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 \(f \left(\right. x \left.\right) = a^{x} .\) 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 \(f \left(\right. x \left.\right) = a^{x}\) in terms of the exponential function \(e^{x} .\) We then examine logarithms with bases other than \(e\) as inverse functions of exponential functions.

Definition

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

\[y = a^{x} = e^{x \text{ln} a} .\]

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

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

The corresponding integration formula follows immediately.

Theorem 6.20

Derivatives and Integrals Involving General Exponential Functions

Let \(a > 0 .\) Then,

\[\frac{d}{d x} a^{x} = a^{x} \text{ln} a\]

and

\[\int a^{x} d x = \frac{1}{\text{ln} a} a^{x} + C .\]

If \(a \neq 1 ,\) 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,

\[y = \text{log}_{a} x \text{if and only if} x = a^{y} .\]

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

\[\begin{aligned} \text{ln} x & = & \text{ln} \left(\right. a^{y} \left.\right) \\ \text{ln} x & = & y \text{ln} a \\ y & = & \frac{\text{ln} x}{\text{ln} a} \\ \text{log}_{a} x & = & \frac{\text{ln} x}{\text{ln} a} . \end{aligned}\]

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,

\[\begin{aligned} \frac{d y}{d x} & = \frac{d}{d x} \left(\right. \text{log}_{a} x \left.\right) \\ & = \frac{d}{d x} \left(\right. \frac{\text{ln} x}{\text{ln} a} \left.\right) \\ & = \left(\right. \frac{1}{\text{ln} a} \left.\right) \frac{d}{d x} \left(\right. \text{ln} x \left.\right) \\ & = \frac{1}{\text{ln} a} \cdot \frac{1}{x} \\ & = \frac{1}{x \text{ln} a} . \end{aligned}\]

Theorem 6.21

Derivatives of General Logarithm Functions

Let \(a > 0 .\) Then,

\[\frac{d}{d x} \text{log}_{a} x = \frac{1}{x \text{ln} a} .\]

Example 6.40

Calculating Derivatives of General Exponential and Logarithm Functions

Evaluate the following derivatives:

\(\frac{d}{d t} \left(\right. 4^{t} \cdot 2^{t^{2}} \left.\right)\)
\(\frac{d}{d x} \text{log}_{8} \left(\right. 7 x^{2} + 4 \left.\right)\)

Solution

We need to apply the chain rule as necessary.

\(\frac{d}{d t} \left(\right. 4^{t} \cdot 2^{t^{2}} \left.\right) = \frac{d}{d t} \left(\right. 2^{2 t} \cdot 2^{t^{2}} \left.\right) = \frac{d}{d t} \left(\right. 2^{2 t + t^{2}} \left.\right) = 2^{2 t + t^{2}} \text{ln} \left(\right. 2 \left.\right) \left(\right. 2 + 2 t \left.\right)\)
\(\frac{d}{d x} \text{log}_{8} \left(\right. 7 x^{2} + 4 \left.\right) = \frac{1}{\left(\right. 7 x^{2} + 4 \left.\right) \left(\right. \text{ln} 8 \left.\right)} \left(\right. 14 x \left.\right)\)

Checkpoint 6.40

Evaluate the following derivatives:

\(\frac{d}{d t} 4^{t^{4}}\)
\(\frac{d}{d x} \text{log}_{3} \left(\right. \sqrt{x^{2} + 1} \left.\right)\)

Example 6.41

Integrating General Exponential Functions

Evaluate the following integral: \(\int \frac{3}{2^{3 x}} d x .\)

Solution

Use \(u -\text{substitution}\) and let \(u = −3 x .\) Then \(d u = −3 d x\) and we have

\[\int \frac{3}{2^{3 x}} d x = \int 3 \cdot 2^{−3 x} d x = − \int 2^{u} d u = - \frac{1}{\text{ln} 2} 2^{u} + C = - \frac{1}{\text{ln} 2} 2^{−3 x} + C .\]

Checkpoint 6.41

Evaluate the following integral: \(\int x^{2} 2^{x^{3}} d x .\)

This lesson is part of:

Applications of Integration

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