Natural Logarithm of Absolute Value, foundations for calculus lesson

The natural logarithm refuses negative inputs, yet calculus keeps meeting quantities like $\frac{1}{x}$ that make sense on both sides of zero. Putting an absolute value inside the logarithm gives a logarithm that also accepts negative inputs. The input first becomes its distance from zero, and only that positive distance reaches the logarithm. The result is a graph with two separate pieces, one on each side of a forbidden point.

Key terms

The function ln of absolute x

Definition The function $f(x) = \ln \lvert x \rvert$ is the composition of the absolute value with the natural logarithm. To evaluate it at a real number $x$ , first compute $\lvert x \rvert$ , then apply $\ln$ . It is defined exactly for those $x$ with $\lvert x \rvert > 0$ .

In plain words Strip the sign off the input, then take the natural logarithm of what is left.

Example Two inputs with the same distance from zero give the same output.

\begin{gathered} \ln \lvert -e^3 \rvert = \ln\left(e^3\right) = 3, \\[0.4em] \ln \lvert e^3 \rvert = 3. \end{gathered}

The order matters. The expression $\lvert \ln x \rvert$ applies the logarithm first and is a different function, defined only for $x > 0$ .

Positive argument and the nonzero domain

Definition The natural logarithm $\ln u$ is defined exactly when its argument satisfies $u > 0$ . For $\ln \lvert x \rvert$ the argument is $\lvert x \rvert$ , and $\lvert x \rvert > 0$ holds precisely when $x \ne 0$ . So the domain of $\ln \lvert x \rvert$ is

(-\infty, 0) \cup (0, \infty).

In plain words The absolute value makes every negative input positive, so negatives are now allowed. Zero stays zero, and $\ln 0$ is undefined, so $x = 0$ is still excluded.

Example At $x = -2$ the argument is $\lvert -2 \rvert = 2 > 0$ , so $\ln \lvert -2 \rvert = \ln 2 \approx 0.693$ is defined. At $x = 0$ the argument is $0$ and there is no output. Near zero the outputs fall without bound. For instance $\ln \lvert \pm 0.001 \rvert \approx -6.91$ , and $\ln \lvert x \rvert \to -\infty$ as $x \to 0$ from either side, so the line $x = 0$ is a vertical asymptote.

Piecewise description

Definition Since $\lvert x \rvert = x$ for $x > 0$ and $\lvert x \rvert = -x$ for $x < 0$ ,

\ln \lvert x \rvert = \begin{cases} \ln x, & x > 0, \\ \ln(-x), & x < 0. \end{cases}

No case is written for $x = 0$ , because the function is not defined there.

In plain words On the right of zero it is the ordinary natural logarithm. On the left it is the natural logarithm of the positive number $-x$ .

Example At $x = -\frac{1}{e}$ use the second case.

\ln\left(-\left(-\frac{1}{e}\right)\right) = \ln\left(\frac{1}{e}\right) = -1.

The second case never feeds a negative number to the logarithm, because $-x > 0$ whenever $x < 0$ .

Two branches

Definition The graph of $y = \ln \lvert x \rvert$ consists of two disconnected branches, one over $(0, \infty)$ and one over $(-\infty, 0)$ . The right branch is the graph of $\ln x$ and is increasing on $(0, \infty)$ . The left branch is the graph of $\ln(-x)$ and is decreasing on $(-\infty, 0)$ . The zeros are $x = 1$ and $x = -1$ . The range is all real numbers, because for every real $y$ the two inputs $x = e^y$ and $x = -e^y$ both give output $y$ .

In plain words You draw the familiar logarithm curve on the right, then a second curve on the left that falls as it moves toward zero. The pen has to lift at $x = 0$ .

Example Exact points are easy to build from powers of $e$ .

$x$	$\lvert x \rvert$	$\ln \lvert x \rvert$
$-e^2$	$e^2$	$2$
$-e$	$e$	$1$
$-1$	$1$	$0$
$-\frac{1}{e}$	$\frac{1}{e}$	$-1$
$\frac{1}{e}$	$\frac{1}{e}$	$-1$
$1$	$1$	$0$
$e$	$e$	$1$
$e^2$	$e^2$	$2$

Reading the left half from top to bottom, $x$ moves toward zero and the output drops from $2$ to $-1$ , so the left branch decreases. On the right half $x$ moves away from zero and the output climbs.

Even symmetry

Definition A function $f$ is even when $-x$ is in its domain whenever $x$ is, and $f(-x) = f(x)$ for every $x$ in the domain. The function $\ln \lvert x \rvert$ is even, because its domain is symmetric about $0$ and

\ln \lvert -x \rvert = \ln \lvert x \rvert \qquad \text{for all } x \ne 0.

Its graph is symmetric about the $y$ -axis.

In plain words The left branch is the mirror image of the right branch across the $y$ -axis.

Example Reflecting the point $(e, 1)$ on the right branch gives $(-e, 1)$ , and indeed $\ln \lvert -e \rvert = 1$ . This also explains why one branch increases and the other decreases. A mirror reverses left and right, so a curve that rises as you move right becomes one that rises as you move left, which means it falls as $x$ increases.

Absolute value inside a logarithm

Definition For a function $g$ , the expression $\ln \lvert g(x) \rvert$ is defined exactly at the inputs $x$ in the domain of $g$ with $g(x) \ne 0$ . In particular, for a real number $h$ the function $\ln \lvert x - h \rvert$ is the graph of $\ln \lvert x \rvert$ shifted $h$ units to the right. Its domain is all $x \ne h$ , its vertical asymptote is $x = h$ , its zeros are $x = h - 1$ and $x = h + 1$ , and its graph is symmetric about the line $x = h$ .

In plain words Find where the inside is zero and remove that point. Everywhere else the absolute value turns the inside into a positive distance.

Example Consider $f(x) = \ln \lvert x - 3 \rvert$ . The inside is zero at $x = 3$ , so the domain is $x \ne 3$ and the line $x = 3$ is the asymptote. The zeros solve $\lvert x - 3 \rvert = 1$ , so

x = 2 \quad \text{or} \quad x = 4.

Points at equal distance from $3$ match, for example $f(3 + e) = f(3 - e) = 1$ .

Logarithmic identity with absolute values

Definition For every real $x \ne 0$ ,

\ln\left(x^2\right) = 2 \ln \lvert x \rvert.

This follows from the power law applied to the positive number $\lvert x \rvert$ , since $x^2 = \lvert x \rvert^2$ . The shorter form $2 \ln x$ is correct only for $x > 0$ . More generally, for nonzero real numbers $u$ and $v$ ,

\ln \lvert uv \rvert = \ln \lvert u \rvert + \ln \lvert v \rvert.

In plain words When you pull an even power out of a logarithm, an absolute value has to appear, or you lose the negative half of the domain.

Example At $x = -e$ the left side is $\ln\left(e^2\right) = 2$ . The correct right side is $2 \ln \lvert -e \rvert = 2 \cdot 1 = 2$ . The careless form $2 \ln(-e)$ is undefined, so writing $\ln\left(x^2\right) = 2 \ln x$ would wrongly throw away every negative input. Looking ahead, this matters in calculus. There you will learn that the slope of $\ln \lvert x \rvert$ at any $x \ne 0$ is $\frac{1}{x}$ , on both sides of zero, so the absolute value is exactly what keeps the negative inputs.

Check yourself

Find the domain and the zeros of $f(x) = \ln \lvert x + 2 \rvert$ .

Show the answer

The inside $x + 2$ is zero at $x = -2$ , so the domain is all $x \ne -2$ . The zeros solve $\lvert x + 2 \rvert = 1$ , which gives $x + 2 = 1$ or $x + 2 = -1$ , so $x = -1$ or $x = -3$ .

Evaluate $\ln \lvert -e^3 \rvert$ and $\ln \left\lvert -\frac{1}{e^2} \right\rvert$ exactly.

Show the answer

The absolute values are $e^3$ and $e^{-2}$ , so the outputs are $3$ and $-2$ .

A student simplifies $\ln\left(x^2\right)$ to $2 \ln x$ and then says the expression is undefined at $x = -4$ . What went wrong, and what is the correct value?

Show the answer

The rewrite $2 \ln x$ is valid only for $x > 0$ . The original $\ln\left(x^2\right)$ is defined at $x = -4$ , and the correct identity gives

\begin{aligned} \ln\left((-4)^2\right) &= 2 \ln \lvert -4 \rvert \\[0.4em] &= 2 \ln 4 \approx 2.773. \end{aligned}

Natural Logarithm of Absolute Value

Key terms

The function ln of absolute x

Positive argument and the nonzero domain

Piecewise description

Two branches

Even symmetry

Absolute value inside a logarithm

Logarithmic identity with absolute values

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus