&+&4)^3]^{\\frac{1}{3}}&=&[-125]^{\\frac{1}{3}} \\\\\r\nx&+&4&=&-5 \\\\\r\n&-&4&&-4 \\\\\r\n\\midrule\r\n&&x&=&-9\r\n\\end{array}\\]\r\n\r\nChecking this solution in the original equation indicates that it is a valid solution.\r\n\r\n<\/div>\r\n<\/div>\r\nSince you are solving for an odd root, there is only one solution to the cube root of \u2212125. It is only even-powered roots that have both a positive and a negative solution.\r\n

Questions<\/h1>\r\nSolve.\r\n
    \r\n \t
  1. \\(x^2=75\\)<\/li>\r\n \t
  2. \\(x^3=-8\\)<\/li>\r\n \t
  3. \\(x^2+5=13\\)<\/li>\r\n \t
  4. \\(4x^3-2=106\\)<\/li>\r\n \t
  5. \\(3x^2+1=73\\)<\/li>\r\n \t
  6. \\((x-4)^2=49\\)<\/li>\r\n \t
  7. \\((x+2)^5=-243\\)<\/li>\r\n \t
  8. \\((5x+1)^4=16\\)<\/li>\r\n \t
  9. \\((2x+5)^3-6=21\\)<\/li>\r\n \t
  10. \\((2x+1)^2+3=21\\)<\/li>\r\n \t
  11. \\((x-1)^{\\frac{2}{3}}=16\\)<\/li>\r\n \t
  12. \\((x-1)^{\\frac{3}{2}}=8\\)<\/li>\r\n \t
  13. \\((2-x)^{\\frac{3}{2}}=27\\)<\/li>\r\n \t
  14. \\((2x+3)^{\\frac{4}{3}}=16\\)<\/li>\r\n \t
  15. \\((2x-3)^{\\frac{2}{3}}=4\\)<\/li>\r\n \t
  16. \\((3x-2)^{\\frac{4}{5}}=16\\)<\/li>\r\n<\/ol>\r\nAnswer Key 10.2<\/a>","rendered":"

    Exponential equations are often reduced by using radicals\u2014similar to using exponents to solve for radical equations. There is one caveat, though: while odd index roots can be solved for either negative or positive values, even-powered roots can only be taken for even values, but have both positive and negative solutions. This is shown below:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{l}<\/p>\n

    \n
    \n

    Example 10.2.1<\/p>\n<\/header>\n

    Solve for \"x\" in the equation \"x^5 = 32\".<\/div>\n
    \n

    The solution for this requires that you take the fifth root of both sides.<\/p>\n

    \"\begin{array}{ccc}<\/p>\n<\/div>\n<\/div>\n

    When taking a positive root, there will be two solutions. For example:<\/p>\n

    \n
    \n

    Example 10.2.2<\/p>\n<\/header>\n

    \n

    Solve for \"x\" in the equation \"x^4 = 16\".<\/p>\n

    The solution for this requires that the fourth root of both sides is taken.<\/p>\n

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

    The answer is \"\pm 2\" because \"(2)^4=16\" and \"(-2)^4=16\".<\/p>\n<\/div>\n<\/div>\n

    When encountering more complicated problems that require radical solutions,\u00a0 work the problem so that there is a single power to reduce as the starting point of the solution. This strategy makes for an easier solution.<\/p>\n

    \n
    \n

    Example 10.2.3<\/p>\n<\/header>\n

    \n

    Solve for \"x\" in the equation \"2(2x + 4)^2 = 72\".<\/p>\n

    The first step should be to isolate \"(2x+4)^2\", which is done by dividing both sides by 2. This results in \"(2x + 4)^2 = 36\".<\/p>\n

    Once isolated,\u00a0 take the square root of both sides of this equation:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrcrrrr}<\/p>\n

    Checking these solutions in the original equation indicates that both work.<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 10.2.4<\/p>\n<\/header>\n

    \n

    Solve for \"x\" in the equation \"(x + 4)^3 + 6 = -119\".<\/p>\n

    First, isolate \"(x + 4)^3\" by subtracting 6 from both sides. This results in \"(x + 4)^3 = -125\".<\/p>\n

    Now,\u00a0 take the cube root of both sides, which leaves:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrl}<\/p>\n

    Checking this solution in the original equation indicates that it is a valid solution.<\/p>\n<\/div>\n<\/div>\n

    Since you are solving for an odd root, there is only one solution to the cube root of \u2212125. It is only even-powered roots that have both a positive and a negative solution.<\/p>\n

    Questions<\/h1>\n

    Solve.<\/p>\n

      \n
    1. \"x^2=75\"<\/li>\n
    2. \"x^3=-8\"<\/li>\n
    3. \"x^2+5=13\"<\/li>\n
    4. \"4x^3-2=106\"<\/li>\n
    5. \"3x^2+1=73\"<\/li>\n
    6. \"(x-4)^2=49\"<\/li>\n
    7. \"(x+2)^5=-243\"<\/li>\n
    8. \"(5x+1)^4=16\"<\/li>\n
    9. \"(2x+5)^3-6=21\"<\/li>\n
    10. \"(2x+1)^2+3=21\"<\/li>\n
    11. \"(x-1)^{\frac{2}{3}}=16\"<\/li>\n
    12. \"(x-1)^{\frac{3}{2}}=8\"<\/li>\n
    13. \"(2-x)^{\frac{3}{2}}=27\"<\/li>\n
    14. \"(2x+3)^{\frac{4}{3}}=16\"<\/li>\n
    15. \"(2x-3)^{\frac{2}{3}}=4\"<\/li>\n
    16. \"(3x-2)^{\frac{4}{5}}=16\"<\/li>\n<\/ol>\n

      Answer Key 10.2<\/a><\/p>\n","protected":false},"author":540,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-728","chapter","type-chapter","status-publish","hentry"],"part":393,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/728","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":7,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/728\/revisions"}],"predecessor-version":[{"id":3649,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/728\/revisions\/3649"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/393"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/728\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=728"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=728"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=728"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=728"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}