{"id":1450,"date":"2021-12-02T19:38:04","date_gmt":"2021-12-03T00:38:04","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/10-2-solving-exponential-equations\/"},"modified":"2023-08-31T18:37:23","modified_gmt":"2023-08-31T22:37:23","slug":"solving-exponential-equations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/solving-exponential-equations\/","title":{"raw":"10.2 Solving Exponential Equations","rendered":"10.2 Solving Exponential Equations"},"content":{"raw":"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:\r\n\r\n[latex]\\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}[\/latex]\r\n
\r\n

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

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

[latex]\\begin{array}{ccc}\r\n(x^5)^{\\frac{1}{5}}&=&(32)^{\\frac{1}{5}} \\\\\r\nx&=&2\r\n\\end{array}[\/latex]<\/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 [latex]x[\/latex] in the equation [latex]x^4 = 16[\/latex].\r\n\r\nThe solution for this requires that the fourth root of both sides is taken.\r\n

[latex]\\begin{array}{rcl}\r\n(x^4)^{\\frac{1}{4}}&=&(16)^{\\frac{1}{4}} \\\\ \\\\\r\nx&=&\\pm 2 \\\\ \\\\\r\n\\end{array}[\/latex]<\/p>\r\nThe answer is [latex]\\pm 2[\/latex] because [latex](2)^4=16[\/latex] and [latex](-2)^4=16[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\nWhen encountering more complicated problems that require radical solutions, 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 [latex]x[\/latex] in the equation [latex]2(2x + 4)^2 = 72[\/latex].\r\n\r\nThe first step should be to isolate [latex](2x+4)^2[\/latex], which is done by dividing both sides by 2. This results in [latex](2x + 4)^2 = 36[\/latex].\r\n\r\nOnce isolated, take the square root of both sides of this equation:\r\n\r\n[latex]\\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}[\/latex]\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 [latex]x[\/latex] in the equation [latex](x + 4)^3 + 6 = -119[\/latex].\r\n\r\nFirst, isolate [latex](x + 4)^3[\/latex] by subtracting 6 from both sides. This results in [latex](x + 4)^3 = -125[\/latex].\r\n\r\nNow, take the cube root of both sides, which leaves:\r\n\r\n[latex]\\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\\hline\r\n&&x&=&-9\r\n\\end{array}[\/latex]\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. [latex]x^2=75[\/latex]<\/li>\r\n \t
  2. [latex]x^3=-8[\/latex]<\/li>\r\n \t
  3. [latex]x^2+5=13[\/latex]<\/li>\r\n \t
  4. [latex]4x^3-2=106[\/latex]<\/li>\r\n \t
  5. [latex]3x^2+1=73[\/latex]<\/li>\r\n \t
  6. [latex](x-4)^2=49[\/latex]<\/li>\r\n \t
  7. [latex](x+2)^5=-243[\/latex]<\/li>\r\n \t
  8. [latex](5x+1)^4=16[\/latex]<\/li>\r\n \t
  9. [latex](2x+5)^3-6=21[\/latex]<\/li>\r\n \t
  10. [latex](2x+1)^2+3=21[\/latex]<\/li>\r\n \t
  11. [latex](x-1)^{\\frac{2}{3}}=16[\/latex]<\/li>\r\n \t
  12. [latex](x-1)^{\\frac{3}{2}}=8[\/latex]<\/li>\r\n \t
  13. [latex](2-x)^{\\frac{3}{2}}=27[\/latex]<\/li>\r\n \t
  14. [latex](2x+3)^{\\frac{4}{3}}=16[\/latex]<\/li>\r\n \t
  15. [latex](2x-3)^{\\frac{2}{3}}=4[\/latex]<\/li>\r\n \t
  16. [latex](3x-2)^{\\frac{4}{5}}=16[\/latex]<\/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

    [latex]\\begin{array}{l} \\text{For odd values of }n,\\text{ then }a^n=b\\text{ and }a=\\sqrt[n]{b} \\\\ \\text{For even values of }n,\\text{ then }a^n=b\\text{ and }a=\\pm \\sqrt[n]{b} \\end{array}[\/latex]<\/p>\n

    \n
    \n

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

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

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

    [latex]\\begin{array}{ccc} (x^5)^{\\frac{1}{5}}&=&(32)^{\\frac{1}{5}} \\\\ x&=&2 \\end{array}[\/latex]<\/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 [latex]x[\/latex] in the equation [latex]x^4 = 16[\/latex].<\/p>\n

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

    [latex]\\begin{array}{rcl} (x^4)^{\\frac{1}{4}}&=&(16)^{\\frac{1}{4}} \\\\ \\\\ x&=&\\pm 2 \\\\ \\\\ \\end{array}[\/latex]<\/p>\n

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

    When encountering more complicated problems that require radical solutions, 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 [latex]x[\/latex] in the equation [latex]2(2x + 4)^2 = 72[\/latex].<\/p>\n

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

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

    [latex]\\begin{array}{rrcrrrr} [(2x&+&4)^2]^{\\frac{1}{2}}&=&36^{\\frac{1}{2}}&& \\\\ 2x&+&4&=&\\pm 6&& \\\\ &&2x&=&-4 &\\pm &6 \\\\ &&x&=&-2 &\\pm& 3 \\\\ \\\\ &&x&=&-5, &1& \\end{array}[\/latex]<\/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 [latex]x[\/latex] in the equation [latex](x + 4)^3 + 6 = -119[\/latex].<\/p>\n

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

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

    [latex]\\begin{array}{rrrrl} [(x&+&4)^3]^{\\frac{1}{3}}&=&[-125]^{\\frac{1}{3}} \\\\ x&+&4&=&-5 \\\\ &-&4&&-4 \\\\ \\hline &&x&=&-9 \\end{array}[\/latex]<\/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. [latex]x^2=75[\/latex]<\/li>\n
    2. [latex]x^3=-8[\/latex]<\/li>\n
    3. [latex]x^2+5=13[\/latex]<\/li>\n
    4. [latex]4x^3-2=106[\/latex]<\/li>\n
    5. [latex]3x^2+1=73[\/latex]<\/li>\n
    6. [latex](x-4)^2=49[\/latex]<\/li>\n
    7. [latex](x+2)^5=-243[\/latex]<\/li>\n
    8. [latex](5x+1)^4=16[\/latex]<\/li>\n
    9. [latex](2x+5)^3-6=21[\/latex]<\/li>\n
    10. [latex](2x+1)^2+3=21[\/latex]<\/li>\n
    11. [latex](x-1)^{\\frac{2}{3}}=16[\/latex]<\/li>\n
    12. [latex](x-1)^{\\frac{3}{2}}=8[\/latex]<\/li>\n
    13. [latex](2-x)^{\\frac{3}{2}}=27[\/latex]<\/li>\n
    14. [latex](2x+3)^{\\frac{4}{3}}=16[\/latex]<\/li>\n
    15. [latex](2x-3)^{\\frac{2}{3}}=4[\/latex]<\/li>\n
    16. [latex](3x-2)^{\\frac{4}{5}}=16[\/latex]<\/li>\n<\/ol>\n

      Answer Key 10.2<\/a><\/p>\n","protected":false},"author":90,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1450","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1446,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1450","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1450\/revisions"}],"predecessor-version":[{"id":2156,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1450\/revisions\/2156"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1446"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1450\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1450"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1450"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1450"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1450"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}