📚 Exercise 6E.4: Solving Logarithmic Equations | 练习6E.4：对数方程求解
Exercise 6E.4 from the IB Mathematics course focuses on solving logarithmic equations, a key skill that combines understanding of logarithmic definitions, laws, and the ability to handle exponential forms. Mastering these problems builds the foundation for calculus, exponential growth, and many real-world applications.
IB数学课程中的练习6E.4专注于求解对数方程，这项关键技能融合了对数定义、法则的理解以及处理指数形式的能力。掌握这些问题为微积分、指数增长和许多实际应用奠定了基础。
1. Understanding Logarithmic Definitions | 理解对数定义
The logarithmic equation logb a = c is equivalent to the exponential form bc gt; 0. This relationship is the starting point for solving any logarithmic equation, because it lets you move the unknown out of the logarithm and into a form you can manipulate algebraically.
对数方程 logb a = c 等价于指数形式 bc = a，前提是 b > 0，b ≠ 1，且 a > 0。这种关系是求解任何对数方程的出发点，因为它能把未知量从对数中移出，转变成可以用代数方法处理的形式。
ecause 2³ = 8. Similarly, log₅ 25 = 2 because 5² = 25. The base of the logarithm always becomes the base of the power, and the value of the logarithm becomes the exponent.
理解这种等价关系可以使你在隔离变量时在对数形式和指数形式之间进行切换。例如，log₂ 8 = 3，因为 2³ = 8。类似地，log₅ 25 = 2，因为 5² = 25。对数的底数总是变成幂的底数，而对数的值则变成指数。
logb a = c ⇔ bc = a
Two special values follow directly from the definition: logb b = 1 and logb instantly in an exam, because they often appear as the final step after simplifying a longer expression.
由定义可直接得到两个特殊值：logb b = 1 和 logb 1 = 0。在考试中你应该能够立即写出它们，因为它们常常作为化简较长表达式后的最后一步出现。
2. Solving Simple Logarithmic Equations | 求解简单对数方程
ed to check for extraneous solutions here because a logarithm of a positive number is defined for every positive input.
当单个对数等于一个常数时，应立即转换为指数形式。求解 log₂ (x) = 5：改写后得到 x = 2⁵，所以 x = 32。这里不需要检查增根，因为正数的对数对每一个正输入都有定义。
₅ (2x + 1) = 2, we get 2x + 1 = 5² = 25, hence 2x = 24 and x = 12. In both cases the equation collapses to a linear one after conversion.
如果真数是一个线性表达式而不是单个变量，方法完全相同。对于 log₃ (x − 1) = 2，得到 x − 1 = 3² = 9，因此 x = 10。对于 log₅ (2x + 1) = 2，得到 2x + 1 = 5² = 25，所以 2x = 24，x = 12。两种情况下，方程在转换后都化为一元一次方程。
The general procedure is: isolate the logarithmic term, write the equation in the form logb(M) = k, convert to M = bk, then solve the resulting algebraic equation. Keeping this order fixed prevents most careless errors.
通用步骤是：先隔离对数项，把方程写成 logb(M) = k 的形式，转换为 M = bk，再求解得到的代数方程。固定这个顺序可以避免大多数粗心错误。
3. Using Logarithm Laws to Combine Terms | 运用对数法则合并各项
Many equations in Exercise 6E.4 contain a sum or difference of logarithms. These must be combined into a single logarithm before conversion is possible. The three laws you need are the product law logb M + logb N = logb(MN), the quotient law logb M − logb N = logb(M / N), and the power law k logb M = logb(Mk).
6E.4 中的许多方程包含对数的和或差。在转换之前，必须先把它们合并成单个对数。你需要掌握三条法则：积法则 logb M + logb N = logb(MN)，商法则 logb M − logb N = logb(M / N)，以及幂法则 k logb M = logb(Mk)。
es as (x − 4)(x + 2) = 0, giving x = 4 or x = −2. Since the argument of a logarithm must be positive, x = −2 is rejected and the only solution is x = 4.
例题：求解 log₂ x + log₂ (x − 2) = 3。合并左边得 log₂ [x(x − 2)] = 3，所以 x(x − 2) = 2³ = 8。由此得到 x² − 2x − 8 = 0，因式分解为 (x − 4)(x + 2) = 0，得 x = 4 或 x = −2。由于对数的真数必须为正，x = −2 被舍去，唯一解是 x = 4。
Note carefully that logb(M + N) is not equal to logb M + logb N. There is no logarithm law for the logarithm of a sum, so an expression such as log₂ (x + 3) must be handled by conversion, never by splitting.
务必注意，logb(M + N) 不等于 logb M + logb N。对数没有关于"和"的法则，因此像 log₂ (x + 3) 这样的表达式只能通过转换处理，绝不能拆分。
4. Domain Restrictions Before You Solve | 求解之前先确定定义域
atisfy. For log₂ x + log₂ (x − 2) = 3, the conditions are x > 0 and x − 2 > 0, which combine to x > 2. The candidate x = −2 is then rejected instantly without substitution.
每个对数的真数都必须为正，因此在任何代数运算之前，你应先写下每个真数必须满足的条件。对于 log₂ x + log₂ (x − 2) = 3，条件是 x > 0 且 x − 2 > 0，合并为 x > 2。于是候选解 x = −2 无需代入即可立即舍去。
This preliminary step is one of the most reliable marks in the whole topic. Examiners frequently award a method mark for stating the domain, and many students lose accuracy marks simply because they never check whether an algebraic solution is admissible.
这个前置步骤是整个专题中最稳的得分点之一。考评分经常为写出定义域给出方法分，而许多学生丢分，仅仅是因为从未检查代数解是否合法。
ndition 2x − 5 > 0 gives x > 2.5, which is satisfied. Checking the domain against the answer is a quick form of verification.
定义域还能告诉你可能答案的范围。若 log₃ (2x − 5) = 1，则 2x − 5 = 3，所以 x = 4，而定义域条件 2x − 5 > 0 给出 x > 2.5，满足条件。用定义域检验答案是种快速的验证方式。
5. Equations with Logarithms on Both Sides | 两边都有对数的方程
If logb M = logb ositive. This is the one-to-one property of the logarithmic function, and it allows you to delete the logarithms and solve the remaining equation directly.
如果 logb M = logb N，那么 M = N，前提是 M 和 N 都为正。这是对数函数的一一对应性质，它允许你消去对数，直接求解剩下的方程。
4, so x = 3. Both arguments equal 7 at x = 3, which is positive, so the solution is valid.
例：求解 log₃ (2x + 1) = log₃ (x + 4)。令真数相等得 2x + 1 = x + 4，所以 x = 3。当 x = 3 时两个真数都等于 7，为正，因此该解有效。
The bases must match before you can equate arguments. If the bases differ, use the change of base formula logb M = logk M / logk b to rewrite one side, or convert both sides to exponentials and compare.
在令真数相等之前，底数必须相同。若底数不同，可用换底公式 logb M = logk M / logk b 改写其中一边，或将两边都化为指数形式再比较。
log₂ x. Hence log₂ x = 2 and x = 4. This type of question appears frequently in IB papers.
底数不同的例子：求解 log₂ x + log₄ x = 3。将 log₄ x 写成 log₂ x / log₂ 4 = (log₂ x) / 2，左边变为 (3 / 2) log₂ x。因此 log₂ x = 2，x = 4。这类题目在 IB 试卷中经常出现。
6. Equations Leading to Quadratic Form | 化为二次形式的方程
as (y − 1)(y − 2) = 0, giving y = 1 or y = 2. Back-substituting gives log₂ x = 1, so x = 2, and log₂ x = 2, so x = 4.
有些方程含有对数的平方，例如 (log₂ x)² − 3 log₂ x + 2 = 0。令 y = log₂ x，方程变为 y² − 3y + 2 = 0，因式分解为 (y − 1)(y − 2) = 0，得 y = 1 或 y = 2。回代得 log₂ x = 1，故 x = 2；以及 log₂ x = 2，故 x = 4。
rns no marks because the question asks for the value of x, not the value of the substitute.
最后一定要回到原变量。常见错误是把 y = 1 和 y = 2 当作答案，这样不得分，因为题目问的是 x 的值，而不是替代变量的值。
x + 6) = 27, leading to x² + 6x − 27 = 0 and (x + 9)(x − 3) = 0. Thus x = 3 or x = −9, and the domain x > 0 rejects x = −9.
另一种二次型模式来自积法则与转换的结合，例如 log₃ x + log₃ (x + 6) = 3。合并得 log₃ [x(x + 6)] = 3，所以 x(x + 6) = 27，得到 x² + 6x − 27 = 0 及 (x + 9)(x − 3) = 0。因此 x = 3 或 x = −9，而定义域 x > 0 舍去 x = −9。
x² + 6x − 27 = 0 ⇒ (x + 9)(x − 3) = 0 ⇒ x = 3
7. Checking and Rejecting Extraneous Solutions | 检验并舍去增根
Extraneous solutions appear whenever the algebraic manipulation enlarges the domain. Combining logarithms, squaring both sides, and multiplying by an expression containing the unknown can all introduce values that satisfy the new equation but not the original one.
当代数变形扩大了定义域时，就会出现增根。合并对数、两边平方、乘以含未知量的表达式，都可能引入满足新方程但不满足原方程的值。
= 3, which matches. For x = −2, the term log₂ (−2) is undefined, so the value is rejected.
代入检验是最稳妥的方法。在 log₂ x + log₂ (x − 2) = 3 中取 x = 4，左边为 log₂ 4 + log₂ 2 = 2 + 1 = 3，与右边相符。取 x = −2 时，log₂ (−2) 无定义，因此舍去该值。
In an exam, write a short sentence stating why a solution is rejected, for example "x = −2 is rejected because the argument of a logarithm must be positive" soning explicit and protects the accuracy mark even if the arithmetic is later questioned.
在考试中，用一句话说明舍去某解的原因，例如"舍去 x = −2，因为对数的真数必须为正"。这样你的推理就清晰明确，即使后续运算被质疑，也能保住准确性得分。
8. Working with Base e and Base 10 | 底数 e 与底数 10 的处理
ys exactly the same laws: ln M + ln N = ln(MN), ln M − ln N = ln(M / N), and k ln M = ln(Mk). The inverse relationship is ln(ex) = x and eln x = x for x > 0.
以 10 为底的对数写作 log x，以 e 为底的对数写作 ln x。自然对数遵循完全相同的法则：ln M + ln N = ln(MN)，ln M − ln N = ln(M / N)，以及 k ln M = ln(Mk)。反函数关系为 ln(ex) = x，且当 x > 0 时 eln x = x。
nless the question asks for an exact answer, round to three significant figures, which is the standard IB expectation.
例：求解 ln(x + 3) = 2。转换得 x + 3 = e² ≈ 7.389，所以 x ≈ 4.389。除非题目要求精确值，否则保留三位有效数字，这是 IB 的标准要求。
Example with a quadratic: solve ex ows how logarithmic equations and exponential equations are two views of the same problem.
含二次式的例子：求解 ex = 5。两边取自然对数得 x = ln 5 ≈ 1.609。这说明对数方程与指数方程是同一问题的两种视角。
logb M = logk M / logk b
9. Applications: Growth, Decay and Finance | 应用：增长、衰减与金融
n appears in an exponent. For compound interest, the balance after n years is A = P(1 + r)n. To find how long it takes for an investment to double, set 2P = P(1 + r)n, cancel P, and take logarithms: n = log 2 / log(1 + r).
只要未知量出现在指数上，就可以用对数方程建模。对于复利，n 年后的余额为 A = P(1 + r)n。要确定投资翻倍所需的时间，令 2P = P(1 + r)n，约去 P 并取对数：n = log 2 / log(1 + r)。
after about 15 years if interest is credited annually. Radioactive decay follows the same pattern with a decay factor below 1, and the half-life equation is obtained by setting the remaining mass equal to half the initial mass.
取 r = 0.05，得 n = log 2 / log 1.05 ≈ 14.2，因此若按年计息，投资约 15 年后翻倍。放射性衰变遵循相同模式，只是衰减因子小于 1，把剩余质量设为初始质量的一半即可得到半衰期方程。
These modelling questions reward clear structure: define the variable, write the model, substitute the known values, take logarithms of both sides, then solve and interpret the answer in context. Interpreting the answer means writing a sentence such as "the population reaches 8000 after approximately 12 years".
这类建模题得分关键在于结构清晰：定义变量、写出模型、代入已知值、两边取对数，然后求解并结合情境解释答案。解释答案就是要写出类似"该种群大约在 12 年后达到 8000"这样的句子。
10. Common Mistakes to Avoid | 常见错误及规避方法
The first mistake is treating log(M + N) as log M + log N. The second is forgetting the domain restriction and accepting a negative solution. The third is dividing both sides by a logarithm, which can silently delete a valid solution; instead, factorise or use substitution.
第一个错误是把 log(M + N) 当作 log M + log N。第二个错误是忘记定义域限制而接受负解。第三个错误是两边同时除以一个对数，这可能悄悄丢掉一个有效解；应改为因式分解或换元。
The fourth mistake is mishandling the base condition b ≠ 1, which matters when a question asks for the range of values of a parameter. The fifth is rounding too early in a multi-step calculation, which propagates error into the final answer.
第四个错误是忽视底数条件 b ≠ 1，当题目要求参数的取值范围时这一点很重要。第五个错误是在多步计算中过早取近似值，这会把误差传递到最终答案。
A quick error table for revision:
复习用的错误速查表：
| Incorrect | 错误 | Correct | 正确 |
| log(M + N) = log M + log N | log(MN) = log M + log N |
| log₂ 8 = 4 | log₂ 8 = 3 |
| Accepting x = −2 from x(x − 2) = 8 | Rejecting x = −2 by the domain x > 2 |
| Dividing by log x without checking log x = 0 | Factorising so that x = 1 is not lost |
11. Exam Strategy and Practice Questions | 考试策略与练习题
In an exam, start by stating the domain conditions, then combine logarithms, then convert to exponential form, then solve, then check. This five-step routine takes about thirty seconds and prevents most lost marks. If a calculator is permitted, use it to verify each candidate solution numerically.
考试时，先写出定义域条件，再合并对数，然后化为指数形式，接着求解，最后检验。这套五步流程只需约三十秒，可避免大部分失分。若允许使用计算器，可用它数值验证每个候选解。
x + 3 = 0. Fourth, solve ln(2x − 1) = 3, giving your answer to three significant figures.
练习如下。第一，求解 log₄ (3x − 2) = 2。第二，求解 log₂ x + log₂ (x + 6) = 4。第三，求解 (log₃ x)² − 4 log₃ x + 3 = 0。第四，求解 ln(2x − 1) = 3，答案保留三位有效数字。
ny of these by a different but valid route, that is fine, as long as the domain check appears in your working.
答案：x = 6；x = 2（舍去 x = −8）；x = 3 或 x = 27；x ≈ 10.5。如果你用其他同样有效的方法得到这些结果也可以，只要在解题过程中写出了定义域检验。
Finally, remember that logarithmic equations are the natural bridge to differentiation and integration of logarithmic functions later in the course. The fluency you build in Exercise 6E.4 pays off directly in calculus, in differential equations, and in any topic where growth or decay must be quantified.
最后要记住，对数方程是后续课程中对数函数微分与积分的天然桥梁。你在练习6E.4 中培养的熟练度，会直接在微积分、微分方程以及任何需要量化增长或衰减的专题中带来回报。
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