Chapter 6 · Application of Derivatives

Exercise 6.2 · Q10: $\log x$ is increasing on $(0,\infty)$

Exercise 6.2Increasing and decreasing functions · Section 6.3

The question

Prove that the logarithmic function is increasing on $(0, \infty)$ ⁠.

Exam answer

Let $f(x) = \log x$ ⁠, $x > 0$ ⁠. Then $f'(x) = \frac{1}{x}$ ⁠.

For every $x$ in $(0, \infty)$ ⁠, $f'(x) = \frac{1}{x} > 0$ ⁠.

Hence the logarithmic function is increasing on $(0, \infty)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The slope of $\log x$ is $\frac{1}{x}$ ⁠, and on $(0, \infty)$ the number $x$ is positive, so the slope is positive everywhere. Positive slope on an interval means the function rises there (Theorem 1). Chapter 5 described $\log x$ as "ever increasing" from its graph; now we prove it.

Given, to find, and the plan

Given: The logarithmic function $f(x) = \log x$ ⁠, $x > 0$ ⁠. In this book $\log$ means the natural logarithm, to base $e$ (Section 5.4).
To prove: $f$ is increasing on $(0, \infty)$ ⁠: $0 < x_1 < x_2 \Rightarrow \log x_1 \le \log x_2$ ⁠.
Plan: (1) Derivative $\frac{1}{x}$ ⁠. (2) Its sign for $x > 0$ ⁠. (3) Theorem 1 on $[x_1, x_2]$ for any $0 < x_1 < x_2$ ⁠. (4) Conclude, noting strictness; remark on other bases.

Basics you need

1Increasing and strictly increasing (Definition 1)

On an interval I: increasing means $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ⁠; strictly increasing means $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ ⁠, for all $x_1, x_2$ in I. Lesson: increasing and decreasing functions

2Derivative of $\log x$

$\frac{d}{dx}(\log x) = \frac{1}{x}$ for $x > 0$ (Theorem 5 (2), Section 5.4). A function with a derivative at a point is continuous there (Theorem 3, Chapter 5).

3Theorem 1 and the Mean Value step of its proof

Theorem 1 (a): if $f$ is continuous on $[a, b]$ ⁠, differentiable on $(a, b)$ and $f'(x) \ge 0$ there, then $f$ is increasing on $[a, b]$ ⁠. Its proof uses the Mean Value Theorem: $f(b) - f(a) = f'(c)(b - a)$ for some $c$ in $(a, b)$ ⁠; so $f' > 0$ on $(a, b)$ gives $f(a) < f(b)$ ⁠. (The book cites the Mean Value Theorem as "Theorem 8 in Chapter 5", but this reprint's Chapter 5 does not contain it; it is used here as the book uses it, without proof.) Basics: the derivative as a rate of change

4Interval notation

$(0, \infty)$ is the set of all positive real numbers, which is exactly the domain of $\log x$ ⁠. Basics: interval notation

Solution

Derivative

For every $x > 0$ ⁠,
$f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}.$
So $f$ is differentiable at every point of $(0, \infty)$ ⁠, and therefore continuous there.

Theorem 1 needs continuity on a closed interval and differentiability inside. Both come from the fact that $\log x$ has a derivative at every positive $x$ (differentiable implies continuous).

Theorem 5 (2): $\frac{d}{dx}\log x = \frac1x$ ⁠. Theorem 3: differentiable at a point implies continuous there.
The derivative is positive on $(0, \infty)$
$x > 0 \;\Rightarrow\; f'(x) = \frac{1}{x} > 0.$
The reciprocal of a positive number is positive. On the domain of $\log x$ there is not a single point where the slope is zero or negative.

Positive divided by positive is positive. Basics: sign of a product
Apply Theorem 1 between any two positive numbers

Let $0 < x_1 < x_2$ ⁠. Then $[x_1, x_2] \subset (0, \infty)$ ⁠, $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$ ⁠, with $f' > 0$ there. By Theorem 1 (a), $f$ is increasing on $[x_1, x_2]$ ⁠. The Mean Value step of the proof gives, for some $c$ between $x_1$ and $x_2$ ⁠,
$\log x_2 - \log x_1 = \frac{1}{c}\,(x_2 - x_1) > 0.$
So $\log x_1 < \log x_2$ ⁠.

$(0, \infty)$ is open and unbounded, so Theorem 1 cannot be applied to it in one go; but every pair of points sits in a closed interval $[x_1, x_2]$ inside the domain, and that is all we need. Both factors $\frac1c$ and $x_2 - x_1$ are positive because $c > 0$ and $x_2 > x_1$ ⁠.

Mean Value step of the proof of Theorem 1: $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ ⁠. Lesson: increasing and decreasing functions
Conclude

For all $0 < x_1 < x_2$ we have $\log x_1 < \log x_2$ ⁠. So $\log x$ is strictly increasing on $(0, \infty)$ (Definition 1 (iv)), and in particular the logarithmic function is increasing on $(0, \infty)$ .

$y = \log x$ ⁠: defined only for $x > 0$ ⁠, crossing the $x$ -axis at $(1, 0)$ and reaching 1 at $x = e$ ⁠. It rises everywhere, steeply near 0 (slope $\frac1x$ large) and gently for large $x$ (slope $\frac1x$ small but still positive).

A remark on other bases. For $\log_a x = \frac{\log x}{\log a}$ ⁠, the derivative is $\frac{1}{x\log a}$ ⁠. If $a > 1$ ⁠, then $\log a > 0$ and the same proof shows $\log_a x$ is increasing. If $0 < a < 1$ ⁠, then $\log a < 0$ and $\log_a x$ is decreasing. The book's "logarithmic function" is the natural one (base $e > 1$ ⁠), and Section 5.4 discusses bases $b > 1$ ⁠, so it is increasing.

Result

$f'(x) = \frac{1}{x} > 0$ for every $x > 0$ ⁠. By Theorem 1, applied on $[x_1, x_2]$ for any $0 < x_1 < x_2$ ⁠, $\log x_1 < \log x_2$ ⁠. Hence the logarithmic function is increasing on $(0, \infty)$ ⁠, in fact strictly increasing. (The answer key gives no answer for this proof.)

Check it with numbers

$\log 0.5 \approx -0.693147$ ⁠, $\log 1 = 0$ ⁠, $\log 2 \approx 0.693147$ ⁠, $\log e = 1$ ⁠, $\log 10 \approx 2.302585$ ⁠: increasing inputs give increasing outputs. Mean Value check on $[1, e]$ ⁠: $\frac{\log e - \log 1}{e - 1} = \frac{1}{e - 1} \approx 0.581977$ ⁠, which equals $\frac1c$ for $c = e - 1 \approx 1.718282$ ⁠, a point inside $(1, e)$ ⁠, exactly as the theorem promises.

Common mistakes

Writing " $f'(x) = \frac1x > 0$ for all $x$ ". It is positive only for $x > 0$ ⁠; for $x < 0$ the function $\log x$ is not even defined. Always say where.
Proving it from a few values such as $\log 1 < \log 2 < \log 10$ ⁠. That is a check, not a proof; the argument must work for arbitrary $0 < x_1 < x_2$ ⁠.
Applying Theorem 1 directly to the open interval $(0, \infty)$ ⁠. Apply it on $[x_1, x_2]$ instead.

What to take away

$\log x$ is strictly increasing on its whole domain $(0, \infty)$ ⁠, because its slope $\frac1x$ is always positive there.
Consequence used everywhere: for positive $a, b$ ⁠, $a < b \iff \log a < \log b$ ⁠, so taking logarithms keeps inequalities.
For a base $a$ ⁠, $\log_a x$ increases if $a > 1$ and decreases if $0 < a < 1$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing and strictly increasing	Step 4	Definition 1 (i), (iv), Section 6.3 (Lesson: increasing and decreasing functions)
$\log$ means base $e$ ⁠; domain $(0, \infty)$ ⁠; "ever increasing" observed from the graph	Given, Step 4	Section 5.4 and its observations on the log function
$\frac{d}{dx}\log x = \frac1x$	Step 1	Theorem 5 (2), Section 5.4
Differentiable implies continuous	Step 1	Theorem 3, Section 5.3
$f' > 0$ inside gives increase; Mean Value step	Step 3	Theorem 1 (a) and its proof, Section 6.3

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Exercise 6.2 · Q10: log⁡x\log xlogx is increasing on (0,∞)(0,\infty)(0,∞)

Basics you need

Solution

Derivative

The derivative is positive on (0,∞)(0, \infty)(0,∞)

Apply Theorem 1 between any two positive numbers

Conclude

Every fact used, and where it comes from

Exercise 6.2 · Q10: $\log x$ is increasing on $(0,\infty)$

The derivative is positive on $(0, \infty)$