{"id":1509,"date":"2021-12-02T19:38:20","date_gmt":"2021-12-03T00:38:20","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/11-4-exponential-functions\/"},"modified":"2023-08-31T19:26:22","modified_gmt":"2023-08-31T23:26:22","slug":"exponential-functions","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/exponential-functions\/","title":{"raw":"11.4 Exponential Functions","rendered":"11.4 Exponential Functions"},"content":{"raw":"As our study of algebra gets more advanced, we begin to study more involved functions. One pair of inverse functions you will look at are exponential and logarithmic functions.\r\n\r\nExponential functions are functions in which the variable is in the exponent, such as [latex]f(x) = a^x.[\/latex]\r\n\r\nSolving exponential equations cannot be done using the techniques used prior. For example, if [latex]3^x = 9[\/latex], one cannot take the [latex]x^{\\text{th}}[\/latex] root of 9 because we do not know what the index is. However, if you notice that 9 is 32<\/sup>, you can then conclude that, if [latex]3^x = 3^2[\/latex], then [latex]x = 2[\/latex]. This is the process is used to solve exponential functions.\r\n\r\nIf the problem is rewritten so the bases are the same, then the exponents must also equal each other.\r\n
\r\n

Example 11.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for the exponent [latex]x[\/latex] in the equation [latex]5^5 = 5^{x+2}[\/latex].\r\n\r\nSince the bases for these exponents are the same, then the exponents must equal each other. Thus:\r\n

[latex]\\begin{array}{rrl}\r\n5&=&x+2 \\\\\r\nx&=&3\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nGenerally, manipulating the bases on each side of an exponential function to make them equal. These types of problems are as follows:\r\n

\r\n

Example 11.4.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for the exponent [latex]x[\/latex] in the equation [latex]5^{2x+1} = 125[\/latex].\r\n\r\nSince the bases for these exponents are not equal, then the first challenge is to find the lowest common base. For this problem, 125 is the same as 53<\/sup>.\r\n\r\nTherefore, the equation is rewritten as [latex]5^{2x+1} = 5^3[\/latex]. Thus:\r\n

[latex]\\begin{array}{rrrrr}\r\n2x&+&1&=&3 \\\\\r\n&-&1&&-1 \\\\\r\n\\hline\r\n&&2x&=&2 \\\\\r\n&&x&=&1\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 11.4.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for the exponent [latex]x[\/latex] in the equation [latex]8^{3x} = 32[\/latex].\r\n\r\nFinding the common base is a bit more complicated for this problem, but this issue is easily resolved if terms are reduced to their prime factorization of [latex]8 = 2^3[\/latex] and [latex]32 = 2^5[\/latex]. Use this to rewrite the original equation as [latex](2^3)^{3x} = 2^5[\/latex].\r\n\r\nWith identical bases, now solve for the exponents:\r\n

[latex]\\begin{array}{rrr}\r\n3(3x)&=&5 \\\\ \\\\\r\n9x&=&5 \\\\ \\\\\r\nx&=&\\dfrac{5}{9}\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 11.4.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for the exponent [latex]x[\/latex] in the equation [latex]5^{4x}\\cdot 5^{2x-1} = 5^{3x+11}[\/latex].\r\n\r\nSince the bases already equal each other, simplify both sides before beginning to solve this problem. [latex]5^{4x}\\cdot 5^{2x-1}[\/latex] reduces to [latex]5^{6x-1}[\/latex], and [latex]5^{3x+11}[\/latex] is already reduced.\r\n\r\nWith the bases simplified, now solve:\r\n

[latex]\\begin{array}{rrrrrrr}\r\n&&5^{6x-1}&=&5^{3x+11}&& \\\\ \\\\\r\n6x&-&1&=&3x&+&11 \\\\\r\n-3x&+&1&&-3x&+&1 \\\\\r\n\\hline\r\n&&3x&=&12&& \\\\\r\n&&x&=&4&&\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 11.4.5<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for the exponent [latex]x[\/latex] in the equation [latex]\\left(\\dfrac{1}{9}\\right)^{2x} = 3^{7x-1}[\/latex].\r\n\r\nFirst, since [latex]\\dfrac{1}{9} = 3^{-2}[\/latex], the common base is 3. Rewriting the equation in the base of 3 yields:\r\n

[latex]\\begin{array}{rrl}\r\n(3^{-2})^{2x}&=&3^{7x-1} \\\\ \\\\\r\n(-2)2x&=&7x-1 \\\\ \\\\\r\n-4x&=&7x-1 \\\\\r\n-7x&&-7x \\\\\r\n\\hline\r\n-11x&=&-1 \\\\ \\\\\r\nx&=&\\dfrac{1}{11}\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nSolve each equation.\r\n
    \r\n \t
  1. [latex]3^{1-2n}=3^{1-3n}[\/latex]<\/li>\r\n \t
  2. [latex]4^{2x}=\\dfrac{1}{16}[\/latex]<\/li>\r\n \t
  3. [latex]4^{2a}=1[\/latex]<\/li>\r\n \t
  4. [latex]16^{-3p}=64^{-3p}[\/latex]<\/li>\r\n \t
  5. [latex]\\left(\\dfrac{1}{25}\\right)^{-k}=125^{-2k-2}[\/latex]<\/li>\r\n \t
  6. [latex]625^{-n-2}=\\dfrac{1}{125}[\/latex]<\/li>\r\n \t
  7. [latex]6^{2m+1}=\\dfrac{1}{36}[\/latex]<\/li>\r\n \t
  8. [latex]6^{2r-3}=6^{r-3}[\/latex]<\/li>\r\n \t
  9. [latex]6^{-3x}=36[\/latex]<\/li>\r\n \t
  10. [latex]5^{2n}=5^{-n}[\/latex]<\/li>\r\n \t
  11. [latex]64^b=2^5[\/latex]<\/li>\r\n \t
  12. [latex]216^{-3v}=36^{3v}[\/latex]<\/li>\r\n \t
  13. [latex]\\left(\\dfrac{1}{4}\\right)^x=16[\/latex]<\/li>\r\n \t
  14. [latex]27^{-2n-1}=9[\/latex]<\/li>\r\n \t
  15. [latex]4^{3a}=4^3[\/latex]<\/li>\r\n \t
  16. [latex]4^{-3v}=64[\/latex]<\/li>\r\n \t
  17. [latex]36^{3x}=216^{2x+1}[\/latex]<\/li>\r\n \t
  18. [latex]64^{x+2}=16[\/latex]<\/li>\r\n \t
  19. [latex]9^{2n+3}=243[\/latex]<\/li>\r\n \t
  20. [latex]16^{2k}=\\dfrac{1}{64}[\/latex]<\/li>\r\n \t
  21. [latex]3^{3x-2}=3^{3x+1}[\/latex]<\/li>\r\n \t
  22. [latex]243^p=27^{-3p}[\/latex]<\/li>\r\n \t
  23. [latex]3^{-2x}=3^3[\/latex]<\/li>\r\n \t
  24. [latex]4^{2n}=4^{2-3n}[\/latex]<\/li>\r\n \t
  25. [latex]5^{m+2}=5^{-m}[\/latex]<\/li>\r\n \t
  26. [latex]625^{2x}=25[\/latex]<\/li>\r\n \t
  27. [latex]\\left(\\dfrac{1}{36}\\right)^{b-1}=216[\/latex]<\/li>\r\n \t
  28. [latex]216^{2n}=36[\/latex]<\/li>\r\n \t
  29. [latex]6^{2-2x}=6^2[\/latex]<\/li>\r\n \t
  30. [latex]\\left(\\dfrac{1}{4}\\right)^{3v-2}=64^{1-v}[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 11.4<\/a>","rendered":"

    As our study of algebra gets more advanced, we begin to study more involved functions. One pair of inverse functions you will look at are exponential and logarithmic functions.<\/p>\n

    Exponential functions are functions in which the variable is in the exponent, such as [latex]f(x) = a^x.[\/latex]<\/p>\n

    Solving exponential equations cannot be done using the techniques used prior. For example, if [latex]3^x = 9[\/latex], one cannot take the [latex]x^{\\text{th}}[\/latex] root of 9 because we do not know what the index is. However, if you notice that 9 is 32<\/sup>, you can then conclude that, if [latex]3^x = 3^2[\/latex], then [latex]x = 2[\/latex]. This is the process is used to solve exponential functions.<\/p>\n

    If the problem is rewritten so the bases are the same, then the exponents must also equal each other.<\/p>\n

    \n
    \n

    Example 11.4.1<\/p>\n<\/header>\n

    \n

    Solve for the exponent [latex]x[\/latex] in the equation [latex]5^5 = 5^{x+2}[\/latex].<\/p>\n

    Since the bases for these exponents are the same, then the exponents must equal each other. Thus:<\/p>\n

    [latex]\\begin{array}{rrl} 5&=&x+2 \\\\ x&=&3 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    Generally, manipulating the bases on each side of an exponential function to make them equal. These types of problems are as follows:<\/p>\n

    \n
    \n

    Example 11.4.2<\/p>\n<\/header>\n

    \n

    Solve for the exponent [latex]x[\/latex] in the equation [latex]5^{2x+1} = 125[\/latex].<\/p>\n

    Since the bases for these exponents are not equal, then the first challenge is to find the lowest common base. For this problem, 125 is the same as 53<\/sup>.<\/p>\n

    Therefore, the equation is rewritten as [latex]5^{2x+1} = 5^3[\/latex]. Thus:<\/p>\n

    [latex]\\begin{array}{rrrrr} 2x&+&1&=&3 \\\\ &-&1&&-1 \\\\ \\hline &&2x&=&2 \\\\ &&x&=&1 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.4.3<\/p>\n<\/header>\n

    \n

    Solve for the exponent [latex]x[\/latex] in the equation [latex]8^{3x} = 32[\/latex].<\/p>\n

    Finding the common base is a bit more complicated for this problem, but this issue is easily resolved if terms are reduced to their prime factorization of [latex]8 = 2^3[\/latex] and [latex]32 = 2^5[\/latex]. Use this to rewrite the original equation as [latex](2^3)^{3x} = 2^5[\/latex].<\/p>\n

    With identical bases, now solve for the exponents:<\/p>\n

    [latex]\\begin{array}{rrr} 3(3x)&=&5 \\\\ \\\\ 9x&=&5 \\\\ \\\\ x&=&\\dfrac{5}{9} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.4.4<\/p>\n<\/header>\n

    \n

    Solve for the exponent [latex]x[\/latex] in the equation [latex]5^{4x}\\cdot 5^{2x-1} = 5^{3x+11}[\/latex].<\/p>\n

    Since the bases already equal each other, simplify both sides before beginning to solve this problem. [latex]5^{4x}\\cdot 5^{2x-1}[\/latex] reduces to [latex]5^{6x-1}[\/latex], and [latex]5^{3x+11}[\/latex] is already reduced.<\/p>\n

    With the bases simplified, now solve:<\/p>\n

    [latex]\\begin{array}{rrrrrrr} &&5^{6x-1}&=&5^{3x+11}&& \\\\ \\\\ 6x&-&1&=&3x&+&11 \\\\ -3x&+&1&&-3x&+&1 \\\\ \\hline &&3x&=&12&& \\\\ &&x&=&4&& \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.4.5<\/p>\n<\/header>\n

    \n

    Solve for the exponent [latex]x[\/latex] in the equation [latex]\\left(\\dfrac{1}{9}\\right)^{2x} = 3^{7x-1}[\/latex].<\/p>\n

    First, since [latex]\\dfrac{1}{9} = 3^{-2}[\/latex], the common base is 3. Rewriting the equation in the base of 3 yields:<\/p>\n

    [latex]\\begin{array}{rrl} (3^{-2})^{2x}&=&3^{7x-1} \\\\ \\\\ (-2)2x&=&7x-1 \\\\ \\\\ -4x&=&7x-1 \\\\ -7x&&-7x \\\\ \\hline -11x&=&-1 \\\\ \\\\ x&=&\\dfrac{1}{11} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Solve each equation.<\/p>\n

      \n
    1. [latex]3^{1-2n}=3^{1-3n}[\/latex]<\/li>\n
    2. [latex]4^{2x}=\\dfrac{1}{16}[\/latex]<\/li>\n
    3. [latex]4^{2a}=1[\/latex]<\/li>\n
    4. [latex]16^{-3p}=64^{-3p}[\/latex]<\/li>\n
    5. [latex]\\left(\\dfrac{1}{25}\\right)^{-k}=125^{-2k-2}[\/latex]<\/li>\n
    6. [latex]625^{-n-2}=\\dfrac{1}{125}[\/latex]<\/li>\n
    7. [latex]6^{2m+1}=\\dfrac{1}{36}[\/latex]<\/li>\n
    8. [latex]6^{2r-3}=6^{r-3}[\/latex]<\/li>\n
    9. [latex]6^{-3x}=36[\/latex]<\/li>\n
    10. [latex]5^{2n}=5^{-n}[\/latex]<\/li>\n
    11. [latex]64^b=2^5[\/latex]<\/li>\n
    12. [latex]216^{-3v}=36^{3v}[\/latex]<\/li>\n
    13. [latex]\\left(\\dfrac{1}{4}\\right)^x=16[\/latex]<\/li>\n
    14. [latex]27^{-2n-1}=9[\/latex]<\/li>\n
    15. [latex]4^{3a}=4^3[\/latex]<\/li>\n
    16. [latex]4^{-3v}=64[\/latex]<\/li>\n
    17. [latex]36^{3x}=216^{2x+1}[\/latex]<\/li>\n
    18. [latex]64^{x+2}=16[\/latex]<\/li>\n
    19. [latex]9^{2n+3}=243[\/latex]<\/li>\n
    20. [latex]16^{2k}=\\dfrac{1}{64}[\/latex]<\/li>\n
    21. [latex]3^{3x-2}=3^{3x+1}[\/latex]<\/li>\n
    22. [latex]243^p=27^{-3p}[\/latex]<\/li>\n
    23. [latex]3^{-2x}=3^3[\/latex]<\/li>\n
    24. [latex]4^{2n}=4^{2-3n}[\/latex]<\/li>\n
    25. [latex]5^{m+2}=5^{-m}[\/latex]<\/li>\n
    26. [latex]625^{2x}=25[\/latex]<\/li>\n
    27. [latex]\\left(\\dfrac{1}{36}\\right)^{b-1}=216[\/latex]<\/li>\n
    28. [latex]216^{2n}=36[\/latex]<\/li>\n
    29. [latex]6^{2-2x}=6^2[\/latex]<\/li>\n
    30. [latex]\\left(\\dfrac{1}{4}\\right)^{3v-2}=64^{1-v}[\/latex]<\/li>\n<\/ol>\n

      Answer Key 11.4<\/a><\/p>\n","protected":false},"author":90,"menu_order":4,"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-1509","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1491,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1509","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\/1509\/revisions"}],"predecessor-version":[{"id":2175,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1509\/revisions\/2175"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1491"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1509\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1509"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1509"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1509"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}