📚 Logarithms | 对数
Logarithms are one of the most important algebraic tools in Edexcel A-Level Pure Mathematics. They allow you to solve exponential equations, analyse rapid growth or decay, and understand the inverse relationship between powers and indices.
对数是爱德思 A-Level 纯数学中最重要的代数工具之一。掌握对数可以帮助你求解指数方程、分析快速增长或衰减，并理解幂与指数之间的逆运算关系。
1. What Is a Logarithm? | 什么是对数
A logarithm answers the question: "To what power must the base be raised to produce a given number?" re a ≠ 1, and a positive number x, the expression logₐ(x) means the exponent y such that aʸ = x.
对数回答的问题是：“底数 a 要乘方多少次才能得到给定的数？”对于正底数 a（且 a ≠ 1）和正数 x，logₐ(x) 表示满足 aʸ = x 的指数 y。
logₐ(x) = y ⇔ aʸ = x
This equivalence is the foundation of every logarithm problem. You must be able to switch quickly between logarithmic form and exponential form.
这种等价关系是所有对数问题的基础。你必须能够在对数形式和指数形式之间快速转换。
8. Likewise, log₁₀1000 = 3 because 10³ = 1000. The base is always written as a subscript.
例如，log₂8 = 3，因为 2³ = 8。同样，log₁₀1000 = 3，因为 10³ = 1000。底数总是写成下标。
The logarithm is only defined for positive arguments. This means logₐ(x) exists only when x > 0. The base must also be positive and cannot equal 1.
对数只对正的真数有定义。也就是说，logₐ(x) 只有在 x > 0 时才有意义。底数也必须为正，并且不能等于 1。
2. Converting Between Exponential and Logarithmic Form | 指数式与对数式互化
Exam questions often test the ability to rewrite an exponential statement as a logarithm, or vice versa. The base stays the same, the exponent becomes the answer, and the result becomes the input of the logarithm.
考试题目经常考查将指数式改写为对数式，或将对数式改写为指数式的能力。底数保持不变，指数变成对数的结果，而幂的结果变成对数的真数。
| Exponential form | 指数形式 | Logarithmic form | 对数形式 |
|---|---|
| 2³ = 8 | log₂8 = 3 |
| 10² = 100 | log₁₀100 = 2 |
| 5⁰ = 1 | log₅1 = 0 |
| a¹ = a | logₐa = 1 |
ase a, and logₐa = 1. These special values are frequently used in simplifying expressions.
从这些例子中可以得出两个始终成立的结论：对于任意合法的底数 a，logₐ1 = 0；同时 logₐa = 1。这些特殊值经常用于化简表达式。
1 = -1. This shows that logarithms can output any real number, not just positive integers.
对数也可以取负值。例如，10⁻¹ = 0.1，所以 log₁₀0.1 = -1。这说明对数的结果可以是任意实数，而不仅限于正整数。
3. The Laws of Logarithms | 对数运算法则
Logarithms turn multiplication into addition, division into subtraction, and powers into multiplication. These laws are essential for simplifying and solving logarithmic and exponential equations.
对数能把乘法转化为加法、除法转化为减法、乘方转化为乘法。这些法则对于化简和求解对数方程、指数方程至关重要。
- Product rule | 乘法法则： logₐ(xy) = logₐx + logₐy
- Quotient rule | 除法法则： logₐ(x ÷ y) = logₐx − logₐy
- Power rule | 乘方法则： logₐ(xⁿ) = n logₐx
- Reciprocal rule | 倒数法则： logₐ(1 ÷ x) = −logₐx
, log₂(8 × 4) = log₂8 + log₂4 = 3 + 2 = 5. This matches the direct calculation log₂32 = 5.
例如，log₂(8 × 4) = log₂8 + log₂4 = 3 + 2 = 5。这与直接计算 log₂32 = 5 的结果一致。
tient rule, log₁₀(1000 ÷ 10) = log₁₀1000 − log₁₀10 = 3 − 1 = 2, which is exactly log₁₀100.
使用除法法则，log₁₀(1000 ÷ 10) = log₁₀1000 − log₁₀10 = 3 − 1 = 2，这正好等于 log₁₀100。
The power rule is especially useful when an unknown exponent appears. For instance, log₃(9ˣ) = x log₃9 = 2x, because log₃9 = 2.
乘方法则在出现未知指数时尤其有用。例如，log₃(9ˣ) = x log₃9 = 2x，因为 log₃9 = 2。
Remember that logₐ(x + y) cannot be simplified using these laws. Only multiplication, division and powers inside a logarithm follow the rules above.
请记住，logₐ(x + y) 不能用这些法则化简。只有对数内部的乘法、除法和乘方可以按照上述法则拆分。
4. Solving Exponential Equations | 用对数解指数方程
When the unknown is in the exponent, taking logarithms of both sides is usually the fastest method. You may use base 10 logarithms or natural logarithms, as long as you apply the same operation to both sides.
当未知数出现在指数上时，通常最快的方法是对两边同时取对数。你可以使用常用对数或自然对数，但必须对两边施加完全相同的运算。
Solve 3ˣ = 20. Taking log base 10 of both sides gives:
求解 3ˣ = 20。对两边同时取以 10 为底的对数：
log 3ˣ = log 20
x log 3 = log 20
x = log 20 ÷ log 3 ≈ 2.727
The exact form is log 20 ÷ log 3. In an exam, give the exact answer first, then a decimal approximation if requested.
精确形式是 log 20 ÷ log 3。在考试中，应先写出精确答案，如果题目要求再给出小数近似值。
For equations involving e, use the natural logarithm. Solve e2x−1 = 7:
对于含有 e 的方程，使用自然对数。求解 e2x−1 = 7：
2x − 1 = ln 7
x = (1 + ln 7) ÷ 2
e ln(eᵏ) = k for any real number k. Taking logs removes the base and exposes the exponent.
这个方法之所以有效，是因为 ln(eᵏ) = k 对任意实数 k 都成立。取对数可以消去底数，把指数暴露出来。
5. Solving Logarithmic Equations | 解对数方程
To solve equations with more than one logarithm, first combine the terms into a single logarithm using the laws. Then convert to exponential form and solve the resulting algebraic equation.
求解含有多个对数的方程时，应先用对数法则把各项合并成一个对数。然后转化为指数形式，再解得到的代数方程。
Solve log₂(x + 6) − log₂(x + 2) = 1.
求解 log₂(x + 6) − log₂(x + 2) = 1。
log₂[(x + 6) ÷ (x + 2)] = 1
(x + 6) ÷ (x + 2) = 2¹ = 2
x + 6 = 2x + 4 ⇒ x = 2
he original logarithms are defined. Here x + 6 > 0 and x + 2 > 0, so x = 2 is valid.
一定要检查原对数是否有定义。在这里 x + 6 > 0 且 x + 2 > 0，因此 x = 2 是有效解。
A common trap is producing a solution that makes the argument of a logarithm negative or zero. Such solutions must be rejected, even if they satisfy the transformed equation.
一个常见陷阱是求出的解使对数的真数为负数或零。这类解即使满足变形后的方程，也必须舍去。
so x = ±3. However, x = -3 is invalid because x − 1 < 0, so the only solution is x = 3.
例如，log₂(x − 1) + log₂(x + 1) = 3 可化为 x² − 1 = 8，所以 x = ±3。但 x = -3 无效，因为 x − 1 < 0，所以唯一解是 x = 3。
6. Change of Base Formula | 换底公式
Calculators usually only provide log base 10 and natural log. To evaluate logarithms with other bases, use the change of base formula:
计算器通常只提供以 10 为底的对数和自然对数。要计算其他底数的对数，需要使用换底公式：
logₐb = log_c b ÷ log_c a
= 10 or c = e is most convenient. This formula also appears in proof-style exam questions.
这里的 c 可以是任何合法的底数，但实际中 c = 10 或 c = e 最方便。这个公式也经常出现在证明类考题中。
For example, to evaluate log₂5, use natural logarithms:
例如，要计算 log₂5，可以使用自然对数：
log₂5 = ln 5 ÷ ln 2 ≈ 2.3219
The change of base formula is also useful when solving equations where the bases are different. It lets you rewrite all logarithms in one common base before simplifying.
当方程中的底数不同时，换底公式也很有用。它可以让你先把所有对数改写为相同的底数，再进行化简。
a. They are reciprocals when written via the change of base formula: logₐb = 1 ÷ log_b a.
不要把 logₐb 和 log_b a 混淆。通过换底公式可以看出它们互为倒数：logₐb = 1 ÷ log_b a。
7. The Natural Logarithm and e | 自然对数与 e
2.71828. It arises naturally in calculus, compound interest, population growth and decay problems.
自然对数写作 ln x，是以 e 为底的对数，其中 e ≈ 2.71828。它在微积分、复利、人口增长和衰减问题中自然出现。
ln x = logₑx
The key cancellation identities are:
关键的抵消恒等式为：
eln x = x and ln(eˣ) = x
= 1. These values are used constantly when simplifying expressions and solving equations.
还要记住 ln 1 = 0 和 ln e = 1。在化简表达式和求解方程时，这些值会被频繁使用。
g natural logs is the most direct method. For example, if P = Aeᵏᵗ, then ln P = ln A + kt.
当指数模型以 e 为底时，取自然对数是最直接的方法。例如，若 P = Aeᵏᵗ，则 ln P = ln A + kt。
8. Graphs of Logarithmic Functions | 对数函数图像
shape. It passes through (1, 0), increases slowly, and has a vertical asymptote at x = 0.
当 a > 1 时，y = logₐx 的图像形状很有特点。它经过点 (1, 0)，缓慢上升，并以 x = 0 为竖直渐近线。
- Domain: x > 0 | 定义域：x > 0
- Range: all real numbers | 值域：全体实数
- x-intercept: (1, 0) | x 轴截距：(1, 0)
- Vertical asymptote: x = 0 | 竖直渐近线：x = 0
The logarithmic graph is the reflection of the exponential graph y = aˣ in the line y = x. This is because logarithms and exponentials are inverse functions.
对数函数的图像是指数函数 y = aˣ 关于直线 y = x 的对称图形。这是因为对数函数与指数函数互为反函数。
y = ln(x − 2) + 3 is translated 2 units to the right and 3 units up, with asymptote x = 2.
图像变换与通常一致。例如，y = ln(x − 2) + 3 表示向右平移 2 个单位、向上平移 3 个单位，渐近线变为 x = 2。
9. Exponential Growth and Decay Modelling | 指数增长与衰减建模
Edexcel exam questions often use logarithms to analyse exponential models such as y = abˣ or P = Aeᵏᵗ. Taking logarithms can turn these curves into straight lines.
爱德思考试题经常使用对数来分析指数模型，如 y = abˣ 或 P = Aeᵏᵗ。取对数可以将这些曲线转化为直线。
If y = abˣ, then:
如果 y = abˣ，那么：
ln y = ln a + x ln b
This means that plotting ln y against x gives a straight line with gradient ln b and vertical intercept ln a. You can then estimate a and b from experimental data.
这意味着以 ln y 对 x 作图会得到一条直线，其斜率为 ln b，纵截距为 ln a。这样你就可以根据实验数据估计 a 和 b。
For a decay model such as m = m₀e−kt, taking logarithms gives ln m = ln m₀ − kt, where k is a positive decay constant.
对于衰减模型，如 m = m₀e−kt，取对数得到 ln m = ln m₀ − kt，其中 k 是正的衰减常数。
This linear form is especially useful when you need to find half-life, doubling time, or unknown constants from a table of values.
当你需要根据数据表求半衰期、倍增时间或未知常数时，这种线性形式特别有用。
10. Differentiation Involving Natural Logarithms | 含自然对数的求导
The natural logarithm has a simple derivative that appears frequently in A-Level calculus.
自然对数的导数形式简单，在 A-Level 微积分中经常出现。
d/dx (ln x) = 1 ÷ x, for x > 0
Using the chain rule, this generalises to:
使用链式法则，可以推广为：
d/dx [ln(f(x))] = f '(x) ÷ f(x)
For example, if y = ln(3x² + 1), then dy/dx = 6x ÷ (3x² + 1). The derivative of eˣ is eˣ itself, so logarithmic differentiation often pairs with exponential differentiation.
例如，若 y = ln(3x² + 1)，则 dy/dx = 6x ÷ (3x² + 1)。eˣ 的导数仍是 eˣ，因此对数求导经常与指数求
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