Properties of the Natural Logarithm

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.

Theorem 6.15

Derivative of the Natural Logarithm

For \(x > 0 ,\) the derivative of the natural logarithm is given by

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

Theorem 6.16

Corollary to the Derivative of the Natural Logarithm

The function \(\text{ln} x\) is differentiable; therefore, it is continuous.

A graph of \(\text{ln} x\) is shown in Figure 6.76. Notice that it is continuous throughout its domain of \(\left(\right. 0 , \infty \left.\right) .\)

This figure is a graph. It is an increasing curve labeled f(x)=lnx. The curve is increasing with the y-axis as an asymptote. The curve intersects the x-axis at x=1.

Figure 6.76 The graph of \(f \left(\right. x \left.\right) = \text{ln} x\) shows that it is a continuous function.

Example 6.35

Calculating Derivatives of Natural Logarithms

Calculate the following derivatives:

\(\frac{d}{d x} \text{ln} \left(\right. 5 x^{3} - 2 \left.\right)\)
\(\frac{d}{d x} \left(\right. \text{ln} \left(\right. 3 x \left.\right) \left.\right)^{2}\)

Solution

We need to apply the chain rule in both cases.

\(\frac{d}{d x} \text{ln} \left(\right. 5 x^{3} - 2 \left.\right) = \frac{15 x^{2}}{5 x^{3} - 2}\)
\(\frac{d}{d x} \left(\right. \text{ln} \left(\right. 3 x \left.\right) \left.\right)^{2} = \frac{2 \left(\right. \text{ln} \left(\right. 3 x \left.\right) \left.\right) \cdot 3}{3 x} = \frac{2 \left(\right. \text{ln} \left(\right. 3 x \left.\right) \left.\right)}{x}\)

Checkpoint 6.35

Calculate the following derivatives:

\(\frac{d}{d x} \text{ln} \left(\right. 2 x^{2} + x \left.\right)\)
\(\frac{d}{d x} \left(\right. \text{ln} \left(\right. x^{3} \left.\right) \left.\right)^{2}\)

Note that if we use the absolute value function and create a new function \(\text{ln} \left|\right. x \left|\right. ,\) we can extend the domain of the natural logarithm to include \(x < 0 .\) Then \(\left(\right. d / \left(\right. d x \left.\right) \left.\right) \text{ln} \left|\right. x \left|\right. = 1 / x .\) This gives rise to the familiar integration formula.

Theorem 6.17

Integral of (1/u) du

The natural logarithm is the antiderivative of the function \(f \left(\right. u \left.\right) = 1 / u :\)

\[\int \frac{1}{u} d u = \text{ln} \left|\right. u \left|\right. + C .\]

Example 6.36

Calculating Integrals Involving Natural Logarithms

Calculate the integral \(\int \frac{x}{x^{2} + 4} d x .\)

Solution

Using \(u\)-substitution, let \(u = x^{2} + 4 .\) Then \(d u = 2 x d x\) and we have

\[\int \frac{x}{x^{2} + 4} d x = \frac{1}{2} \int \frac{1}{u} d u = \frac{1}{2} \text{ln} \left|\right. u \left|\right. + C = \frac{1}{2} \text{ln} \left|\right. x^{2} + 4 \left|\right. + C = \frac{1}{2} \text{ln} \left(\right. x^{2} + 4 \left.\right) + C .\]

Checkpoint 6.36

Calculate the integral \(\int \frac{x^{2}}{x^{3} + 6} d x .\)

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

Theorem 6.18

Properties of the Natural Logarithm

If \(a , b > 0\) and \(r\) is a rational number, then

\(\text{ln} 1 = 0\)
\(\text{ln} \left(\right. a b \left.\right) = \text{ln} a + \text{ln} b\)
\(\text{ln} \left(\right. \frac{a}{b} \left.\right) = \text{ln} a - \text{ln} b\)
\(\text{ln} \left(\right. a^{r} \left.\right) = r \text{ln} a\)

Proof

i. By definition, \(\text{ln} 1 = \int_{1}^{1} \frac{1}{t} d t = 0 .\)

ii. We have

\[\text{ln} \left(\right. a b \left.\right) = \int_{1}^{a b} \frac{1}{t} d t = \int_{1}^{a} \frac{1}{t} d t + \int_{a}^{a b} \frac{1}{t} d t .\]

Use \(u -\text{substitution}\) on the last integral in this expression. Let \(u = t / a .\) Then \(d u = \left(\right. 1 / a \left.\right) d t .\) Furthermore, when \(t = a , u = 1 ,\) and when \(t = a b , u = b .\) So we get

\[\text{ln} \left(\right. a b \left.\right) = \int_{1}^{a} \frac{1}{t} d t + \int_{a}^{a b} \frac{1}{t} d t = \int_{1}^{a} \frac{1}{t} d t + \int_{a}^{a b} \frac{a}{t} \cdot \frac{1}{a} d t = \int_{1}^{a} \frac{1}{t} d t + \int_{1}^{b} \frac{1}{u} d u = \text{ln} a + \text{ln} b .\]

iv. Note that

\[\frac{d}{d x} \text{ln} \left(\right. x^{r} \left.\right) = \frac{r x^{r - 1}}{x^{r}} = \frac{r}{x} .\]

Furthermore,

\[\frac{d}{d x} \left(\right. r \text{ln} x \left.\right) = \frac{r}{x} .\]

Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have

\[\text{ln} \left(\right. x^{r} \left.\right) = r \text{ln} x + C\]

for some constant \(C .\) Taking \(x = 1 ,\) we get

\[\begin{aligned} \text{ln} \left(\right. 1^{r} \left.\right) & = & r \text{ln} \left(\right. 1 \left.\right) + C \\ 0 & = & r \left(\right. 0 \left.\right) + C \\ C & = & 0 . \end{aligned}\]

Thus \(\text{ln} \left(\right. x^{r} \left.\right) = r \text{ln} x\) and the proof is complete. Note that we can extend this property to irrational values of \(r\) later in this section.
Part iii. follows from parts ii. and iv. and the proof is left to you.

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Example 6.37

Using Properties of Logarithms

Use properties of logarithms to simplify the following expression into a single logarithm:

\[\text{ln} 9 - 2 \text{ln} 3 + \text{ln} \left(\right. \frac{1}{3} \left.\right) .\]

Solution

We have

\[\text{ln} 9 - 2 \text{ln} 3 + \text{ln} \left(\right. \frac{1}{3} \left.\right) = \text{ln} \left(\right. 3^{2} \left.\right) - 2 \text{ln} 3 + \text{ln} \left(\right. 3^{−1} \left.\right) = 2 \text{ln} 3 - 2 \text{ln} 3 - \text{ln} 3 = − \text{ln} 3 .\]

Checkpoint 6.37

Use properties of logarithms to simplify the following expression into a single logarithm:

\[\text{ln} 8 - \text{ln} 2 - \text{ln} \left(\right. \frac{1}{4} \left.\right) .\]

This lesson is part of:

Applications of Integration

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