Integrals, Exponential Functions, and Logarithms: Learn It 3

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Integrals, Exponential Functions, and Logarithms: Learn It 3

General Logarithmic and Exponential Functions

We close this section by looking at exponential functions and logarithms with bases other than $e.$

Exponential functions are functions of the form $f(x)={a}^{x}.$ Note that unless $a=e,$ we still do not have a mathematically rigorous definition of these functions for irrational exponents.

Let’s rectify that here by defining the function $f(x)={a}^{x}$ in terms of the exponential function ${e}^{x}.$

definition of general exponential functions

For any $a>0,$ and for any real number $x,$ define $y={a}^{x}$ as follows:

y = a^{x} = e^{x ln a}

$y={a}^{x}={e}^{x\text{ln }a}$

Now ${a}^{x}$ is defined rigorously for all values of $x$ .

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 $r.$ It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.

Let’s now apply this definition to calculate a differentiation formula for ${a}^{x}.$ We have

\frac{d}{d x} a^{x} = \frac{d}{d x} e^{x ln a} = e^{x ln a} ln a = a^{x} ln a .

$\frac{d}{dx}{a}^{x}=\frac{d}{dx}{e}^{x\text{ln }a}={e}^{x\text{ln }a}\text{ln}a={a}^{x}\text{ln }a.$

The corresponding integration formula follows immediately.

derivatives and integrals involving general exponential functions

Let $a>0.$ Then,

\frac{d}{d x} a^{x} = a^{x} ln a

$\frac{d}{dx}{a}^{x}={a}^{x}\text{ln }a$

and

\int a^{x} d x = \frac{1}{ln a} a^{x} + C

$\displaystyle\int {a}^{x}dx=\frac{1}{\text{ln }a}{a}^{x}+C$

If $a\ne 1,$ then the function ${a}^{x}$ is one-to-one and has a well-defined inverse. Its inverse is denoted by ${\text{log}}_{a}x.$ Then,

y = {log}_{a} x if and only if x = a^{y}

$y={\text{log}}_{a}x\text{if and only if}x={a}^{y}$

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

\begin{array}{ccc} ln x & = & ln (a^{y}) \\ ln x & = & y ln a \\ y & = & \frac{ln x}{ln a} \\ {log}_{} x & = & \frac{ln x}{ln a} . \end{array}

$\begin{array}{ccc}\hfill \text{ln}x& =\hfill & \text{ln}({a}^{y})\hfill \\ \hfill \text{ln}x& =\hfill & y\text{ln}a\hfill \\ \hfill y& =\hfill & \frac{\text{ln}x}{\text{ln}a}\hfill \\ \hfill {\text{log}}_{}x& =\hfill & \frac{\text{ln}x}{\text{ln}a}.\hfill \end{array}$

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 $a.$ Again, let $y={\text{log}}_{a}x.$ Then,

\begin{array}{cc} \frac{d y}{d x} & = \frac{d}{d x} ({log}_{a} x) \\ = \frac{d}{d x} (\frac{ln x}{ln a}) \\ = (\frac{1}{ln a}) \frac{d}{d x} (ln x) \\ = \frac{1}{ln a} \cdot \frac{1}{x} \\ = \frac{1}{x ln a} . \end{array}

$\begin{array}{cc}\hfill \frac{dy}{dx}& =\frac{d}{dx}({\text{log}}_{a}x)\hfill \\ & =\frac{d}{dx}(\frac{\text{ln}x}{\text{ln}a})\hfill \\ & =(\frac{1}{\text{ln}a})\frac{d}{dx}(\text{ln}x)\hfill \\ & =\frac{1}{\text{ln}a}·\frac{1}{x}\hfill \\ & =\frac{1}{x\text{ln}a}.\hfill \end{array}$

derivatives of general logarithm functions

Let $a>0.$ Then,

\frac{d}{d x} {log}_{a} x = \frac{1}{x ln a}

$\frac{d}{dx}{\text{log}}_{a}x=\frac{1}{x\text{ln}a}$

Evaluate the following derivatives:

$\frac{d}{dt}({4}^{t}·{2}^{{t}^{2}})$
$\frac{d}{dx}{\text{log}}_{8}(7{x}^{2}+4)$

We need to apply the chain rule as necessary.

$\frac{d}{dt}({4}^{t}·{2}^{{t}^{2}})=\frac{d}{dt}({2}^{2t}·{2}^{{t}^{2}})=\frac{d}{dt}({2}^{2t+{t}^{2}})={2}^{2t+{t}^{2}}\text{ln}(2)(2+2t)$
$\frac{d}{dx}{\text{log}}_{8}(7{x}^{2}+4)=\frac{1}{(7{x}^{2}+4)(\text{ln}8)}(14x)$

Evaluate the following integral: $\displaystyle\int {x}^{2}{2}^{{x}^{3}}dx.$

$\displaystyle\int {x}^{2}{2}^{{x}^{3}}dx=\frac{1}{3\text{ln}2}{2}^{{x}^{3}}+C$

Watch the following video to see the worked solution to the above Try It.

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