Laws of Logarithms | 对数运算法则

📚 Laws of Logarithms | 对数运算法则

Logarithms are a core part of the Edexcel A-Level Mathematics specification. They reverse the operation of exponentiation and allow us to solve equations where the unknown appears in the power. The laws of logarithms turn multiplication into addition, division into subtraction and powers into multiplication, making them essential tools for simplifying algebraic expressions and tackling exponential models.

对数是 Edexcel A-Level 数学大纲的核心内容。它们是指数运算的逆运算，能帮助我们求解未知数出现在指数中的方程。对数运算法则能把乘法转化为加法、除法转化为减法、乘方转化为乘法，因此是化简代数表达式和处理指数模型的重要工具。

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1. Definition and Notation | 定义与记号

If a > 0, a ≠ 1 and x > 0, then y = logₐ x means aʸ = x. We read this as ‘log base a of x’. The base a is written as a subscript and must always be positive and not equal to 1.

如果 a > 0，a ≠ 1 且 x > 0，那么 y = logₐ x 表示 aʸ = x。我们读作“以 a 为底 x 的对数”。底数 a 写在下标位置，它必须始终为正且不能等于 1。

The logarithm is the exponent to which the base must be raised to produce the number. For example, log₂ 8 = 3 because 2³ = 8.

对数就是为使底数变成某个数而必须使用的指数。例如，log₂ 8 = 3，因为 2³ = 8。

aʸ = x ⇔ y = logₐ x

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2. The Product Rule | 乘法法则

The product rule states that the logarithm of a product is the sum of the logarithms. For any positive values M and N and any valid base a, we have logₐ(MN) = logₐ M + logₐ N.

乘法法则指出，乘积的对数等于各对数之和。对于任意正数 M、N 和有效底数 a，我们有 logₐ(MN) = logₐ M + logₐ N。

logₐ(MN) = logₐ M + logₐ N

For example, log₂(8 × 4) = log₂ 8 + log₂ 4 = 3 + 2 = 5. This matches the direct check: 8 × 4 = 32 = 2⁵.

例如，log₂(8 × 4) = log₂ 8 + log₂ 4 = 3 + 2 = 5。这与直接验算一致：8 × 4 = 32 = 2⁵。

This rule is useful both for expanding a single logarithm into a sum and for combining several logarithmic terms into one.

这个法则既可用于把单个对数展开为和，也可用于把多个对数项合并为一个。

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3. The Quotient Rule | 除法法则

The quotient rule states that the logarithm of a quotient is the difference of the logarithms. For positive M and N, logₐ(M/N) = logₐ M − logₐ N.

除法法则指出，商的对数等于各对数之差。对于正数 M 和 N，有 logₐ(M/N) = logₐ M − logₐ N。

logₐ(M/N) = logₐ M − logₐ N

For example, log₃(81/9) = log₃ 81 − log₃ 9 = 4 − 2 = 2. This is correct because 81/9 = 9 = 3².

例如，log₃(81/9) = log₃ 81 − log₃ 9 = 4 − 2 = 2。这是正确的，因为 81/9 = 9 = 3²。

Be careful to keep the order: the numerator’s logarithm comes first, followed by the denominator’s logarithm.

注意保持顺序：分子的对数在前，分母的对数在后。

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4. The Power Rule | 幂法则

The power rule states that the logarithm of a power is the exponent multiplied by the logarithm of the base. For any real exponent n, logₐ(xⁿ) = n logₐ x.

幂法则指出，幂的对数等于指数乘以底数的对数。对于任意实数指数 n，有 logₐ(xⁿ) = n logₐ x。

logₐ(xⁿ) = n logₐ x

This rule also works for roots because they can be written as fractional powers. For instance, logₐ √x = logₐ x^(1/2) = (1/2) logₐ x.

这个法则也适用于根式，因为根式可以写成分数指数。例如，logₐ √x = logₐ x^(1/2) = (1/2) logₐ x。

Example: log₅ 25³ = 3 log₅ 25 = 3 × 2 = 6. The power rule is especially powerful when solving equations where the variable is in an exponent.

示例：log₅ 25³ = 3 log₅ 25 = 3 × 2 = 6。当变量位于指数中时，幂法则在解方程时尤其有用。

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5. Special Cases and Law Summary | 特殊值与法则汇总

Two special values follow directly from the definition. Since a⁰ = 1, we have logₐ 1 = 0. Since a¹ = a, we have logₐ a = 1. These simple results appear very often in simplification and equation work.

两个特殊值直接来自定义。因为 a⁰ = 1，所以 logₐ 1 = 0。因为 a¹ = a，所以 logₐ a = 1。这些简单结果在化简和方程中经常出现。

logₐ 1 = 0 and logₐ a = 1

The inverse properties are also essential: logₐ(aˣ) = x and a^(logₐ x) = x. They allow a logarithm and an exponential with the same base to cancel each other.

反函数性质也至关重要：logₐ(aˣ) = x 以及 a^(logₐ x) = x。它们使同底的对数和指数可以相互抵消。

Law | 法则	Formula | 公式
Product | 乘法	logₐ(MN) = logₐ M + logₐ N
Quotient | 除法	logₐ(M/N) = logₐ M − logₐ N
Power | 幂	logₐ(xⁿ) = n logₐ x
Special values | 特殊值	logₐ 1 = 0, logₐ a = 1

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6. Change of Base Formula | 换底公式

Calculators usually provide logarithms only in base 10 or base e. To evaluate a logarithm with a different base, we use the change of base formula: logₐ b = log_c b / log_c a, where c is any valid base.

计算器通常只提供以 10 或 e 为底的对数。要计算其他底数的对数，我们使用换底公式：logₐ b = log_c b / log_c a，其中 c 是任意有效底数。

logₐ b = log_c b / log_c a

For example, log₂ 7 = ln 7 / ln 2 ≈ 1.9459 / 0.6931 ≈ 2.807. The choice of base c does not affect the answer as long as the same base is used in both numerator and denominator.

例如，log₂ 7 = ln 7 / ln 2 ≈ 1.9459 / 0.6931 ≈ 2.807。只要分子和分母使用相同的底数，底数 c 的选择不会影响结果。

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7. Natural Logarithms and Base e | 自然对数与底 e

The natural logarithm, written as ln x, is simply logₑ x. It obeys the same laws: ln(MN) = ln M + ln N, ln(M/N) = ln M − ln N, and ln(xⁿ) = n ln x.

自然对数写作 ln x，它就是 logₑ x。它遵循相同的法则：ln(MN) = ln M + ln N，ln(M/N) = ln M − ln N，ln(xⁿ) = n ln x。

ln(MN) = ln M + ln N, ln(M/N) = ln M − ln N, ln(xⁿ) = n ln x

The special values are ln 1 = 0 and ln e = 1. The inverse properties are e^(ln x) = x and ln(eˣ) = x. Natural logarithms are especially common in calculus and in modelling exponential growth or decay.

特殊值是 ln 1 = 0 和 ln e = 1。反函数性质为 e^(ln x) = x 和 ln(eˣ) = x。自然对数在微积分以及指数增长或衰减模型中尤其常见。

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8. Solving Exponential Equations | 解指数方程

When the unknown is in the exponent, take logarithms of both sides and apply the power rule. For example, to solve 3ˣ = 20, take log of both sides: log 3ˣ = log 20. This gives x log 3 = log 20, so x = log 20 / log 3 ≈ 2.727.

当未知数在指数中时，对两边取对数并应用幂法则。例如，要解 3ˣ = 20，对两边取对数：log 3ˣ = log 20。由此得到 x log 3 = log 20，所以 x = log 20 / log 3 ≈ 2.727。

x = log 20 / log 3

You can use any base, but base 10 or base e is most common. For a more involved equation such as 5^(2x−1) = 12, take natural logs to get (2x − 1)ln 5 = ln 12, then solve for x.

你可以使用任意底数，但以 10 或 e 为底最常用。对于更复杂的方程，例如 5^(2x−1) = 12，取自然对数得到 (2x − 1)ln 5 = ln 12，然后解出 x。

Always write the exact expression first and only round at the final step if a decimal answer is required.

始终先写出精确表达式，只有在题目要求小数答案时才在最后一步取近似值。

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9. Solving Logarithmic Equations | 解对数方程

To solve equations with logarithms, first use the laws to combine or simplify terms. If two logarithms with the same base are equal, their arguments must be equal: if logₐ M = logₐ N, then M = N.

解含对数的方程时，先用运算法则合并或化简各项。如果两个同底对数相等，则它们的真数必须相等：若 logₐ M = logₐ N，则 M = N。

logₐ M = logₐ N ⇒ M = N

Example: log₂(x + 1) = log₂(3x − 5) gives x + 1 = 3x − 5, so x = 3. Since x = 3 makes both arguments positive, it is valid.

示例：log₂(x + 1) = log₂(3x − 5) 得到 x + 1 = 3x − 5，所以 x = 3。由于 x = 3 使两个真数都为正，该解有效。

Another type is log(x + 6) + log(x − 2) = 1. Use the product rule to write log[(x + 6)(x − 2)] = 1, then convert to (x + 6)(x − 2) = 10 and solve the quadratic. Always check that no solution makes a logarithm undefined.

另一种类型是 log(x + 6) + log(x − 2) = 1。使用乘法法则写成 log

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