Logarithmic and Exponential Functions

Lessons

Senior 6 · Mathematics

Logarithmic and Exponential Functions

4 lessons

Lesson 1 of 4

The number e and the natural logarithm function

25 min

A bank that never stops paying interest

Suppose a bank pays 100% interest a year on 1 Frw (an illustrative rate, far above any real one). Paid once at the end of the year, you have 2 Frw. Paid in two halves, each of 50%, you have $\left(1 + \frac{1}{2}\right)^2 = 2.25$ Frw. Paid $n$ times, each at $\frac{100}{n}$ %, you have $\left(1 + \frac{1}{n}\right)^n$ Frw. In S5 (Logarithmic and Exponential Equations) you used this compound-interest formula; now let $n$ grow:

Payments per year, $n$	1	2	12	365	8760 (hourly)	1 000 000
$\left(1 + \frac{1}{n}\right)^n$	2	2.25	2.613 04	2.714 57	2.718 13	2.718 28

The amounts grow, but they do not grow without limit: they settle towards one number, about 2.718 28. That number is the base of the natural logarithm, and of the exponential function that describes every process growing at a rate proportional to its size.

Definition

The number $e$ : the limit

e = \lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^n \approx 2.718\,281\,828 .

It is irrational. Putting $h = \frac{1}{n}$ , the same number is $e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$ ; Lesson 2 uses this form (with $h$ allowed to be any small real number, positive or negative) to differentiate $\ln x$ .

Definition

Natural logarithm: for $x \gt 0$ , $\ln x$ is the logarithm of $x$ to base $e$ :

y = \ln x \iff e^y = x .

Natural logarithmic function: $f(x) = \ln x$ , with domain $(0, +\infty)$ and range $\mathbb{R}$ . Note $\ln 1 = 0$ and $\ln e = 1$ ; $\ln x \lt 0$ for $0 \lt x \lt 1$ and $\ln x \gt 0$ for $x \gt 1$ .

Example: $\ln e^5 = 5$ , $e^{\ln 7} = 7$ and $\ln 20 \approx 2.996$ . Some books start instead from "the function whose derivative is $\frac{1}{x}$ and which is 0 at $x = 1$ "; Lesson 2 shows that this is the same function.

Important

Laws of natural logarithms. For $a \gt 0$ , $b \gt 0$ and any real $r$ :

\ln(ab) = \ln a + \ln b, \qquad \ln\frac{a}{b} = \ln a - \ln b, \qquad \ln a^r = r\ln a .

Derivation of the first: $e^{\ln a + \ln b} = e^{\ln a}\,e^{\ln b} = ab$ by the laws of indices, so $\ln a + \ln b$ is the power of $e$ that gives $a b$ , which is $\ln(ab)$ . In the same way $e^{r\ln a} = \left(e^{\ln a}\right)^r = a^r$ gives the third law, and the second follows from the first two with $\frac{a}{b} = ab^{-1}$ . In particular $\ln\frac{1}{a} = -\ln a$ .

Warning

$\ln(a + b) \ne \ln a + \ln b$ . There is no law for the logarithm of a sum: for example, $\ln(1 + 1) = \ln 2 \approx 0.693$ , but $\ln 1 + \ln 1 = 0$ .
Watch the domain. $\ln(x^2)$ is defined for every $x \ne 0$ , but $2\ln x$ only for $x \gt 0$ . The correct identity is $\ln(x^2) = 2\ln|x|$ .

Worked example 1: domains

Find the domain of (a) $f(x) = \ln(x^2 - 4)$ (b) $g(x) = \ln\frac{x - 1}{3 - x}$ (c) $h(x) = \ln(\ln x)$ .

Solution The argument of $\ln$ must be strictly positive.

$x^2 - 4 \gt 0 \iff (x - 2)(x + 2) \gt 0 \iff x \lt -2$ or $x \gt 2$ . Domain: $(-\infty, -2) \cup (2, +\infty)$ .
$\frac{x - 1}{3 - x} \gt 0$ : the numerator and denominator must have the same sign. A sign table gives $1 \lt x \lt 3$ . Domain: $(1, 3)$ .
First $x \gt 0$ for $\ln x$ , then $\ln x \gt 0$ , so $x \gt 1$ . Domain: $(1, +\infty)$ .

Limits and the asymptote

Because $\ln x$ is increasing and $\ln(e^N) = N$ for every $N$ , $\ln x$ exceeds any number once $x$ is large enough, and falls below any number once $x$ is close enough to 0:

\lim_{x \to +\infty}\ln x = +\infty, \qquad \lim_{x \to 0^+}\ln x = -\infty .

So the line $x = 0$ (the $y$ -axis) is a vertical asymptote of $y = \ln x$ ; there is no horizontal asymptote. Although $\ln x \to +\infty$ , it grows more slowly than any positive power of $x$ . Lesson 2 proves the two standard limits

\lim_{x \to +\infty}\frac{\ln x}{x} = 0, \qquad \lim_{x \to 0^+} x\ln x = 0 .

$y = \ln x$ : domain $(0, +\infty)$ , range $\mathbb{R}$ . The curve crosses the $x$ -axis at $x = 1$ and reaches $y = 1$ at $x = e \approx 2.718$ .

Worked example 2: limits

Find (a) $\lim_{x \to 0^+}\frac{\ln x}{x}$ (b) $\lim_{x \to +\infty}\ln\frac{2x + 1}{x + 3}$ (c) $\lim_{x \to +\infty}(x - \ln x)$ .

Solution

The numerator tends to $-\infty$ and the denominator to $0^{+}$ . This is not an indeterminate form: the quotient tends to $-\infty$ .
$\frac{2x + 1}{x + 3} \to 2$ and $\ln$ is continuous at 2, so the limit is $\ln 2 \approx 0.693$ .
" $\infty - \infty$ " is indeterminate, so factorise: $x - \ln x = x\left(1 - \frac{\ln x}{x}\right)$ . The bracket tends to $1 - 0 = 1$ , so the limit is $+\infty$ .

Worked example 3: an equation and an inequality

Solve (a) $\ln(x + 1) + \ln(x - 1) = \ln 3$ (b) $\ln(2x - 1) \lt 0$ .

Solution

Domain first: $x + 1 \gt 0$ and $x - 1 \gt 0$ , so $x \gt 1$ . Then $\ln\left[(x + 1)(x - 1)\right] = \ln 3$ , so $x^{2} - 1 = 3$ and $x = \pm 2$ . Only $x = 2$ is in the domain.
Domain: $x \gt \frac{1}{2}$ . Since $\ln$ is increasing and $\ln 1 = 0$ , $\ln(2x - 1) \lt 0 \iff 2x - 1 \lt 1 \iff x \lt 1$ . Solution: $\frac{1}{2} \lt x \lt 1$ .

Tip

In every logarithm question, write the domain down before you start. The algebra often produces an extra "solution" (like $x = - 2$ above) that the domain throws out, and examiners award a mark for rejecting it with a reason.

Numberphile: e (Euler's Number). Watch for the compound-interest story in the first minutes, the same limit as the table at the top of this lesson, and for where else e turns up.

✦

Practice

State the domain in every equation before solving it.

Simplify: (a) $\ln e^3$ (b) $e^{\ln 5}$ (c) $\ln\sqrt{e}$ (d) $\ln 12 - \ln 3 + \ln\frac{1}{4}$ .
Write in terms of $\ln 2$ and $\ln 3$ : (a) $\ln 72$ (b) $\ln\frac{4}{27}$ .
Find the domain of (a) $\ln(3 - x)$ (b) $\ln(x^2 - 5x + 6)$ (c) $\ln|x - 1|$ .
Find (a) $\lim_{x \to 0^+}(\ln x)^2$ (b) $\lim_{x \to +\infty}\ln\left(1 + \frac{1}{x}\right)$ (c) $\lim_{x \to +\infty}\frac{\ln x}{x^2}$ .
Solve $\ln(2x + 3) = 2\ln x$ .
Solve $\ln(x - 2) + \ln(x + 1) = \ln 4$ .
Solve the inequality $\ln(x + 3) \le \ln(5 - x)$ .
Show that " $\ln(x^2) = 2\ln x$ " fails for $x = - 1$ , and state the identity that is true for every $x \ne 0$ .

Show answersHide answers

(a) 3 (b) 5 (c) $\frac{1}{2}$ (d) $\ln\left(\frac{12}{3} \times \frac{1}{4}\right) = \ln 1 = 0$ .
(a) $72 = 2^3 \times 3^2$ , so $3\ln 2 + 2\ln 3$ (b) $2\ln 2 - 3\ln 3$ .
(a) $x \lt 3$ (b) $(x - 2)(x - 3) \gt 0$ : $x \lt 2$ or $x \gt 3$ (c) all real $x \ne 1$ .
(a) $+\infty$ (b) $\ln 1 = 0$ (c) $\frac{\ln x}{x^2} = \frac{\ln x}{x} \times \frac{1}{x} \to 0 \times 0 = 0$ .
Domain $x \gt 0$ . $2 x + 3 = x^{2}$ , so $x^{2} - 2 x - 3 = 0$ , $x = 3$ or $x = - 1$ . Answer: $x = 3$ .
Domain $x \gt 2$ . $(x - 2) (x + 1) = 4$ , so $x^{2} - x - 6 = 0$ , $x = 3$ or $x = - 2$ . Answer: $x = 3$ .
Domain $-3 \lt x \lt 5$ . $\ln$ is increasing, so $x + 3 \le 5 - x$ , $x \le 1$ . Answer: $-3 \lt x \le 1$ .
$\ln\left((-1)^2\right) = \ln 1 = 0$ , but $2\ln(-1)$ is not defined. For $x \ne 0$ , $\ln(x^2) = 2\ln|x|$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the graph (plotted from computed points), the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S6 Core Mathematics (2026–2027); no REB text is reproduced.
Video: e (Euler's Number) - Numberphile, Numberphile, embedded from YouTube.

1A bank that never stops paying interest

2Limits and the asymptote

A bank that never stops paying interest

Limits and the asymptote