📚 Solving Equations Using Logarithms | 利用对数求解方程
Equations in which the unknown sits in an exponent cannot be solved by ordinary algebraic rearrangement alone, because the usual operations of adding, multiplying or factoring do not reach inside an index. The logarithm is the inverse of the exponential function, so applying it to both sides of an exponential equation lifts the unknown down from the exponent and turns the problem into a linear, quadratic or otherwise familiar equation. This article sets out the Edexcel A-level Pure Mathematics treatment of such equations: the laws you must know, the standard solution methods, the role of domain restrictions, and the exam habits that protect your marks.
当未知数位于指数位置时，单靠常规的代数变形无法求解，因为加、乘或因式分解这些运算都触及不到指数内部。对数函数是指数函数的反函数，因此对方程两边同时取对数，就能把未知数从指数上"放下来"，把问题转化为一次方程、二次方程或其他熟悉的形式。本文按照 Edexcel A-level Pure Mathematics 的要求，系统讲解：必须掌握的对数法则、标准解法、定义域限制的作用，以及能保住分数的应试习惯。
1. Why Exponential Equations Need Logarithms | 为什么指数方程需要对数
Consider the equation 3ˣ ction that works, and you cannot isolate x by dividing both sides by 3, because the 3 is the base of a power, not a coefficient. What you need is an operation that reverses exponentiation, and that operation is the logarithm. Because logₐ(aˣ ithms of both sides converts an equation about powers into an equation about the unknown itself.
以方程 3ˣ = 20 为例。没有任何整数或简单分数能满足它，你也不能通过两边同除以 3 来孤立 x，因为这里的 3 是幂的底数而不是系数。你需要的是一种能"抵消"指数运算的运算，那就是对数。由于对任意合法底数 a 都有 logₐ(aˣ) = x，所以对两边同时取对数，就能把关于幂的方程变成关于未知数本身的方程。
The same idea applies whether the base is 10, 2, 5 or the constant e. The structure of the method never changes; only the choice of logarithm does. This is why the topic sits at the junction of index laws, logarithm laws and domain restrictions, and why examiners return to it so often.
无论底数是 10、2、5 还是常数 e，思路完全一样。方法的结构从不变，变化的只是取哪种对数。这正是本主题处在指数律、对数法则和定义域限制三者交汇处的原因，也是考官反复考查它的原因。
2. Core Logarithm Laws You Must Know | 必须掌握的对数核心法则
Before you solve anything, you must be able to apply the main laws of logarithms fluently in both directions. They hold for any base a with a > 0 and a ≠ 1, and for arguments x > 0 and y > 0.
在动手解题之前，你必须能够熟练地双向使用对数的主要法则。这些法则对任意底数 a（a > 0 且 a ≠ 1）成立，并要求真数 x > 0、y > 0。
| Law | 法则 | Form | 形式 |
| Product rule | 积法则 | logₐ(xy) = logₐ x + logₐ y |
| Quotient rule | 商法则 | logₐ(x ÷ y) = logₐ x − logₐ y |
| Power rule | 幂法则 | logₐ(xⁿ) = n logₐ x |
| Zero and one | 零与一 | logₐ 1 = 0, logₐ a = 1 |
| Reciprocal | 倒数 | logₐ(1 ÷ x) = −logₐ x |
| Change of base | 换底公式 | logₐ x = logᵦ x ÷ logᵦ a |
| Inverse pair | 反函数配对 | logₐ(aˣ) = x and alogₐ x = x |
The power rule is the engine of the whole topic: it is the law that pulls the exponent to the front. Change of base is what lets a calculator with only log and ln keys evaluate a logarithm to any base.
幂法则是整个主题的发动机：正是它把指数"拉到前面"。换底公式则让你只用计算器上的 log 和 ln 键，就能求出任意底数的对数值。
3. The Standard Method: Solving aˣ = b | 标准方法：求解 aˣ = b
For an equation of the form aˣ on is obtained in three steps. First, take logarithms of both sides using any convenient base. Second, apply the power rule to bring the exponent down as a multiplier. Third, divide by that multiplier to isolate x.
对于形如 aˣ = b 的方程（a > 0、a ≠ 1、b > 0），解法分三步。第一步，用任意方便的底数对两边取对数；第二步，用幂法则把指数降下来变成乘数；第三步，除以这个乘数，解出 x。
aˣ = b ⟹ x = log b ÷ log a = logₐ b
Worked example: solve 3ˣ e decimal places. Taking logarithms gives x log 3 = log 20, so x = log 20 ÷ log 3 ≈ 2.727.
例题：解 3ˣ = 20，答案保留三位小数。两边取对数得 x log 3 = log 20，所以 x = log 20 ÷ log 3 ≈ 2.727。
A slightly harder version has a linear expression in the index, such as 52x − 1 et must be treated as a single quantity: expand only after the logarithm has been applied.
稍难一点的版本是指数上带有一次式，例如 52x − 1 = 40。两边取对数得 (2x − 1) log 5 = log 40，所以 2x − 1 = log 40 ÷ log 5 ≈ 2.292，于是 2x ≈ 3.292，x ≈ 1.646。注意括号必须当作一个整体处理：只有在取完对数之后才能展开。
4. Choosing Your Base: log₁₀ or ln | 选择底数：常用对数还是自然对数
e you choose does not affect the final answer. Any base gives the same value of x; only the intermediate numbers differ. In practice, use base 10 (the log key) or base e (the ln key) on your calculator, and pick whichever makes the numbers tidier.
由于 logₐ x = logᵦ x ÷ logᵦ a，你选择的底数不会影响最终答案。任何底数得到的 x 都相同，只是中间过程的数字不同。实际做题时，用计算器上的 log 键（底数 10）或 ln 键（底数 e），选让数字更整洁的那个即可。
x = log b ÷ log a = ln b ÷ ln a
If the original equation involves eˣ, always use ln, because ln(eˣ) = x immediately. If the original equation involves 10ˣ works; ln is usually marginally more accurate on a calculator, but log is easier to read.
如果原方程含有 eˣ，一律用 ln，因为 ln(eˣ) = x 一步到位。如果原方程含有 10ˣ，就用 log。如果底数像 7 这样，两种键都可以；在计算器上 ln 通常精度略高一点，但 log 更易读。
5. Different Bases and Comparing Indices | 不同底数与指数比较
on can be written as powers of the same number, you do not need logarithms at all. Solve 2ˣ⁺¹ = 8ˣ⁻² by rewriting 8 as 2³, so the right side becomes 2³ˣ⁻⁶. Equating indices gives x + 1 = 3x − 6, hence x = 3.5.
当方程两边都能写成同一个数的幂时，根本不需要用对数。解 2ˣ⁺¹ = 8ˣ⁻²，先把 8 改写为 2³，右边变成 2³ˣ⁻⁶。令指数相等得 x + 1 = 3x − 6，所以 x = 3.5。
When the bases are genuinely unrelated, such as 2ˣ⁺¹ = 5ˣ e compulsory. Take logarithms of both sides, use the power rule, then collect the x terms:
当底数确实毫无关联时，例如 2ˣ⁺¹ = 5ˣ，比较指数就行不通了，必须取对数。两边取对数，用幂法则，然后合并含 x 的项：
(x + 1) log 2 = x log 5 ⟹ x = log 2 ÷ (log 5 − log 2) ≈ 0.756
The key skill here is algebra rather than logarithms: expand the bracket, gather all x terms on one side, and factorise x out before dividing.
这里的关键技能其实是代数而不是对数：展开括号、把所有含 x 的项移到一边、先提出公因式 x 再相除。
6. Quadratics in Disguise | 伪装的二次方程
n exponential equation. The clue is that one power is exactly double the other, because a²ˣ = (aˣ)². Take 2²ˣ − 5(2ˣ) + 4 = 0. Substitute y = 2ˣ − 5y + 4 = 0, so (y − 1)(y − 4) = 0 and y = 1 or y = 4. Reversing the substitution gives 2ˣ = 1, so x = 0, or 2ˣ = 4, so x = 2.
许多考题把二次方程藏在指数方程里。线索就是其中一个幂正好是另一个的两倍，因为 a²ˣ = (aˣ)²。以 2²ˣ − 5(2ˣ) + 4 = 0 为例。令 y = 2ˣ，得 y² − 5y + 4 = 0，即 (y − 1)(y − 4) = 0，y = 1 或 y = 4。回代得 2ˣ = 1，故 x = 0；或 2ˣ = 4，故 x = 2。
The same pattern with base e: e²ˣ − 3eˣ + 2 = 0 becomes y² − 3y + 2 = 0 with y = eˣ erse the substitution at the end, and remember that a solution y must be positive, since aˣ > 0 for every real x.
底数为 e 时是同一套路：e²ˣ − 3eˣ + 2 = 0 令 y = eˣ 得 y² − 3y + 2 = 0，解得 y = 1 或 y = 2，故 x = 0 或 x = ln 2。最后一定要回代，并记住 y 必须为正，因为对任意实数 x 都有 aˣ > 0。
7. Equations Built on e and ln | 以 e 与 ln 为核心的方程
e inverse functions, equations involving them are usually the quickest of all. To solve e²ˣ lve ln(3x − 1) = 2, rewrite in exponential form: 3x − 1 = e², so x = (e² + 1) ÷ 3 ≈ 2.796.
由于 ln 与 e 互为反函数，含它们的方程通常是最快的。解 e²ˣ = 7，两边取自然对数：2x = ln 7，故 x = ln 7 ÷ 2 ≈ 0.973。解 ln(3x − 1) = 2，改写为指数形式：3x − 1 = e²，故 x = (e² + 1) ÷ 3 ≈ 2.796。
The two directions of conversion are the whole trick. An exponential statement aˣ x; switching between them is legal at any point and often removes the logarithm entirely.
两个方向之间的互换就是全部诀窍。指数式 aˣ = b 与对数式 logₐ b = x 是同一件事；任何时刻在两者之间切换都是合法的，而且常常能直接消掉对数。
8. Log Equations that Need Condensing First | 需要先合并的对数方程
. The product rule gives log₂[x(x − 2)] = 3, so x(x − 2) = 2³ = 8, giving x² − 2x − 8 = 0 and (x − 4)(x + 2) = 0.
当一侧出现多个对数时，先把它们合并成一个对数，再化为指数形式。解 log₂ x + log₂(x − 2) = 3。用积法则得 log₂[x(x − 2)] = 3，所以 x(x − 2) = 2³ = 8，即 x² − 2x − 8 = 0，(x − 4)(x + 2) = 0。
1 = 3x − 3, hence x = 4. Cancelling or combining logarithms without checking is the easiest way to lose marks here.
遇到对数相减，就用商法则。解 log₃(2x + 1) − log₃(x − 1) = 1，得 log₃[(2x + 1) ÷ (x − 1)] = 1，所以 (2x + 1) ÷ (x − 1) = 3，即 2x + 1 = 3x − 3，故 x = 4。在这里不检验就约分或随意合并对数，是最容易丢分的做法。
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