Euler's Number, e^x and ln x

The number $e \approx 2.71828$ is the one base for which the exponential function is its own gradient: the curve $y = e^x$ climbs at exactly the rate $e^x$. That single property makes $e^x$ and its inverse, the natural logarithm $\ln x$, the functions that calculus, growth and decay are built on. The syllabus asks you to understand the definition and properties of $e^x$ and $\ln x$, including their relationship as inverse functions and their graphs; in the exam they appear in almost every question, from solving equations to differentiation, integration and differential equations.

Where e comes from

Gradients of exponential curves

Every exponential curve $y = a^x$ (with $a > 0$) has a gradient that is proportional to its height. To see why, look at the gradient of the chord from $x$ to $x + h$:

$$\frac{a^{x + h} - a^x}{h} = a^x \cdot \frac{a^h - 1}{h}.$$

The factor $a^x$ comes straight out. As $h \to 0$, the second factor tends to a fixed number that depends only on $a$. Call it $k_a$. Then

$$\frac{d}{dx}\left(a^x\right) = k_a\, a^x.$$

Evaluating $\dfrac{a^h - 1}{h}$ for a tiny $h$ (say $h = 10^{-7}$) gives the constant:

Base $a$	$2$	$2.5$	$3$
$k_a \approx$	$0.693$	$0.916$	$1.099$

For $a = 2$ the gradient is a bit less than the height; for $a = 3$ it is a bit more. Somewhere between $2$ and $3$ there is a base for which $k_a = 1$ exactly, so the gradient equals the height at every point. That base is $e$.

A second route: compound growth

The same number appears when growth is compounded more and more often. Put $1 in an account paying $100\%$ interest a year. Compounded once, you have $2$ at the end; compounded $n$ times, you have $\left(1 + \tfrac{1}{n}\right)^n$.

$n$	$1$	$10$	$100$	$1000$	$1\,000\,000$
$\left(1 + \frac{1}{n}\right)^n$	$2$	$2.5937$	$2.7048$	$2.7169$	$2.71828$

The values settle on $e$. You do not need this limit for the exam, but it explains why $e$ turns up in every model of continuous growth.

The natural logarithm

Because $\ln$ is a logarithm, every law from logarithms and the laws of logarithms holds for it unchanged:

$$\ln(pq) = \ln p + \ln q, \qquad \ln\frac{p}{q} = \ln p - \ln q, \qquad \ln(p^k) = k\ln p \qquad (p, q > 0).$$

The last identity follows from $a = e^{\ln a}$, so $a^x = \left(e^{\ln a}\right)^x = e^{x\ln a}$. It shows that $2^x$, $10^x$ and $\left(\tfrac{1}{2}\right)^x$ are all of the form $e^{kx}$ with $k = \ln a$, and it explains the table above: the constant $k_a$ is exactly $\ln a$ (indeed $\ln 2 = 0.693$ and $\ln 3 = 1.099$).

Simplifying with e and ln

The skill is to bring every power of $e$ and every $\ln$ into a form where they can cancel. Move coefficients inside the log first, then cancel:

$$e^{2\ln 3} = e^{\ln 9} = 9, \qquad e^{-\ln 4} = e^{\ln(1/4)} = \tfrac{1}{4}, \qquad \ln\left(\frac{1}{e^2}\right) = -2.$$

Graphs and properties

The graphs of $y = e^x$ and $y = \ln x$ are reflections of each other in the line $y = x$, because the two functions are inverses: swapping $x$ and $y$ in $y = e^x$ gives $x = e^y$, which is $y = \ln x$.

The curve through $(0, 1)$ is $y = e^x$; the curve through $(1, 0)$ is $y = \ln x$; the straight line is the mirror line $y = x$.

	$y = e^x$	$y = \ln x$
Domain	all real $x$	$x > 0$
Range	$y > 0$	all real $y$
Key point	$(0, 1)$	$(1, 0)$
Asymptote	$y = 0$ (as $x \to -\infty$)	$x = 0$ (as $x \to 0^+$)
Gradient at key point	$1$	$1$
Behaviour	increasing, grows faster than any power of $x$	increasing, grows more slowly than any power of $x$

Points to notice:

• $e^x > 0$ for every $x$. An equation such as $e^x = -3$ or $e^{2x} = 0$ has no solution.
• $\ln x$ is negative for $0 < x < 1$, zero at $x = 1$ and positive for $x > 1$.
• $\ln x$ is undefined for $x \le 0$. This is why every log equation needs its domain checked.
• Both functions are one-one (each output comes from exactly one input), so each has an inverse, and taking $\ln$ or $e$ of both sides of an equation is always reversible.
• Both functions are increasing, so they preserve inequalities: $a < b \iff e^a < e^b$, and for positive $a, b$, $a < b \iff \ln a < \ln b$.

The sketching of transformed versions such as $y = 3 - 2e^{-x}$ and $y = \ln(2x - 1)$ is covered in exponential graphs.

Solving equations with e and ln

Two moves do almost all the work.

• To free $x$ from an exponent, isolate the exponential, then take $\ln$ of both sides: $e^{\text{stuff}} = c \Rightarrow \text{stuff} = \ln c$ (only possible if $c > 0$).
• To free $x$ from a logarithm, combine into a single $\ln$, then raise $e$ to both sides: $\ln(\text{stuff}) = c \Rightarrow \text{stuff} = e^c$.

Inverse functions involving e and ln

Because $e^x$ and $\ln x$ undo each other, a function built from one has an inverse built from the other. Find it exactly as in P1 (one-one and inverse functions): write $y = f(x)$, make $x$ the subject, then swap the letters. The domain of $f^{-1}$ is the range of $f$, so always find the range of $f$ first.

Exponential growth and decay

A quantity that changes at a rate proportional to its current size follows

$$N = N_0e^{kt},$$

where $N_0$ is the value when $t = 0$. If $k > 0$ it is growth; if $k < 0$ (often written $N_0e^{-kt}$ with $k > 0$) it is decay. You will derive this from a differential equation in differential equations; here the skill is to use the model.

• Two data values fix the two constants. Divide one equation by the other to eliminate $N_0$, then take logs to find $k$.
• Halving or doubling time does not depend on the starting value: $N_0e^{-kT} = \tfrac{1}{2}N_0$ gives $T = \dfrac{\ln 2}{k}$.
• Long-term behaviour: as $t \to \infty$, $e^{-kt} \to 0$ for $k > 0$. A model like $N = 5000 - 3000e^{-0.2t}$ therefore tends to $5000$.

Worked examples

Practice

Answers

1. (a) $e^{\ln 9^{1/2}} = e^{\ln 3} = 3$. (b) $\ln\left(e^{x}\right) = x$. (c) $e^{\ln(5/2)} = \tfrac{5}{2}$. (d) $\ln e^{3/2} = \tfrac{3}{2}$.

2. $e^x = 7$, so $x = \ln 7$.

3. Domain $x > 1$. $\ln\big(x(x - 1)\big) = \ln 6 \Rightarrow x^2 - x - 6 = 0 \Rightarrow (x - 3)(x + 2) = 0$. Reject $x = -2$ (outside the domain), so $x = 3$.

4. Let $u = e^x$: $2u^2 + u - 6 = 0 \Rightarrow (2u - 3)(u + 2) = 0$. $u = -2$ is impossible since $e^x > 0$. $e^x = \tfrac{3}{2}$, so $x = \ln\tfrac{3}{2}$.

5. $y = \ln(2x - 1) \Rightarrow 2x - 1 = e^y \Rightarrow x = \tfrac{1}{2}(e^y + 1)$. So $g^{-1}(x) = \tfrac{1}{2}(e^x + 1)$ for $x \in \mathbb{R}$. Its range is the domain of $g$: $g^{-1}(x) > \tfrac{1}{2}$.

6. Domain $x > 0$. $\ln(2x + 3) = \ln e + \ln x = \ln(ex)$, so $2x + 3 = ex$, giving $x(e - 2) = 3$ and $x = \dfrac{3}{e - 2} = 4.18$ (3 s.f.). It is positive, so valid.

7. At $t = 0$: $T = 80\ ^\circ\text{C}$. As $t \to \infty$, $e^{-0.05t} \to 0$, so $T \to 20\ ^\circ\text{C}$. For $T = 40$: $60e^{-0.05t} = 20 \Rightarrow e^{-0.05t} = \tfrac{1}{3} \Rightarrow -0.05t = -\ln 3 \Rightarrow t = 20\ln 3 = 22.0$ minutes (3 s.f.).

8. Multiply by $e^x$: $3e^{2x} - 5e^x - 2 = 0 \Rightarrow (3e^x + 1)(e^x - 2) = 0$. $e^x = -\tfrac{1}{3}$ has no solution because $e^x > 0$ for all $x$. So $e^x = 2$ and $x = \ln 2$ is the only solution.

9. (a) $500e^{6k} = 1000 \Rightarrow e^{6k} = 2 \Rightarrow k = \dfrac{\ln 2}{6}$. (b) $500e^{kt} = 2000e^{-0.1t} \Rightarrow e^{(k + 0.1)t} = 4 \Rightarrow t = \dfrac{\ln 4}{k + 0.1} = \dfrac{1.3863}{0.21552} = 6.43$ years (3 s.f.).