Exponentials and Logarithms: A Complete Guide for AQA International A-Level P2 | 指数与对数：AQA 国际 A-Level P2 完全指南

📚 Exponentials and Logarithms: A Complete Guide for AQA International A-Level P2 | 指数与对数：AQA 国际 A-Level P2 完全指南

Exponentials and logarithms form one of the most important building blocks of Pure Mathematics. In Unit P2 of the AQA International A-Level, you are expected to understand the natural exponential function eˣ, its inverse the natural logarithm ln x, and how to apply these to equations, graphs, differentiation and integration. This guide breaks down every examinable skill into clear steps with worked examples.

指数与对数是纯数学中最重要的基石之一。在 AQA 国际 A-Level 的 P2 单元中，你需要理解自然指数函数 eˣ、它的反函数自然对数 ln x，以及如何将这些知识应用于方程、图像、求导和积分。本指南将每一个可考的技能分解为清晰的步骤，并配有完整的例题。

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

The number e ≈ 2.71828 is defined as the base of the natural logarithm. The function f(x) = eˣ is called the natural exponential function. It is the unique function whose derivative is equal to itself.

常数 e ≈ 2.71828 被定义为自然对数的底数。函数 f(x) = eˣ 称为自然指数函数，它是唯一一个导数等于自身的函数。

The domain of eˣ is all real numbers and its range is all positive real numbers.

eˣ 的定义域为全体实数，值域为全体正实数。
The inverse of eˣ is the natural logarithm ln x, defined only for x > 0.

eˣ 的反函数是自然对数 ln x，其定义域仅为 x > 0。
The key inverse property is ln(eˣ) = x and e^(ln x) = x.

关键的反函数性质是 ln(eˣ) = x 以及 e^(ln x) = x。

These two functions undo each other, just as addition undoes subtraction and multiplication undoes division.

这两个函数互为逆运算，就像加法抵消减法、乘法抵消除法一样。

2. Laws of Logarithms | 对数运算法则

All logarithm laws follow directly from the laws of indices. You must be able to apply them fluently in both directions.

所有对数运算法则都直接来源于指数运算法则。你必须能够熟练地在两个方向上运用它们。

ln(xy) = ln x + ln y, ln(x/y) = ln x − ln y, ln(xⁿ) = n ln x

The multiplication law turns products into sums.

乘法法则将乘积转化为和。
The division law turns quotients into differences.

除法法则将商转化为差。
The power law brings exponents down as coefficients.

幂法则将对数中的指数提到前面作为系数。

Example: Simplify ln(a²) + ln(b) − ln(c³). Using the laws, combine the sum first and then the difference.

例题：化简 ln(a²) + ln(b) − ln(c³)。先使用加法法则合并和，再使用减法法则合并差。

ln(a²) + ln(b) − ln(c³) = ln(a²b) − ln(c³) = ln(a²b / c³)

3. Solving Exponential Equations | 解指数方程

To solve an equation of the form aˣ = b, take the natural logarithm of both sides and use the power law.

要解形如 aˣ = b 的方程，对两边取自然对数并使用幂法则。

ln(aˣ) = ln b → x ln a = ln b → x = ln b / ln a

Example: Solve 3ˣ = 20. Taking logs gives x ln 3 = ln 20, so x = ln 20 / ln 3 ≈ 2.727.

例题：解方程 3ˣ = 20。两边取对数得 x ln 3 = ln 20，因此 x = ln 20 / ln 3 ≈ 2.727。

When the unknown appears in two different exponential or logarithmic terms, collect like terms before taking logs, or use a substitution such as y = eˣ.

当未知量出现在两个不同的指数项或对数项中时，先合并同类项再取对数，或者使用代换，例如令 y = eˣ。

Example: Solve e²ˣ − 5eˣ + 6 = 0. Let y = eˣ, giving y² − 5y + 6 = 0, so y = 2 or y = 3. Hence eˣ = 2 or eˣ = 3, and x = ln 2 or x = ln 3.

例题：解方程 e²ˣ − 5eˣ + 6 = 0。令 y = eˣ，得到 y² − 5y + 6 = 0，因此 y = 2 或 y = 3。于是 eˣ = 2 或 eˣ = 3，所以 x = ln 2 或 x = ln 3。

4. Change of Base | 换底公式

For any positive bases a and b, the change of base formula allows us to convert between logarithmic bases.

对于任意正底数 a 和 b，换底公式允许我们在不同对数底数之间进行转换。

logₐx = ln x / ln a

This is particularly useful when solving equations or when a calculator only has ln and log₁₀ functions.

这在解方程或当计算器只有 ln 和 log₁₀ 功能时特别有用。

Example: Evaluate log₂7. Using the formula, log₂7 = ln 7 / ln 2 ≈ 2.807.

例题：计算 log₂7。使用换底公式，log₂7 = ln 7 / ln 2 ≈ 2.807。

5. Graphs of Exponential and Logarithmic Functions | 指数函数与对数函数的图像

The graph of y = eˣ passes through (0, 1), has a horizontal asymptote at y = 0, and increases without bound as x increases.

y = eˣ 的图像经过点 (0, 1)，以 y = 0 为水平渐近线，并且随着 x 增大而无界增加。

The graph of y = ln x passes through (1, 0), has a vertical asymptote at x = 0, and increases without bound as x increases.

y = ln x 的图像经过点 (1, 0)，以 x = 0 为垂直渐近线，并且随着 x 增大而无界增加。

Function 函数	Domain 定义域	Range 值域	Asymptote 渐近线
y = eˣ	ℝ	y > 0	y = 0
y = ln x	x > 0	ℝ	x = 0

The two graphs are reflections of each other in the line y = x, because each function is the inverse of the other.

这两个图像关于直线 y = x 互为镜像，因为这两个函数互为反函数。

6. Modelling with Exponentials and Logarithms | 指数与对数建模

Exponential models describe quantities that grow or decay at a rate proportional to their current size, such as population, radioactive decay, and cooling.

指数模型描述以与当前大小成正比的速度增长或衰减的量，例如人口、放射性衰变和冷却过程。

The general model is N = N₀eᵏᵗ, where N₀ is the initial amount and k is the growth (k > 0) or decay (k < 0) constant.

一般模型为 N = N₀eᵏᵗ，其中 N₀ 是初始量，k 是增长（k > 0）或衰减（k < 0）常数。

To find the constant k, substitute a known data point and solve using natural logarithms.

要求常数 k，代入已知的数据点并使用自然对数求解。

Example: A population doubles in 10 years. Find k. If N = 2N₀ when t = 10, then 2N₀ = N₀e^(10k), so 2 = e^(10k), giving k = ln 2 / 10 ≈ 0.0693.

例题：某人口在 10 年内翻倍。求 k。当 t = 10 时 N = 2N₀，则 2N₀ = N₀e^(10k)，即 2 = e^(10k)，所以 k = ln 2 / 10 ≈ 0.0693。

7. Differentiation of eˣ and ln x | eˣ 与 ln x 的求导

The two most important derivative results in P2 are the following.

P2 中两个最重要的求导结果是以下公式。

d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x

For composite functions, apply the chain rule.

对于复合函数，应用链式法则。

d/dx e^(3x) = 3e^(3x), d/dx ln(5x + 2) = 5/(5x + 2)

In general, d/dx e^(kx) = ke^(kx) and d/dx ln(kx) = 1/x.

一般地，d/dx e^(kx) = ke^(kx)，且 d/dx ln(kx) = 1/x。

Remember that when differentiating e^(f(x)), you must multiply by f'(x). For example, the derivative of e^(x²) is 2xe^(x²).

记住，对 e^(f(x)) 求导时，必须乘以 f'(x)。例如，e^(x²) 的导数是 2xe^(x²)。

8. Natural Logarithm in Integration | 自然对数在积分中的应用

Since the derivative of ln x is 1/x, integration of reciprocal functions produces natural logarithms.

由于 ln x 的导数是 1/x，因此对倒数函数积分会产生自然对数。

∫ 1/x dx = ln |x| + c, ∫ 1/(ax + b) dx = (1/a) ln |ax + b| + c

When the numerator is the derivative of the denominator, the integral is a natural logarithm.

当分子是分母的导数时，其积分就是自然对数。

Example: Evaluate ∫ (6x²)/(2x³ + 1) dx. The derivative of the denominator is 6x², so the integral is ln(2x³ + 1) + c.

例题：计算 ∫ (6x²)/(2x³ + 1) dx。分母的导数是 6x²，因此积分为 ln(2x³ + 1) + c。

Use the modulus sign in ∫ 1/x dx when x could take negative values within the integration interval.

当 x 在积分区间内可能取负值时，在 ∫ 1/x dx 中使用绝对值符号。

9. Exponential Growth and Decay Problems | 指数增长与衰减问题

These problems often require you to rearrange N = N₀eᵏᵗ to find the time t.

这类问题通常要求你通过变换 N = N₀eᵏᵗ 来求时间 t。

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Exponentials and Logarithms: A Complete Guide for AQA International A-Level P2 | 指数与对数：AQA 国际 A-Level P2 完全指南