📚 Exponentials and Logarithms: A Complete A-Level Guide | 指数与对数：A-Level 完整指南
Exponential and logarithmic functions are central to the A-Level Mathematics syllabus, especially for Edexcel, AQA and OCR specifications. They describe rapid growth and decay, provide the tools for solving equations in which the unknown sits in a power, and underpin essential calculus techniques such as differentiating eˣ and integrating 1/x. Understanding their definitions, graphs and laws, together with the relationship between them, is the key to scoring well in this topic.
指数函数与对数函数是 A-Level 数学大纲的核心内容，对 Edexcel、AQA 与 OCR 考试局而言尤其如此。它们刻画了快速增长与衰减过程，为求解"未知量位于指数位置"的方程提供了工具，同时也支撑了诸如 eˣ 的求导、1/x 的积分等关键微积分技巧。理解它们的定义、图像与运算法则，以及两者之间的互逆关系，是掌握该主题并取得高分的关键。
1. Introduction to Exponential Functions | 指数函数简介
An exponential function has the form f(x) = aˣ ether the curve rises or falls, how steep it is, and where it sits relative to the x-axis.
指数函数具有形式 f(x) = aˣ，其中 a 是一个不为 1 的正常数，即 a > 0 且 a ≠ 1。底数 a 决定了函数的全部行为：它决定了曲线是上升还是下降、有多陡峭，以及它相对于 x 轴的位置。
strictly decreasing and models exponential decay. The case a = 1 is excluded because y = 1ˣ = 1 is simply a horizontal line, not an exponential curve at all.
若 a > 1，函数严格递增，用于模拟指数增长；若 0 < a < 1，函数严格递减，用于模拟指数衰减。底数 a = 1 被排除在外，因为 y = 1ˣ = 1 只是一条水平直线，根本算不上指数曲线。
The graph of y = aˣ xis (y = 0) which the curve approaches but never touches. The domain is all real numbers x ∈ ℝ, and the range is y > 0. This means aˣ is never zero and never negative for real x.
y = aˣ 的图像总是经过点 (0, 1)，因为对任何合法底数都有 a⁰ = 1。它有一条水平渐近线即 x 轴（y = 0），曲线无限接近却永不相交。其定义域为全体实数 x ∈ ℝ，值域为 y > 0。这意味着对实数 x 而言，aˣ 既不为零，也不为负。
aᵐ)ⁿ = aᵐⁿ, a⁰ = 1 and a⁻ⁿ = 1 ÷ aⁿ. For example, 2³ × 2⁴ = 2⁷ = 128 and (3²)³ = 3⁶ = 729.
指数运算法则依然适用且经常需要用到：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁰ = 1，a⁻ⁿ = 1 ÷ aⁿ。例如 2³ × 2⁴ = 2⁷ = 128，(3²)³ = 3⁶ = 729。
2. The Number e and the Natural Exponential Function | 数 e 与自然指数函数
most important exponential function in calculus is the natural exponential function y = eˣ, where e ≈ 2.718281828… turally whenever a quantity grows continuously at a rate proportional to its current size.
微积分中最重要的指数函数是自然指数函数 y = eˣ，其中 e ≈ 2.718281828…。这个无理数被定义为极限 e = lim (n → ∞) (1 + 1 ÷ n)ⁿ。只要一个量以正比于其当前大小的速率连续增长，e 就会自然出现。
Because e > 1, y = eˣ h passes through (0, 1) and (1, e), has the x-axis as a horizontal asymptote, has domain x ∈ ℝ and range y > 0. Its defining calculus property is unique: the gradient of y = eˣ at any point equals the value of the function there, so dy/dx = eˣ.
因为 e > 1，所以 y = eˣ 严格递增，图像经过点 (0, 1) 与 (1, e)，以 x 轴为水平渐近线，定义域 x ∈ ℝ，值域 y > 0。它有一个独一无二的微积分性质：y = eˣ 在任意点处的切线斜率都等于该点的函数值，即 dy/dx = eˣ。
In exam questions you will often see the more general form y = ekx . If k > 0 the curve grows, if k < 0 it decays, and the chain rule gives dy/dx = k·ekx. This single family of curves covers almost all growth-and-decay modelling in A-Level.
在考题中，你会经常见到更一般的形式 y = ekx，其中 k 为常数。若 k > 0，曲线增长；若 k < 0，曲线衰减；由链式法则可得 dy/dx = k·ekx。这一族曲线几乎涵盖了 A-Level 中所有的增长与衰减建模问题。
3. Graphs and Transformations of Exponential Curves | 指数曲线的图像与变换
ble to sketch exponential graphs quickly and state their asymptotes. The basic shape y = aˣ rizontal asymptote; all transformations follow the standard function-transformation rules.
你需要能够快速画出指数函数草图并写出其渐近线。基本形状 y = aˣ 是一条经过 (0, 1) 的光滑曲线，只有一条水平渐近线；所有变换都遵循标准的函数变换法则。
A vertical translation y = aˣ es the whole curve up by c, so the asymptote becomes y = c. A horizontal translation y = ax + d or right (if d < 0) but keeps the asymptote y = 0, because it is applied to the input.
竖直平移 y = aˣ + c 会把整条曲线向上移动 c 个单位，渐近线随之变为 y = c。水平平移 y = ax + d 会把曲线向左（d > 0）或向右（d < 0）平移，但渐近线仍为 y = 0，因为变换作用在自变量上。
A vertical stretch y = k·aˣ te is still y = 0 but the intercept becomes (0, k). A reflection in the y-axis turns y = aˣ into y = a⁻ˣ = (1 ÷ a)ˣ, converting growth into decay. Reflections in the x-axis produce y = −aˣ, which has range y < 0.
竖直伸缩 y = k·aˣ 把每个函数值乘以 k，因此渐近线仍为 y = 0，但截距变为 (0, k)。关于 y 轴的反射把 y = aˣ 变为 y = a⁻ˣ = (1 ÷ a)ˣ，使增长变为衰减。关于 x 轴的反射得到 y = −aˣ，其值域为 y < 0。
| Transformation 变换 | Equation 方程 | Asymptote 渐近线 |
| Basic 基本 | y = aˣ | y = 0 |
| Vertical shift 竖直平移 | y = aˣ + c | y = c |
| Vertical stretch 竖直伸缩 | y = k·aˣ | y = 0 |
| Reflection in y-axis 关于 y 轴反射 | y = a⁻ˣ | y = 0 |
4. What Is a Logarithm? | 对数的定义
A logarithm answers the question "what power do I raise the base to in order to get this number?" Formally, if a > 0 and a ≠ 1, then logₐ x = n is defined by the equivalence logₐ x = n ⇌ aⁿ = x. In words, the logarithm is the inverse operation of exponentiation.
对数回答的问题是"要把底数提高到多少次方才能得到这个数？"形式上，若 a > 0 且 a ≠ 1，则 logₐ x = n 由等价关系定义：logₐ x = n ⇌ aⁿ = x。换言之，对数是指数运算的逆运算。
Because aⁿ is always positive for real n, the logarithm logₐ x is only defined for x > 0. This domain restriction is the single most common source of lost marks: whenever you solve a logarithmic equation, you must check that your answers keep every logarithm's argument positive.
由于对实数 n 而言 aⁿ 恒为正，对数 logₐ x 只在 x > 0 时有定义。这一"定义域限制"是失分最常见的来源：每当你解对数方程时，都必须检验所得到的解是否使每一个对数的真数都为正。
dels. logₑ x is written ln x and is called the natural logarithm. Key values follow directly from the definition: logₐ a = 1, logₐ 1 = 0, logₐ aⁿ = n and alogₐ x = x.
两种特殊记号会反复出现。log₁₀ x 常简写为 log x，称为常用对数，广泛用于计算器和指数模型中。logₑ x 写作 ln x，称为自然对数。由定义可直接得到关键值：logₐ a = 1，logₐ 1 = 0，logₐ aⁿ = n，以及 alogₐ x = x。
5. The Laws of Logarithms | 对数运算法则
The laws of logarithms mirror the index laws and are the working tools for every log question. They hold for any valid base a > 0, a ≠ 1, and for positive arguments x > 0, y > 0.
对数运算法则与指数运算法则一一对应，是解决所有对数问题的基本工具。它们对任意合法底数 a > 0, a ≠ 1 以及正数真数 x > 0, y > 0 均成立。
ee laws allow you to break complicated expressions apart or to combine several logarithms into one.
乘法法则：logₐ(xy) = logₐ x + logₐ y。除法法则：logₐ(x ÷ y) = logₐ x − logₐ y。幂法则：logₐ(xⁿ) = n·logₐ x。这三条法则可以让你把复杂表达式拆开，也可以把若干对数合并成一个。
A very useful special case of the power law is logₐ(1 ÷ x) = logₐ x⁻¹ = −logₐ x, and logₐ √x = logₐ x1/2 logₐ x + logₐ y. This is a classic trap in multiple-choice and structured questions alike.
幂法则有两个极为常用的特例：logₐ(1 ÷ x) = logₐ x⁻¹ = −logₐ x，以及 logₐ √x = logₐ x1/2 = ½·logₐ x。要注意 logₐ(x + y) 无法化简，它并不等于 logₐ x + logₐ y。这是选择题与结构题中都常见的陷阱。
The change-of-base formula allows you to evaluate any logarithm on a calculator that only has log and ln keys. For any valid bases a and b:
换底公式让你可以在只有 log 键与 ln 键的计算器上求出任意对数的值。对任意合法底数 a、b 有：
logₐ x = log_b x ÷ log_b a = ln x ÷ ln a
the bridge used to solve equations where the unknown is an exponent with an awkward base.
例如，log₂ 7 = ln 7 ÷ ln 2 ≈ 1.9459 ÷ 0.6931 ≈ 2.807。这个公式也是求解"未知量位于不便处理的底数的指数上"这类方程的桥梁。
6. Solving Exponential Equations | 求解指数方程
To solve an equation of the form ax , take logarithms of both sides using the same base. Taking base-10 logarithms gives log(ax Alternatively use natural logarithms: x = ln b ÷ ln a. Both routes give identical answers.
求解形如 ax = b 的方程时，两边取同底对数。取常用对数得 log(ax) = log b，由幂法则得 x·log a = log b，故 x = log b ÷ log a。也可以取自然对数：x = ln b ÷ ln a。两种途径结果完全相同。
Worked example: solve 32x + 1 lways keep at least four significant figures during the working and round only at the end.
例题：解 32x + 1 = 20。取自然对数得 (2x + 1)·ln 3 = ln 20，故 2x + 1 = ln 20 ÷ ln 3 ≈ 2.7268，于是 2x = 1.7268，x ≈ 0.8634。运算过程中应至少保留四位有效数字，只在最后一步四舍五入。
Some exponential equations are hidden quadratics. For example, e2x − 5eˣ + 6 = 0 can be written as (eˣ)² − 5eˣ + 6 = 0. Substituting u = eˣ gives u² − 5u + 6 = 0, so u = 2 or u = 3, giving eˣ = 2 or eˣ = 3, hence x = ln 2 ≈ 0.693 or x = ln 3 ≈ 1.099.
有些指数方程是"伪装的二次方程"。例如 e2x − 5eˣ + 6 = 0 可写成 (eˣ)² − 5eˣ + 6 = 0。令 u = eˣ，得 u² − 5u + 6 = 0，解出 u = 2 或 u = 3，即 eˣ = 2 或 eˣ = 3，故 x = ln 2 ≈ 0.693 或 x = ln 3 ≈ 1.099。
Remember that eˣ . In the example above, if the quadratic had produced u = −1, the corresponding equation eˣ = −1 would have no real solution and must be discarded.
记住对一切实数 x 有 eˣ > 0，因此任何 u ≤ 0 的换元解都必须舍去。在上例中，若二次方程解出 u = −1，则对应的 eˣ = −1 无实数解，必须舍弃。
7. Solving Logarithmic Equations | 求解对数方程
o combine terms into a single logarithm, then convert to exponential form using logₐ x = n ⇌ ter obtaining candidate solutions, always substitute them back into the original equation.
求解含对数的方程时，先用对数运算法则把各项合并成一个对数，再利用 logₐ x = n ⇌ aⁿ = x 转化为指数形式。得到候选解后，务必代回原方程检验。
1)(x − 1)] = 3, so (x + 1)(x − 1) = 2³ = 8. Hence x² − 1 = 8, x² = 9, so x = 3 or x = −3.
例题：解 log₂(x + 1) + log₂(x − 1) = 3。用乘法法则合并得 log₂[(x + 1)(x − 1)] = 3，故 (x + 1)(x − 1) = 2³ = 8。于是 x² − 1 = 8，x² = 9，得 x = 3 或 x = −3。
o the solution is valid. The only solution is x = 3. This rejection step is worth an explicit mark in most mark schemes.
接下来检验定义域。若 x = −3，则 x + 1 = −2，x − 1 = −4，两者皆为负，两个对数都无定义。若 x = 3，则 x + 1 = 4，x − 1 = 2，两者皆为正，解有效。因此唯一的解是 x = 3。在多数评分标准中，这一步"舍去"是单独给分的。
d logarithm on one side and raise the base to both sides: log₂[(x + 1)(x − 1)] = 3 means 23 = (x + 1)(x − 1). Whichever route you choose, the domain check remains compulsory.
另一种方法是保留一侧的合并对数，两边同时以底数为底取幂：log₂[(x + 1)(x − 1)] = 3 即 2³ = (x + 1)(x − 1)。无论采用哪种途径，定义域检验都不可省略。
8. Graphs of Logarithms and the Inverse Relationship | 对数图像与互逆关系
The function y = logₐ x is the inverse of y = aˣ he two graphs are reflections of each other in the line y = x. Every point (p, q) on y = aˣ corresponds to the point (q, p) on y = logₐ x.
函数 y = logₐ x 是 y = aˣ 的反函数，这意味着两条图像关于直线 y = x 互为镜像。y = aˣ 上的每一点 (p, q) 都对应 y = logₐ x 上的点 (q, p)。
al asymptote x = 0. The domain and range swap over: y = logₐ x has domain x > 0 and range y ∈ ℝ.
由于指数图像经过 (0, 1) 且以 y = 0 为水平渐近线，对数图像便经过 (1, 0) 并以 x = 0 为竖直渐近线。定义域与值域互换：y = logₐ x 的定义域为 x > 0，值域为 y ∈ ℝ。
r y = ln x passes through (1, 0) and (e, 1), and the natural logarithm is the inverse of eˣ, so ln(eˣ) = x for all real x and eln x = x for all x > 0.
当 a > 1 时，对数曲线缓慢上升，并在 (1, 0) 处穿过 x 轴；当 0 < a < 1 时曲线下降。特别地，y = ln x 经过 (1, 0) 和 (e, 1)，且自然对数与 eˣ 互为反函数，因此对一切实数 x 有 ln(eˣ) = x，对一切 x > 0 有 eln x = x。
9. Differentiation and Integration with e and ln | 含 e 与 ln 的微分与积分
The natural exponential and natural logarithm give the cleanest results in calculus, which is precisely why they are singled out in the syllabus. You must know these standard results by heart, together with the chain rule versions.
自然指数与自然对数在微积分中给出的结果最为简洁，这正是大纲将它们单列出来的原因。你必须熟记以下标准结果，以及配合链式法则的推广形式。
d/dx (eˣ) = eˣ d/dx (ekx) = k·ekx d/dx (ln x) = 1 ÷ x d/dx (ln(kx)) = 1 ÷ x
For composite functions, the chain rule gives d/dx (e^(f(x))) = f'(x)·e^(f(x)) and d/dx (ln f(x)) =
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图