{"id":485,"date":"2025-02-13T19:45:23","date_gmt":"2025-02-13T19:45:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-exponential-functions-and-logarithms-learn-it-2\/"},"modified":"2025-02-13T19:45:23","modified_gmt":"2025-02-13T19:45:23","slug":"integrals-exponential-functions-and-logarithms-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-exponential-functions-and-logarithms-learn-it-2\/","title":{"raw":"Integrals, Exponential Functions, and Logarithms: Learn It 2","rendered":"Integrals, Exponential Functions, and Logarithms: Learn It 2"},"content":{"raw":"\n

Defining the Number [latex]e[\/latex]<\/h2>\n

Now that we have the natural logarithm defined, we can use that function to define the number [latex]e.[\/latex]<\/p>\n

\n

the number [latex]e[\/latex]<\/h3>\n

The number [latex]e[\/latex] is defined to be the real number such that<\/p>\n

[latex]\\text{ln}e=1[\/latex]<\/div>\n<\/section>\n
\n

To put it another way, the area under the curve [latex]y=\\frac{1}{t}\/latex] between [latex]t=1[\/latex] and [latex]t=e[\/latex] is [latex]1[\/latex] (Figure 3).<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"304\"]\"This Figure 3. The area under the curve from 1 to [latex]e[\/latex] is equal to one.[\/caption]\n\n

The proof that such a number exists and is unique is left to you. (Hint<\/em>: Use the Intermediate Value Theorem to prove existence and the fact that [latex]\\text{ln}x[\/latex] is increasing to prove uniqueness.)<\/p>\n<\/section>\n

The number [latex]e[\/latex] can be shown to be irrational, although we won\u2019t do so here. Its approximate value is given by<\/p>\n

[latex]e\\approx 2.71828182846[\/latex]<\/div>\n

The Exponential Function<\/h2>\n

We now turn our attention to the function [latex]{e}^{x}.[\/latex] Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by [latex]\\text{exp }x.[\/latex] Then,<\/p>\n

[latex]\\text{exp}(\\text{ln }x)=x\\text{ for }x>0\\text{ and }\\text{ln}(\\text{exp }x)=x\\text{ for all }x.[\/latex]<\/div>\n

The following figure shows the graphs of [latex]\\text{exp }x[\/latex] and [latex]\\text{ln }x.[\/latex]<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"421\"]\"This Figure 4. The graphs of [latex]\\text{ln}x[\/latex] and [latex]\\text{exp}x.[\/latex][\/caption]\n\n

We hypothesize that [latex]\\text{exp }x={e}^{x}.[\/latex]<\/p>\n

For rational values of [latex]x,[\/latex] this is easy to show. If [latex]x[\/latex] is rational, then we have [latex]\\text{ln }({e}^{x})=x\\text{ln }e=x.[\/latex] Thus, when [latex]x[\/latex] is rational, [latex]{e}^{x}=\\text{exp }x.[\/latex]<\/p>\n

For irrational values of [latex]x,[\/latex] we simply define [latex]{e}^{x}[\/latex] as the inverse function of [latex]\\text{ln }x.[\/latex]<\/p>\n

\n

defining the exponential function<\/h3>\n

For any real number [latex]x,[\/latex] define [latex]y={e}^{x}[\/latex] to be the number for which<\/p>\n

[latex]\\text{ln}y=\\text{ln}({e}^{x})=x[\/latex]<\/div>\n<\/section>\n

Then we have [latex]{e}^{x}=\\text{exp}(x)[\/latex] for all [latex]x,[\/latex] and thus<\/p>\n

[latex]{e}^{\\text{ln}x}=x\\text{ for }x>0\\text{ and }\\text{ln}({e}^{x})=x[\/latex]<\/div>\n

for all [latex]x.[\/latex]<\/p>\n

Properties of the Exponential Function<\/h2>\n

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of [latex]e,[\/latex] we must verify that the usual laws of exponents hold for the function [latex]{e}^{x}.[\/latex]<\/p>\n

\n

properties of the exponential function<\/h3>\n

If [latex]p[\/latex] and [latex]q[\/latex] are any real numbers and [latex]r[\/latex] is a rational number, then<\/p>\n

    \n\t
  1. [latex]{e}^{p}{e}^{q}={e}^{p+q}[\/latex]<\/li>\n\t
  2. [latex]\\frac{{e}^{p}}{{e}^{q}}={e}^{p-q}[\/latex]<\/li>\n\t
  3. [latex]{({e}^{p})}^{r}={e}^{pr}[\/latex]<\/li>\n<\/ol>\n<\/section>\n
    \n

    Proof<\/strong><\/p>\n

    \n

    Note that if [latex]p[\/latex] and [latex]q[\/latex] are rational, the properties hold. However, if [latex]p[\/latex] or [latex]q[\/latex] are irrational, we must apply the inverse function definition of [latex]{e}^{x}[\/latex] and verify the properties. Only the first property is verified here; the other two are left to you. We have<\/p>\n

    [latex]\\text{ln}({e}^{p}{e}^{q})=\\text{ln}({e}^{p})+\\text{ln}({e}^{q})=p+q=\\text{ln}({e}^{p+q}).[\/latex]<\/div>\n

     <\/p>\n

    Since [latex]\\text{ln}x[\/latex] is one-to-one, then<\/p>\n

    [latex]{e}^{p}{e}^{q}={e}^{p+q}.[\/latex]<\/div>\n

    [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

    As with part iv. of the logarithm properties, we can extend property iii. to irrational values of [latex]r,[\/latex] and we do so by the end of the section.<\/p>\n

    We also want to verify the differentiation formula for the function [latex]y={e}^{x}.[\/latex]<\/p>\n

    To do this, we need to use implicit differentiation. Let [latex]y={e}^{x}.[\/latex] Then,<\/p>\n

    [latex]\\begin{array}{ccc}\\hfill \\text{ln}y& =\\hfill & x\\hfill \\\\ \\hfill \\frac{d}{dx}\\text{ln}y& =\\hfill & \\frac{d}{dx}x\\hfill \\\\ \\hfill \\frac{1}{y}\\frac{dy}{dx}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{dy}{dx}& =\\hfill & y.\\hfill \\end{array}[\/latex]<\/div>\n

    Thus, we see<\/p>\n

    [latex]\\frac{d}{dx}{e}^{x}={e}^{x}[\/latex]<\/div>\n

    as desired, which leads immediately to the integration formula<\/p>\n

    [latex]\\displaystyle\\int {e}^{x}dx={e}^{x}+C[\/latex]<\/div>\n

    We apply these formulas in the following examples.<\/p>\n

    \n
    \n

    Evaluate the following derivatives:<\/p>\n

      \n\t
    1. [latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}[\/latex]<\/li>\n\t
    2. [latex]\\frac{d}{dx}{e}^{3{x}^{2}}[\/latex]<\/li>\n<\/ol>\n

      [reveal-answer q=\"fs-id1167794293256\"]Show Solution[\/reveal-answer]
      \n[hidden-answer a=\"fs-id1167794293256\"]<\/p>\n

      We apply the chain rule as necessary.<\/p>\n

        \n\t
      1. [latex]\\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)[\/latex]<\/li>\n\t
      2. [latex]\\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x[\/latex]<\/li>\n<\/ol>\n

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

        \n

        Evaluate the following integral: [latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793510888\"]Show Solution[\/reveal-answer]
        \n[hidden-answer a=\"fs-id1167793510888\"]\n\n

        [latex]\\displaystyle\\int \\frac{4}{{e}^{3x}}dx=-\\frac{4}{3}{e}^{-3x}+C[\/latex]<\/p>\n

        Watch the following video to see the worked solution to this example.<\/p>\n