{"id":1151,"date":"2026-01-18T15:02:16","date_gmt":"2026-01-18T05:02:16","guid":{"rendered":"https:\/\/archive4ones.com\/2ndstudy\/?p=1151"},"modified":"2026-01-19T14:48:15","modified_gmt":"2026-01-19T04:48:15","slug":"year11-math-3-1-4-exponential-and-logarithmic-functions","status":"publish","type":"post","link":"https:\/\/archive4ones.com\/2ndstudy\/?p=1151","title":{"rendered":"Year11 MATH 3-1-4"},"content":{"rendered":"\n

Chapter 4: Exponential and Logarithmic Functions<\/h1>\n\n\n\n

In the previous chapters, we focused on polynomial functions like x<\/mi>2<\/mn><\/msup>x^2<\/annotation><\/semantics><\/math> and x<\/mi>3<\/mn><\/msup>x^3<\/annotation><\/semantics><\/math>. In Chapter 4, we explore functions where the variable x<\/mi>x<\/annotation><\/semantics><\/math> is the exponent<\/strong>. These functions are essential for modeling population growth, radioactive decay, and compound interest.<\/p>\n\n\n\n
\n\n\n\n

4.1 Exponential Functions and the Number e<\/mi>e<\/annotation><\/semantics><\/math><\/h2>\n\n\n\n

An exponential function has the form f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>a<\/mi>x<\/mi><\/msup><\/mrow>f(x) = a^x<\/annotation><\/semantics><\/math>. However, in Mathematical Methods, we focus primarily on the natural exponential base<\/strong>, e<\/mi>e<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
    \n
  • The constant <\/strong>e<\/mi>e<\/annotation><\/semantics><\/math>:<\/strong> Approximately 2.71828<\/mn>2.71828<\/annotation><\/semantics><\/math>. It is a unique number because the gradient of the function f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>e<\/mi>x<\/mi><\/msup><\/mrow>f(x) = e^x<\/annotation><\/semantics><\/math> is exactly equal to the value of the function itself at any point.<\/li>\n\n\n\n
  • Asymptotes:<\/strong> The graph of y<\/mi>=<\/mo>e<\/mi>x<\/mi><\/msup><\/mrow>y = e^x<\/annotation><\/semantics><\/math> never touches the x<\/mi>x<\/annotation><\/semantics><\/math>-axis; it has a horizontal asymptote at y<\/mi>=<\/mo>0<\/mn><\/mrow>y = 0<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    4.2 Logarithmic Functions<\/h2>\n\n\n\n

    The logarithm is the inverse<\/strong> of an exponential. If y<\/mi>=<\/mo>e<\/mi>x<\/mi><\/msup><\/mrow>y = e^x<\/annotation><\/semantics><\/math>, then x<\/mi>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>y<\/mi>)<\/mo><\/mrow>x = \\ln(y)<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
      \n
    • ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>\\ln(x)<\/annotation><\/semantics><\/math> (Natural Log):<\/strong> This is the logarithm to the base e<\/mi>e<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
    • Domain and Range:<\/strong> You cannot take the log of a negative number or zero. Therefore, for f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>f(x) = \\ln(x)<\/annotation><\/semantics><\/math>, the domain is x<\/mi>><\/mo>0<\/mn><\/mrow>x > 0<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      4.3 Logarithm Laws<\/h2>\n\n\n\n

      To solve equations involving e<\/mi>e<\/annotation><\/semantics><\/math> and ln<\/mi>\u2061<\/mo><\/mspace><\/mrow>\\ln<\/annotation><\/semantics><\/math>, you must master the log laws. These are the same regardless of the base, but we use them most often with the natural log:<\/p>\n\n\n\n
        \n
      1. Product Law:<\/strong> ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>b<\/mi>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>)<\/mo>+<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>b<\/mi>)<\/mo><\/mrow>\\ln(ab) = \\ln(a) + \\ln(b)<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
      2. Quotient Law:<\/strong> ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>b<\/mi><\/mfrac>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>)<\/mo>\u2212<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>b<\/mi>)<\/mo><\/mrow>\\ln(\\frac{a}{b}) = \\ln(a) – \\ln(b)<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
      3. Power Law:<\/strong> ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>n<\/mi><\/msup>)<\/mo>=<\/mo>n<\/mi><\/mspace>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>a<\/mi>)<\/mo><\/mrow>\\ln(a^n) = n \\ln(a)<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
      4. Inverse Properties:<\/strong> ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>e<\/mi>x<\/mi><\/msup>)<\/mo>=<\/mo>x<\/mi><\/mrow>\\ln(e^x) = x<\/annotation><\/semantics><\/math> and e<\/mi>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow><\/msup>=<\/mo>x<\/mi><\/mrow>e^{\\ln(x)} = x<\/annotation><\/semantics><\/math><\/li>\n<\/ol>\n\n\n\n
        \n

        Worked Example 1: Solving for x<\/mi>x<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Solve 5<\/mn>e<\/mi>2<\/mn>x<\/mi><\/mrow><\/msup>=<\/mo>20<\/mn><\/mrow>5e^{2x} = 20<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
          \n
        1. Divide by 5: e<\/mi>2<\/mn>x<\/mi><\/mrow><\/msup>=<\/mo>4<\/mn><\/mrow>e^{2x} = 4<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
        2. Take the natural log of both sides: ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>e<\/mi>2<\/mn>x<\/mi><\/mrow><\/msup>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>4<\/mn>)<\/mo><\/mrow>\\ln(e^{2x}) = \\ln(4)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
        3. Use the inverse property: 2<\/mn>x<\/mi>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>4<\/mn>)<\/mo><\/mrow>2x = \\ln(4)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
        4. Solve for x<\/mi>x<\/annotation><\/semantics><\/math>: x<\/mi>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>4<\/mn>)<\/mo><\/mrow>2<\/mn><\/mfrac>\u2248<\/mo>0.693<\/mn><\/mrow>x = \\frac{\\ln(4)}{2} \\approx 0.693<\/annotation><\/semantics><\/math>.<\/li>\n<\/ol>\n<\/blockquote>\n\n\n\n
          \n\n\n\n

          4.4 Derivatives of e<\/mi>x<\/mi><\/msup>e^x<\/annotation><\/semantics><\/math> and ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>\\ln(x)<\/annotation><\/semantics><\/math><\/h2>\n\n\n\n

          Calculus becomes very elegant when dealing with base e<\/mi>e<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
            \n
          • The Derivative of <\/strong>e<\/mi>x<\/mi><\/msup>e^x<\/annotation><\/semantics><\/math>:<\/strong> d<\/mi>d<\/mi>x<\/mi><\/mrow><\/mfrac>(<\/mo>e<\/mi>k<\/mi>x<\/mi><\/mrow><\/msup>)<\/mo>=<\/mo>k<\/mi>e<\/mi>k<\/mi>x<\/mi><\/mrow><\/msup><\/mrow>\\frac{d}{dx}(e^{kx}) = ke^{kx}<\/annotation><\/semantics><\/math> (Essentially, the function stays the same, but you multiply by the derivative of the exponent).<\/li>\n\n\n\n
          • The Derivative of ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>\\ln(x)<\/annotation><\/semantics><\/math>: d<\/mi>d<\/mi>x<\/mi><\/mrow><\/mfrac>(<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo>)<\/mo>=<\/mo>1<\/mn>x<\/mi><\/mfrac><\/mrow>\\frac{d}{dx}(\\ln(x)) = \\frac{1}{x}<\/annotation><\/semantics><\/math><\/li>\n<\/ul>\n\n\n\n
            \n

            Worked Example 2: Differentiating<\/p>\n\n\n\n

            Differentiate f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>e<\/mi>5<\/mn>x<\/mi><\/mrow><\/msup>+<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>f(x) = e^{5x} + \\ln(x)<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
              \n
            1. Differentiate e<\/mi>5<\/mn>x<\/mi><\/mrow><\/msup>e^{5x}<\/annotation><\/semantics><\/math>: The derivative of the power (5<\/mn>x<\/mi><\/mrow>5x<\/annotation><\/semantics><\/math>) is 5. Result: 5<\/mn>e<\/mi>5<\/mn>x<\/mi><\/mrow><\/msup><\/mrow>5e^{5x}<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
            2. Differentiate ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>\\ln(x)<\/annotation><\/semantics><\/math>: Result: 1<\/mn>x<\/mi><\/mfrac>\\frac{1}{x}<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
            3. Final Answer:<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>5<\/mn>e<\/mi>5<\/mn>x<\/mi><\/mrow><\/msup>+<\/mo>1<\/mn>x<\/mi><\/mfrac><\/mrow>f'(x) = 5e^{5x} + \\frac{1}{x}<\/annotation><\/semantics><\/math>.<\/li>\n<\/ol>\n<\/blockquote>\n\n\n\n
              \n\n\n\n

              4.5 Practice Problems<\/h2>\n\n\n\n

              Part A: Algebra and Log Laws<\/h3>\n\n\n\n
                \n
              1. Simplify 2<\/mn><\/mspace>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo>+<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>y<\/mi>)<\/mo><\/mrow>2\\ln(x) + \\ln(y)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              2. Solve for x<\/mi>x<\/annotation><\/semantics><\/math>: e<\/mi>x<\/mi>\u2212<\/mo>3<\/mn><\/mrow><\/msup>=<\/mo>10<\/mn><\/mrow>e^{x-3} = 10<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
              3. Solve for x<\/mi>x<\/annotation><\/semantics><\/math>: ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>2<\/mn>x<\/mi>)<\/mo>=<\/mo>5<\/mn><\/mrow>\\ln(2x) = 5<\/annotation><\/semantics><\/math>.<\/li>\n<\/ol>\n\n\n\n

                Part B: Graphs and Features<\/h3>\n\n\n\n
                  \n
                1. State the horizontal asymptote of f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>e<\/mi>x<\/mi><\/msup>+<\/mo>4<\/mn><\/mrow>f(x) = e^x + 4<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                2. Find the x<\/mi>x<\/annotation><\/semantics><\/math>-intercept of g<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>\u2212<\/mo>2<\/mn>)<\/mo><\/mrow>g(x) = \\ln(x – 2)<\/annotation><\/semantics><\/math>. (Hint: Set y<\/mi>=<\/mo>0<\/mn><\/mrow>y=0<\/annotation><\/semantics><\/math> and remember e<\/mi>0<\/mn><\/msup>=<\/mo>1<\/mn><\/mrow>e^0 = 1<\/annotation><\/semantics><\/math>).<\/li>\n<\/ol>\n\n\n\n

                  Part C: Calculus<\/h3>\n\n\n\n
                    \n
                  1. Find the derivative of y<\/mi>=<\/mo>3<\/mn>e<\/mi>2<\/mn>x<\/mi><\/mrow><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>2<\/mn><\/msup><\/mrow>y = 3e^{2x} – 4x^2<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                  2. Find the gradient of the curve f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>)<\/mo><\/mrow>f(x) = \\ln(x)<\/annotation><\/semantics><\/math> at the point where x<\/mi>=<\/mo>5<\/mn><\/mrow>x = 5<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                  3. Challenge:<\/strong> Find the equation of the tangent line to f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>e<\/mi>x<\/mi><\/msup><\/mrow>f(x) = e^x<\/annotation><\/semantics><\/math> at the point where x<\/mi>=<\/mo>0<\/mn><\/mrow>x = 0<\/annotation><\/semantics><\/math>.<\/li>\n<\/ol>\n\n\n\n
                    \n\n\n\n

                    Solutions (Summary)<\/h3>\n\n\n\n
                      \n
                    • 1.<\/strong> ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>2<\/mn><\/msup>y<\/mi>)<\/mo><\/mrow>\\ln(x^2 y)<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 2.<\/strong> x<\/mi>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>10<\/mn>)<\/mo>+<\/mo>3<\/mn>\u2248<\/mo>5.30<\/mn><\/mrow>x = \\ln(10) + 3 \\approx 5.30<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 3.<\/strong> x<\/mi>=<\/mo>e<\/mi>5<\/mn><\/msup>2<\/mn><\/mfrac>\u2248<\/mo>74.21<\/mn><\/mrow>x = \\frac{e^5}{2} \\approx 74.21<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 4.<\/strong> y<\/mi>=<\/mo>4<\/mn><\/mrow>y = 4<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 5.<\/strong> 0<\/mn>=<\/mo>ln<\/mi>\u2061<\/mo><\/mrow>(<\/mo>x<\/mi>\u2212<\/mo>2<\/mn>)<\/mo>\u2192<\/mo>e<\/mi>0<\/mn><\/msup>=<\/mo>x<\/mi>\u2212<\/mo>2<\/mn>\u2192<\/mo>1<\/mn>=<\/mo>x<\/mi>\u2212<\/mo>2<\/mn>\u2192<\/mo>x<\/mi>=<\/mo>3<\/mn><\/mrow>0 = \\ln(x-2) \\rightarrow e^0 = x-2 \\rightarrow 1 = x-2 \\rightarrow x=3<\/annotation><\/semantics><\/math>. Point is (<\/mo>3<\/mn>,<\/mo>0<\/mn>)<\/mo><\/mrow>(3, 0)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
                    • 6.<\/strong> d<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>6<\/mn>e<\/mi>2<\/mn>x<\/mi><\/mrow><\/msup>\u2212<\/mo>8<\/mn>x<\/mi><\/mrow>\\frac{dy}{dx} = 6e^{2x} – 8x<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 7.<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>x<\/mi>)<\/mo>=<\/mo>1<\/mn>x<\/mi><\/mfrac><\/mrow>f'(x) = \\frac{1}{x}<\/annotation><\/semantics><\/math>. At x<\/mi>=<\/mo>5<\/mn><\/mrow>x=5<\/annotation><\/semantics><\/math>, gradient is 1<\/mn>5<\/mn><\/mfrac>\\frac{1}{5}<\/annotation><\/semantics><\/math>.o<\/mi>r<\/mi>0.2<\/mn><\/mrow> or 0.2<\/annotation><\/semantics><\/math><\/li>\n\n\n\n
                    • 8.<\/strong> f<\/mi>\u2032<\/mo><\/msup>(<\/mo>0<\/mn>)<\/mo>=<\/mo>e<\/mi>0<\/mn><\/msup>=<\/mo>1<\/mn><\/mrow>f'(0) = e^0 = 1<\/annotation><\/semantics><\/math> (gradient). Point is (<\/mo>0<\/mn>,<\/mo>1<\/mn>)<\/mo><\/mrow>(0, 1)<\/annotation><\/semantics><\/math>. Tangent: y<\/mi>\u2212<\/mo>1<\/mn>=<\/mo>1<\/mn>(<\/mo>x<\/mi>\u2212<\/mo>0<\/mn>)<\/mo>\u2192<\/mo>y<\/mi>=<\/mo>x<\/mi>+<\/mo>1<\/mn><\/mrow>y – 1 = 1(x – 0) \\rightarrow y = x + 1<\/annotation><\/semantics><\/math>.<\/li>\n<\/ul>\n\n\n\n
                      \n\n\n\n

                      <\/p>\n","protected":false},"excerpt":{"rendered":"

                      Chapter 4: Exponential and Logarithmic Functions In the previous chapters, we focused on polynomial functions like x2x^2 and x3x^3. In Chapter 4, we explore functions where the variable xx is the exponent. These functions are essential for modeling population growth, radioactive decay, and compound interest. 4.1 Exponential Functions and the Number ee An exponential function […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-1151","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1151","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1151"}],"version-history":[{"count":5,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1151\/revisions"}],"predecessor-version":[{"id":1187,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1151\/revisions\/1187"}],"wp:attachment":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}