Logarithmic and Exponential Functions

Lessons

Senior 6 · Mathematics

Logarithmic and Exponential Functions

4 lessons

Lesson 3 of 4

The exponential function eˣ and exponentials to any base

25 min

Undoing the logarithm

Lesson 2 showed that $\ln$ is continuous and strictly increasing on $(0, +\infty)$ , with range $\mathbb{R}$ . So for every real number $y$ there is exactly one $x \gt 0$ with $\ln x = y$ : the function has an inverse (S4, Composite and Inverse Functions). That inverse is the exponential function. It is the function behind every "grows by the same percentage each year" model, and it is the only function (apart from constant multiples of it) that is its own derivative.

Definition

Exponential function (base $e$ ): the inverse of $\ln$ ,

\exp : \mathbb{R} \to (0, +\infty), \qquad y = e^x \iff x = \ln y .

Domain $\mathbb{R}$ , range $(0, +\infty)$ . As inverse functions, $\ln\left(e^x\right) = x$ for every real $x$ and $e^{\ln x} = x$ for every $x \gt 0$ . For rational $x$ , $e^{x}$ is the ordinary power of $e$ , and all the laws of indices hold: $e^{a + b} = e^a e^b$ , $e^{a - b} = \frac{e^a}{e^b}$ , $\left(e^a\right)^r = e^{ar}$ .

Example: $e^{2\ln 3} = \left(e^{\ln 3}\right)^2 = 9$ .

The graphs of $e^{x}$ and $\ln x$

The graph of an inverse function is the reflection of the original graph in the line $y = x$ , because the point $(a, b)$ lies on $y = \ln x$ exactly when $(b, a)$ lies on $y = e^{x}$ .

$y = e^{x}$ (blue) and $y = \ln x$ (orange) are mirror images in $y = x$ , plotted from computed points. Swapping $x$ and $y$ swaps the domain and the range, and turns the horizontal asymptote $y = 0$ into the vertical asymptote $x = 0$ .

Limits

Reflecting the limits of $\ln x$ gives

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

so the $x$ -axis, $y = 0$ , is a horizontal asymptote on the left. The exponential beats every power of $x$ . Put $x = \ln t$ : as $x \to +\infty$ , $t \to +\infty$ and $\frac{e^x}{x} = \frac{t}{\ln t} \to +\infty$ , since $\frac{\ln t}{t} \to 0$ . In the same way,

\lim_{x \to +\infty}\frac{e^x}{x^n} = +\infty \ (n \ge 1), \qquad \lim_{x \to -\infty} x e^x = 0 .

Important

Theorem. $\dfrac{d}{dx}e^x = e^x$ , and by the chain rule $\dfrac{d}{dx}e^{u(x)} = u'(x)\,e^{u(x)}$ .

Proof. Let $y = e^{x}$ , so $\ln y = x$ . Differentiate both sides with respect to $x$ , using Lesson 2 and the chain rule: $\frac{1}{y}\cdot y' = 1$ . Hence $y^{'} = y = e^{x}$ . ∎

Consequences: $e^x \gt 0$ and $\left(e^x\right)'' = e^x \gt 0$ , so $e^{x}$ is strictly increasing and concave up. Its tangent at $(0, 1)$ is $y = x + 1$ , and the curve lies above it: $e^x \ge 1 + x$ for every real $x$ .

Worked example 1: differentiating exponentials

Differentiate (a) $e^{3x}$ (b) $x^2 e^{-x}$ (c) $e^{\sin x}$ (d) $\frac{e^x}{1 + e^x}$ .

Solution

$3e^{3x}$ .
Product rule: $2xe^{-x} - x^2 e^{-x} = x(2 - x)e^{-x}$ .
$\cos x\,e^{\sin x}$ .
Quotient rule: $\frac{e^x(1 + e^x) - e^x \cdot e^x}{(1 + e^x)^2} = \frac{e^x}{(1 + e^x)^2}$ .

Worked example 2: study of $f(x) = xe^{-x}$

Solution Domain $\mathbb{R}$ .

Limits: as $x \to -\infty$ , $x \to -\infty$ and $e^{-x} \to +\infty$ , so $f(x) \to -\infty$ . As $x \to +\infty$ , $f(x) = \frac{x}{e^x} \to 0$ : the $x$ -axis is a horizontal asymptote.
Derivative: $f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$ . Since $e^{-x} \gt 0$ , the sign is that of $1 - x$ : $f$ increases on $(-\infty, 1]$ and decreases on $[1, +\infty)$ , with maximum $f(1) = \frac{1}{e} \approx 0.368$ .
Concavity: $f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$ , which changes sign at $x = 2$ : a point of inflexion at $\left(2, \frac{2}{e^2}\right) \approx (2, 0.271)$ .

$f(x) = xe^{-x}$ , plotted from computed points (the vertical scale is larger than the horizontal one). The $x$ -axis is an asymptote on the right because $e^{x}$ grows faster than $x$ .

Definition

Exponential function to base $a$ ( $a \gt 0$ ): for every real $x$ ,

a^x = e^{x\ln a} .

This agrees with the powers you know, since $e^{x\ln a} = \left(e^{\ln a}\right)^x$ . By the chain rule,

\frac{d}{dx}a^x = a^x\ln a .

For $a \ne 1$ , $a^{x}$ is the inverse of $\log_a x$ . Example: the gradient of $y = 2^{x}$ at $x = 3$ is $2^3\ln 2 = 8\ln 2 \approx 5.55$ .

All three curves pass through $(0, 1)$ , since $a^{0} = 1$ . For $a \gt 1$ , $a^{x}$ increases; for $0 \lt a \lt 1$ it decreases. The graphs of $2^{x}$ and $\left(\frac{1}{2}\right)^x = 2^{-x}$ are reflections of each other in the $y$ -axis.

Worked example 3: exponential equations

Solve (a) $e^{2x} - 5e^x + 6 = 0$ (b) $2^{x + 1} = 5^x$ .

Solution

Put $t = e^x \gt 0$ : $t^{2} - 5 t + 6 = 0$ , so $t = 2$ or $t = 3$ . Both are positive, so $x = \ln 2 \approx 0.693$ or $x = \ln 3 \approx 1.099$ .
Take $\ln$ of both sides: $(x + 1)\ln 2 = x\ln 5$ , so $x(\ln 5 - \ln 2) = \ln 2$ and $x = \frac{\ln 2}{\ln 2.5} \approx 0.756$ . Check: $2^{1.756} \approx 3.38$ and $5^{0.756} \approx 3.38$ ✓.

Warning

$\frac{d}{dx}2^x = 2^x\ln 2$ , not $x \cdot 2^{x - 1}$ : the power rule is for a variable base and a constant exponent.
$e^{a + b} = e^a e^b$ , but $e^{a + b} \ne e^a + e^b$ . In an equation such as $e^{2x} - 5e^x + 6 = 0$ , substitute $t = e^{x}$ instead of taking $\ln$ term by term.

Tip

Inverse functions swap everything: domain and range, $x$ and $y$ in every point, the vertical asymptote $x = 0$ of $\ln x$ and the horizontal asymptote $y = 0$ of $e^{x}$ . Learn the facts for one curve, and reflect them to get the other.

3Blue1Brown: What's so special about Euler's number e? (Chapter 5, Essence of calculus). Watch for why the derivative of 2ˣ is a constant multiple of 2ˣ, and why that constant is exactly 1 for base e.

✦

Practice

Questions 6 to 8 need the derivative.

Simplify: (a) $e^{2\ln 3}$ (b) $\ln\left(e^x \cdot e^2\right)$ (c) $e^{-\ln 4}$ .
Differentiate: (a) $e^{5x - 1}$ (b) $x^3 e^{2x}$ (c) $3^{x}$ (d) $e^{x^2}$ .
Find (a) $\lim_{x \to -\infty}(e^x + 2)$ (b) $\lim_{x \to +\infty} x^2 e^{-x}$ (c) $\lim_{x \to 0}\frac{e^x - 1}{x}$ (hint: a derivative at 0).
Solve $e^{2x} - 4e^x + 3 = 0$ .
Solve $3^{2x} = 7$ , giving $x$ to 3 significant figures.
Find the stationary point of $f (x) = e^{x} - 2 x$ and show that it is a minimum.
Find the equation of the tangent to $y = e^{2x}$ at the point where $x = 0$ .
Using $e^x \ge 1 + x$ , prove that $\left(1 + \frac{1}{n}\right)^n \lt e$ for every positive integer $n$ .

Show answersHide answers

(a) 9 (b) $x + 2$ (c) $\frac{1}{4}$ .
(a) $5e^{5x - 1}$ (b) $3x^2 e^{2x} + 2x^3 e^{2x} = x^2 e^{2x}(3 + 2x)$ (c) $3^x\ln 3$ (d) $2xe^{x^2}$ .
(a) 2 (b) 0, since $e^{x}$ beats $x^{2}$ (c) it is the derivative of $e^{x}$ at 0, which is $e^{0} = 1$ .
$t = e^{x}$ : $(t - 1) (t - 3) = 0$ , so $x = 0$ or $x = \ln 3 \approx 1.10$ .
$2x\ln 3 = \ln 7$ , so $x = \frac{\ln 7}{2\ln 3} \approx 0.886$ .
$f^{'} (x) = e^{x} - 2 = 0$ at $x = \ln 2$ ; $f(\ln 2) = 2 - 2\ln 2 \approx 0.614$ ; $f''(x) = e^x \gt 0$ , so it is a minimum.
$y (0) = 1$ , $y^{'} (0) = 2 e^{0} = 2$ : $y = 2 x + 1$ .
Put $x = \frac{1}{n}$ : $e^{\frac{1}{n}} \gt 1 + \frac{1}{n}$ (strictly, since $x \ne 0$ ). Both sides are positive, so raising to the power $n$ keeps the order: $e \gt \left(1 + \frac{1}{n}\right)^n$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the graphs (plotted from computed points), the worked examples, the practice items and the quiz questions.
Video: What's so special about Euler's number e? | Chapter 5, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1Undoing the logarithm

2The graphs of ex e^x ex and ln⁡x \ln x lnx

3Limits

Undoing the logarithm

The graphs of $e^{x}$ and $\ln x$

Limits