{"id":486,"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-3\/"},"modified":"2025-02-13T19:45:23","modified_gmt":"2025-02-13T19:45:23","slug":"integrals-exponential-functions-and-logarithms-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-exponential-functions-and-logarithms-learn-it-3\/","title":{"raw":"Integrals, Exponential Functions, and Logarithms: Learn It 3","rendered":"Integrals, Exponential Functions, and Logarithms: Learn It 3"},"content":{"raw":"\n

General Logarithmic and Exponential Functions<\/h2>\n

We close this section by looking at exponential functions and logarithms with bases other than [latex]e.[\/latex]<\/p>\n

Exponential functions are functions of the form [latex]f(x)={a}^{x}.[\/latex] Note that unless [latex]a=e,[\/latex] we still do not have a mathematically rigorous definition of these functions for irrational exponents.<\/p>\n

Let\u2019s rectify that here by defining the function [latex]f(x)={a}^{x}[\/latex] in terms of the exponential function [latex]{e}^{x}.[\/latex] <\/p>\n

\n

definition of general exponential functions<\/h3>\n

For any [latex]a>0,[\/latex] and for any real number [latex]x,[\/latex] define [latex]y={a}^{x}[\/latex] as follows:<\/p>\n

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

Now [latex]{a}^{x}[\/latex] is defined rigorously for all values of [latex]x[\/latex].<\/p>\n

This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of [latex]r.[\/latex] It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.<\/p>\n

Let\u2019s now apply this definition to calculate a differentiation formula for [latex]{a}^{x}.[\/latex] We have<\/p>\n

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

The corresponding integration formula follows immediately.<\/p>\n

\n

derivatives and integrals involving general exponential functions<\/h3>\n

Let [latex]a>0.[\/latex] Then,<\/p>\n

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

and<\/p>\n

[latex]\\displaystyle\\int {a}^{x}dx=\\frac{1}{\\text{ln }a}{a}^{x}+C[\/latex]<\/div>\n<\/section>\n

If [latex]a\\ne 1,[\/latex] then the function [latex]{a}^{x}[\/latex] is one-to-one and has a well-defined inverse. Its inverse is denoted by [latex]{\\text{log}}_{a}x.[\/latex] Then,<\/p>\n

[latex]y={\\text{log}}_{a}x\\text{if and only if}x={a}^{y}[\/latex]<\/div>\n

Note that general logarithm functions can be written in terms of the natural logarithm. Let [latex]y={\\text{log}}_{a}x.[\/latex] Then, [latex]x={a}^{y}.[\/latex] Taking the natural logarithm of both sides of this second equation, we get<\/p>\n

[latex]\\begin{array}{ccc}\\hfill \\text{ln}x& =\\hfill & \\text{ln}({a}^{y})\\hfill \\\\ \\hfill \\text{ln}x& =\\hfill & y\\text{ln}a\\hfill \\\\ \\hfill y& =\\hfill & \\frac{\\text{ln}x}{\\text{ln}a}\\hfill \\\\ \\hfill {\\text{log}}_{}x& =\\hfill & \\frac{\\text{ln}x}{\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\n

Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base [latex]a.[\/latex] Again, let [latex]y={\\text{log}}_{a}x.[\/latex] Then,<\/p>\n

[latex]\\begin{array}{cc}\\hfill \\frac{dy}{dx}& =\\frac{d}{dx}({\\text{log}}_{a}x)\\hfill \\\\ & =\\frac{d}{dx}(\\frac{\\text{ln}x}{\\text{ln}a})\\hfill \\\\ & =(\\frac{1}{\\text{ln}a})\\frac{d}{dx}(\\text{ln}x)\\hfill \\\\ & =\\frac{1}{\\text{ln}a}\u00b7\\frac{1}{x}\\hfill \\\\ & =\\frac{1}{x\\text{ln}a}.\\hfill \\end{array}[\/latex]<\/div>\n
\n

derivatives of general logarithm functions<\/h3>\n

Let [latex]a>0.[\/latex] Then,<\/p>\n

[latex]\\frac{d}{dx}{\\text{log}}_{a}x=\\frac{1}{x\\text{ln}a}[\/latex]<\/div>\n<\/section>\n
\n

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

    \n\t
  1. [latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})[\/latex]<\/li>\n\t
  2. [latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)[\/latex]<\/li>\n<\/ol>\n

    [reveal-answer q=\"fs-id1167793298214\"]Show Solution[\/reveal-answer]
    \n[hidden-answer a=\"fs-id1167793298214\"]<\/p>\n

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

      \n\t
    1. [latex]\\frac{d}{dt}({4}^{t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t}\u00b7{2}^{{t}^{2}})=\\frac{d}{dt}({2}^{2t+{t}^{2}})={2}^{2t+{t}^{2}}\\text{ln}(2)(2+2t)[\/latex]<\/li>\n\t
    2. [latex]\\frac{d}{dx}{\\text{log}}_{8}(7{x}^{2}+4)=\\frac{1}{(7{x}^{2}+4)(\\text{ln}8)}(14x)[\/latex]<\/li>\n<\/ol>\n

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

      \n

      Evaluate the following integral: [latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx.[\/latex]<\/p>\n\n[reveal-answer q=\"fs-id1167793979123\"]Show Solution[\/reveal-answer]
      \n[hidden-answer a=\"fs-id1167793979123\"]\n\n\n

      [latex]\\displaystyle\\int {x}^{2}{2}^{{x}^{3}}dx=\\frac{1}{3\\text{ln}2}{2}^{{x}^{3}}+C[\/latex]<\/p>\n

      Watch the following video to see the worked solution to the above Try It.<\/p>\n