AQA Maths: Exponentials and Logarithms Exam Focus | AQA 数学：指数与对数考点精讲

📚 AQA Maths: Exponentials and Logarithms Exam Focus | AQA 数学：指数与对数考点精讲

Exponentials and logarithms are central to AQA A-level Mathematics, bridging algebra, modelling, and calculus. Mastering the relationship between these inverse functions, their algebraic laws, and the art of solving equations is essential for high marks. This revision guide walks through every critical concept, from definitions to hidden quadratics, all aligned with AQA’s specification.

指数与对数是 AQA A-level 数学的核心内容，连接了代数、建模与微积分。掌握这对互逆函数的关系、运算法则以及解方程的技巧是获得高分的关键。本篇复习指南从定义出发，涵盖隐藏二次型等每个关键概念，完全贴合 AQA 考纲。

1. Exponential Functions – Definition and Graphs | 指数函数——定义与图形

An exponential function takes the form y = aˣ where the base a is a positive constant and a ≠ 1. The domain is all real numbers, and the range is y > 0. The curve always passes through (0, 1) because a⁰ = 1. The x-axis is a horizontal asymptote, meaning the function never touches or crosses it.

指数函数的形式为 y = aˣ，其中底数 a 为正的常数且 a ≠ 1。定义域为全体实数，值域为 y > 0。图像始终经过 (0, 1)，因为 a⁰ = 1。x 轴是水平渐近线，函数永远不会触及或穿过它。

When a > 1, the graph rises steeply as x increases and approaches zero as x → –∞. This models exponential growth. For 0 < a < 1, the graph decreases, approaching zero as x → +∞, modelling exponential decay. The steepness depends on the base; a larger base gives a steeper climb.

当 a > 1 时，图像随 x 增加而急剧上升，并在 x → –∞ 时趋近于零，这模拟指数增长。当 0 < a < 1 时，图像下降，在 x → +∞ 时趋近于零，模拟指数衰减。陡峭程度取决于底数；底数越大，上升越陡。

Sketching y = 2ˣ and y = (1/2)ˣ on the same axes reveals they are reflections of each other in the y-axis. Recognising these shapes helps when transforming graphs or solving inequalities.

在同一坐标轴上画出 y = 2ˣ 和 y = (1/2)ˣ 可以看出它们关于 y 轴对称。识别这些形状有助于图像变换和不等式求解。

2. Logarithms – Definition and Connection to Exponentials | 对数——定义及与指数的联系

A logarithm answers the question: ‘To what power must the base be raised to get a given number?’ The statement aˣ = b is equivalent to x = logₐ b. Thus, logarithms are the inverses of exponentials. For example, since 2³ = 8, we write log₂ 8 = 3.

对数回答这样的问题：“底数需要几次幂才能得到给定的数？” 语句 aˣ = b 等价于 x = logₐ b。因此，对数是指数的逆运算。例如，由于 2³ = 8，我们写作 log₂ 8 = 3。

The argument of a logarithm (the number you take the log of) must be strictly positive. The base a must also be positive and cannot equal 1. So logₐ x is defined only for x > 0, a > 0, a ≠ 1.

对数的真数（取对数的数）必须严格为正。底数 a 也必须为正且不能等于 1。因此 logₐ x 仅在 x > 0, a > 0, a ≠ 1 时有定义。

Common bases are 10 (common log, often written as log x) and e (natural log, written as ln x). On your calculator, ‘log’ usually means log₁₀, and ‘ln’ stands for logₑ. The equivalence lets you switch between exponential and logarithmic forms freely.

常用底数为 10（常用对数，常写作 log x）和 e（自然对数，写作 ln x）。计算器上，“log” 通常表示 log₁₀，而 “ln” 表示 logₑ。这种等价性使得你可以在指数形式与对数形式之间自由切换。

3. Laws of Logarithms | 对数运算法则

The three main laws of logarithms allow you to combine and split logarithmic expressions. They are true for any allowed base.

三个主要的对数运算法则允许你合并与拆分对数表达式。它们对任何允许的底数都成立。

Law (Rule)	Formula
Product law	logₐ (xy) = logₐ x + logₐ y
Quotient law	logₐ (x / y) = logₐ x – logₐ y
Power law	logₐ (xⁿ) = n logₐ x

The power law is especially useful: it brings exponents down to become coefficients. For example, log₂ (8⁵) = 5 log₂ 8 = 5 × 3 = 15. Always practise expanding something like logₐ (x²y / z³) into a linear combination of logs.

幂运算法则尤其有用：它把指数“拉下来”变成系数。例如，log₂ (8⁵) = 5 log₂ 8 = 5 × 3 = 15。务必练习将诸如 logₐ (x²y / z³) 的式子展开为对数的线性组合。

You must avoid common misuses: logₐ (x + y) is not logₐ x + logₐ y, and logₐ x ⋅ logₐ y has no simple rule. Stick strictly to the three valid laws when manipulating log expressions.

你必须避免常见误用：logₐ (x + y) 不等于 logₐ x + logₐ y，而 logₐ x ⋅ logₐ y 没有简单法则。处理对数表达式时务必严格遵守这三个有效法则。

4. Solving Exponential Equations Using Logarithms | 使用对数解指数方程

When the unknown is in the exponent, taking logarithms of both sides is the primary technique. For an equation like 5ˣ = 30, apply log (or ln) to both sides: log(5ˣ) = log 30 → x log 5 = log 30 → x = log 30 / log 5. The base of the logarithm doesn’t matter as long as it is the same on both sides.

当未知数出现在指数上时，两边取对数是主要的解法。对于方程 5ˣ = 30，对两边取 log（或 ln）：log(5ˣ) = log 30 → x log 5 = log 30 → x = log 30 / log 5。对数的底数无关紧要，只要两边相同即可。

If the exponential base is e, use natural logarithms for simplicity: e²ˣ = 7 → 2x = ln 7 → x = (ln 7)/2. Always isolate the exponential term before taking logs. For example, in 3·2ˣ + 1 = 25, first rewrite as 2ˣ = 8, then deduce x = 3 directly, or use logs.

如果指数式的底是 e，为简便起见使用自然对数：e²ˣ = 7 → 2x = ln 7 → x = (ln 7)/2。在取对数之前务必先分离出指数项。例如，对于 3·2ˣ + 1 = 25，先化成 2ˣ = 8，然后直接推出 x = 3，或使用对数。

When the equation involves different bases, express them with the same base if possible. For instance, 4ˣ = 8²ˣ⁻¹ → (2²)ˣ = (2³)²ˣ⁻¹ → 2²ˣ = 2⁶ˣ⁻³ → 2x = 6x – 3 → x = 3/4. If equalising bases is not feasible, logs are your tool.

当方程包含不同底数时，尽可能化成同底。例如，4ˣ = 8²ˣ⁻¹ → (2²)ˣ = (2³)²ˣ⁻¹ → 2²ˣ = 2⁶ˣ⁻³ → 2x = 6x – 3 → x = 3/4。如果无法化为同底，对数就是你的工具。

5. Solving Logarithmic Equations | 解对数方程

Logarithmic equations often require using the laws to condense multiple logs into a single log term. For example, log₂ (x + 1) + log₂ (x – 1) = 3 → log₂[(x + 1)(x – 1)] = 3 → log₂ (x² – 1) = 3. Then rewrite in exponential form: x² – 1 = 2³ = 8 → x² = 9 → x = ±3.

对数方程往往需要利用运算法则将多个对数合并成单个对数项。例如，log₂ (x + 1) + log₂ (x – 1) = 3 → log₂[(x + 1)(x – 1)] = 3 → log₂ (x² – 1) = 3。然后化为指数形式：x² – 1 = 2³ = 8 → x² = 9 → x = ±3。

Always check the validity of solutions in the original logarithmic terms. The arguments (x + 1) and (x – 1) must be positive. For x = –3, x + 1 = –2, which is negative, so –3 must be rejected. Hence x = 3 is the sole solution. Missing this check loses marks.

务必在原始对数项中检验解的有效性。真数 (x + 1) 和 (x – 1) 必须为正。当 x = –3 时，x + 1 = –2，为负数，因此 –3 必须舍去。所以 x = 3 是唯一解。漏掉这一步会丢分。

If an equation contains a constant, rewrite it as a logarithm with the same base. For instance, log₃ x = 2 – log₃ (x + 2) → log₃ x + log₃ (x + 2) = 2 → log₃ [x(x + 2)] = log₃ 9 → x(x + 2) = 9. Solve the quadratic and check constraints.

如果方程中含有常数，把它改写成同底的对数。例如，log₃ x = 2 – log₃ (x + 2) → log₃ x + log₃ (x + 2) = 2 → log₃ [x(x + 2)] = log₃ 9 → x(x + 2) = 9。解二次方程并检查约束条件。

6. The Natural Exponential Function eˣ and Natural Logarithm ln x | 自然指数函数 eˣ 与自然对数 ln x

The number e ≈ 2.71828 appears naturally in calculus and continuous growth models. The function y = eˣ has the unique property that its derivative is itself: d/dx (eˣ) = eˣ. Its inverse is the natural logarithm y = ln x, which is the log base e.

数 e ≈ 2.71828 自然地出现在微积分和连续增长模型中。函数 y = eˣ 具有独特的性质：其导数等于它本身，即 d/dx (eˣ) = eˣ。它的反函数是自然对数 y = ln x，即以 e 为底的对数。

The graph of y = eˣ passes through (0, 1) with slope 1, grows extremely fast, and has the x-axis as an asymptote. The graph of y = ln x passes through (1, 0), is defined only for x > 0, and grows slowly. They are reflections in the line y = x.

y = eˣ 的图像经过 (0, 1)，斜率为 1，增长极快，并以 x 轴为渐近线。y = ln x 的图像经过 (1, 0)，仅当 x > 0 时有定义，增长缓慢。它们关于直线 y = x 对称。

Key values: ln 1 = 0, ln e = 1, eˡⁿ ˣ = x for x > 0, and ln(eˣ) = x. The derivative of ln x is 1/x, and integration of 1/x yields ln |x| + C. Many AQA exam questions test these ln properties within calculus contexts.

关键值：ln 1 = 0，ln e = 1，对于 x > 0 有 eˡⁿ ˣ = x，且 ln(eˣ) = x。ln x 的导数是 1/x，而 1/x 的积分是 ln |x| + C。AQA 的许多考题会在微积分情境中测试这些 ln 的性质。

7. Change of Base Formula | 换底公式

The change of base formula allows you to convert a logarithm to a different base, typically base 10 or e for calculator use: logₐ b = log_c b / log_c a. A common choice is logₐ b = log b / log a or logₐ b = ln b / ln a.

换底公式允许你将对数转换为不同的底数，通常转换为底数 10 或 e 以便用计算器计算：logₐ b = log_c b / log_c a。常见的选择是 logₐ b = log b / log a 或 logₐ b = ln b / ln a。

For example, to evaluate log₅ 20, key in log(20) / log(5) or ln(20) / ln(5). The formula also helps solve equations like 3ˣ⁺¹ = 2ˣ. Taking logs gives (x+1) ln 3 = x ln 2, then solve linearly.

例如，计算 log₅ 20，输入 log(20) / log(5) 或 ln(20) / ln(5)。该公式也有助于求解诸如 3ˣ⁺¹ = 2ˣ 的方程：取对数得 (x+1) ln 3 = x ln 2，然后线性求解。

In some proofs and simplifications, change of base reveals hidden relationships. Remember that logₐ b = 1 / log_b a. This reciprocal property is a special case of the change of base rule and can be very handy.

在某些证明和化简中，换底公式能揭示隐藏的关系。记住 logₐ b = 1 / log_b a。这个倒数性质是换底法则的特例，用起来非常方便。

8. Exponential Growth and Decay Models | 指数增长与衰减模型

Many real-world processes are modelled by P = P₀ eᵏᵗ, where P is the quantity at time t, P₀ is the initial quantity, and k is the continuous growth (k > 0) or decay (k < 0) rate. This appears in population, radioactive decay, and compound interest problems.

许多现实过程可用 P = P₀ eᵏᵗ 建模，其中 P 是 t 时刻的量，P₀ 是初始量，k 是连续增长（k > 0）或衰减（k < 0）速率。该模型出现在人口

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AQA Maths: Exponentials and Logarithms Exam Focus | AQA 数学：指数与对数 考点精讲