{"id":166,"date":"2023-09-20T22:47:59","date_gmt":"2023-09-20T22:47:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/logarithmic-functions\/"},"modified":"2024-08-05T12:30:11","modified_gmt":"2024-08-05T12:30:11","slug":"exponential-and-logarithmic-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/exponential-and-logarithmic-functions-learn-it-3\/","title":{"raw":"Exponential and Logarithmic Functions: Learn It 3","rendered":"Exponential and Logarithmic Functions: Learn It 3"},"content":{"raw":"

Logarithmic Functions<\/h2>\r\n

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.\u00a0<\/p>\r\n

\r\n

Inverse Functions<\/strong><\/p>\r\n

For any one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex].\u00a0<\/p>\r\n

The notation [latex]{f}^{-1}[\/latex] is read \"[latex]f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]\".<\/p>\r\n<\/section>\r\n

Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.<\/p>\r\n

The exponential function [latex]f(x)=b^x[\/latex] is one-to-one, with domain [latex](\u2212\\infty ,\\infty)[\/latex] and range [latex](0,\\infty )[\/latex]. Therefore, it has an inverse function, called the logarithmic function<\/strong> with base<\/em> [latex]b[\/latex].<\/p>\r\n

For any [latex]b>0, \\, b \\ne 1[\/latex], the logarithmic function with base [latex]b[\/latex], denoted [latex]\\log_b[\/latex], has domain [latex](0,\\infty )[\/latex] and range [latex](\u2212\\infty ,\\infty )[\/latex], and satisfies<\/p>\r\n

[latex]\\log_b(x)=y[\/latex] if and only if [latex]b^y=x[\/latex].<\/p>\r\n

\r\n

logarithmic functions<\/h3>\r\n

A logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[\/latex]. For a given base [latex]b[\/latex], it tells us the power to which [latex]b[\/latex] must be raised to get [latex]x[\/latex].<\/p>\r\n<\/section>\r\n

\r\n
[latex]\\begin{array}{cccc} \\log_2 (8)=3\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}2^3=8,\\hfill \\\\ \\log_{10} (\\frac{1}{100})=-2\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}10^{-2}=\\frac{1}{10^2}=\\frac{1}{100},\\hfill \\\\ \\log_b (1)=0\\hfill & & & \\text{since}\\phantom{\\rule{3em}{0ex}}b^0=1 \\, \\text{for any base} \\, b>0.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/section>\r\n

The most commonly used logarithmic function is the function [latex]\\log_e (x)[\/latex]. Since this function uses natural [latex]e[\/latex] as its base, it is called the natural logarithm<\/strong>. Here we use the notation [latex]\\ln(x)[\/latex] or [latex]\\ln x[\/latex] to mean [latex]\\log_e (x)[\/latex].<\/p>\r\n

[latex]\\begin{array}{l}\\ln (e)=\\log_e (e)=1 \\\\ \\ln(e^3)=\\log_e (e^3)=3 \\\\ \\ln(1)=\\log_e (1)=0\\end{array}[\/latex]<\/center><\/section>\r\n
\r\n

Euler's number, denoted as [latex]e[\/latex], is a fundamental mathematical constant approximately equal to [latex]2.71828[\/latex]. It is the base of the natural logarithm and the natural exponential function, known for its unique properties in calculus, especially in relation to growth processes and compound interest calculations.<\/p>\r\n<\/section>\r\n

Before solving some equations involving exponential and logarithmic functions, let\u2019s review the basic properties of logarithms.<\/p>\r\n

\r\n

Properties of Logarithms<\/h3>\r\n

If [latex]a,b,c>0, \\, b\\ne 1[\/latex], and [latex]r[\/latex] is any real number, then<\/p>\r\n

[latex]\\begin{array}{cccc}1.\\phantom{\\rule{2em}{0ex}}\\log_b (ac)=\\log_b (a)+\\log_b (c)\\hfill & & & \\text{(Product property)}\\hfill \\\\ 2.\\phantom{\\rule{2em}{0ex}}\\log_b(\\frac{a}{c})=\\log_b (a) -\\log_b (c)\\hfill & & & \\text{(Quotient property)}\\hfill \\\\ 3.\\phantom{\\rule{2em}{0ex}}\\log_b (a^r)=r \\log_b (a)\\hfill & & & \\text{(Power property)}\\hfill \\end{array}[\/latex]<\/p>\r\n<\/section>\r\n

\r\n

Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n

    \r\n\t
  1. [latex]\\ln \\left(\\frac{1}{x}\\right)=4[\/latex]<\/li>\r\n\t
  2. [latex]\\log_{10} \\sqrt{x}+ \\log_{10} x=2[\/latex]<\/li>\r\n\t
  3. [latex]\\ln(2x)-3 \\ln(x^2)=0[\/latex]<\/li>\r\n<\/ol>\r\n

    [reveal-answer q=\"fs-id1170572174799\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1170572174799\"]<\/p>\r\n

      \r\n\t
    1. By the definition of the natural logarithm function,
      \r\n
      [latex]\\ln\\big(\\frac{1}{x}\\big)=4 \\, \\text{ if and only if } \\, e^4=\\frac{1}{x}[\/latex]<\/div>\r\n

      Therefore, the solution is [latex]x=\\frac{1}{e^4}[\/latex].<\/p>\r\n<\/li>\r\n\t

    2. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
      \r\n
      [latex]\\log_{10} \\sqrt{x}+ \\log_{10} x = \\log_{10} x \\sqrt{x} = \\log_{10}x^{3\/2} = \\frac{3}{2} \\log_{10} x[\/latex]<\/div>\r\n

      Therefore, the equation can be rewritten as<\/p>\r\n

      [latex]\\frac{3}{2} \\log_{10} x = 2 \\, \\text{ or } \\, \\log_{10} x = \\frac{4}{3}[\/latex]<\/div>\r\n

      The solution is [latex]x=10^{4\/3}=10\\sqrt[3]{10}[\/latex].<\/p>\r\n<\/li>\r\n\t

    3. Using the power property of logarithmic functions, we can rewrite the equation as [latex]\\ln(2x) - \\ln(x^6) = 0[\/latex].
      \r\nUsing the quotient property, this becomes\r\n\r\n
      [latex]\\ln\\big(\\frac{2}{x^5}\\big)=0[\/latex]<\/div>\r\n

      Therefore, [latex]\\frac{2}{x^5}=1[\/latex], which implies [latex]x=\\sqrt[5]{2}[\/latex].\u00a0<\/p>\r\n<\/li>\r\n<\/ol>\r\n

      [\/hidden-answer]<\/p>\r\n<\/section>\r\n

      \r\n

      Solve each of the following equations for [latex]x[\/latex].<\/p>\r\n

        \r\n\t
      1. [latex]5^x=2[\/latex]<\/li>\r\n\t
      2. [latex]e^x+6e^{\u2212x}=5[\/latex]<\/li>\r\n<\/ol>\r\n

        [reveal-answer q=\"fs-id1170572550555\"]Show Solution[\/reveal-answer]
        \r\n[hidden-answer a=\"fs-id1170572550555\"]<\/p>\r\n

          \r\n\t
        1. Applying the natural logarithm function to both sides of the equation, we have
          \r\n
          [latex]\\ln 5^x=\\ln 2[\/latex]<\/div>\r\n

          Using the power property of logarithms,<\/p>\r\n

          [latex]x \\ln 5=\\ln 2[\/latex]<\/div>\r\n

          Therefore, [latex]x=\\frac{\\ln 2 }{\\ln 5}[\/latex].<\/p>\r\n<\/li>\r\n\t

        2. Multiplying both sides of the equation by [latex]e^x[\/latex], we arrive at the equation
          \r\n
          [latex]e^{2x}+6=5e^x[\/latex]<\/div>\r\n

          Rewriting this equation as<\/p>\r\n

          [latex]e^{2x}-5e^x+6=0[\/latex],<\/div>\r\n

          we can then rewrite it as a quadratic equation in [latex]e^x[\/latex]:<\/p>\r\n

          [latex](e^x)^2-5(e^x)+6=0[\/latex]<\/div>\r\n

          Now we can solve the quadratic equation. Factoring this equation, we obtain<\/p>\r\n

          [latex](e^x-3)(e^x-2)=0[\/latex]<\/div>\r\n

          Therefore, the solutions satisfy [latex]e^x=3[\/latex] and [latex]e^x=2[\/latex]. Taking the natural logarithm of both sides gives us the solutions<\/p>\r\n

          [latex]x=\\ln 3, \\, \\ln 2[\/latex]<\/div>\r\n<\/li>\r\n<\/ol>\r\n\r\nWatch the following video to see the worked solution to this example.