📚 Exercise 7G.2 | 练习7G.2
Exercise 7G.2 in the IB Mathematics: Analysis and Approaches course challenges students to master exponential and logarithmic equations. This set of problems blends algebraic manipulation with the deep properties of exponents and logs, preparing you for both Paper 1 non-calculator work and real-world applications like half-life, population growth, and pH scales. In this article we break down the core techniques, highlight typical pitfalls, and provide worked solutions that mirror the style of IB exam questions.
IB数学分析与方法课程中的练习7G.2旨在让学生精通指数方程与对数方程的求解。这一组题目将代数操作与指数、对数的深层性质结合在一起，为你应对Paper 1非计算器试题以及半衰期、人口增长、pH值等实际应用做好准备。本文拆解核心技法，突出常见陷阱，并提供与IB考试风格相似的标准解答。
1. Review of Exponential Rules | 指数运算法则回顾
e exponents give reciprocals (a⁻ⁿ = 1/aⁿ), and fractional exponents denote roots (a¹/ⁿ = ⁿ√a sides of an equation with the same base, which is the first strategy for equations like 5²ˣ = 5ˣ⁺³.
在开始练习7G.2之前，务必要熟练运用三个基本法则：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，以及 (aᵐ)ⁿ = aᵐⁿ。负指数表示倒数（a⁻ⁿ = 1/aⁿ），分数指数表示方根（a¹/ⁿ = ⁿ√a）。这些法则使你能够将方程两边化为同底数，这是处理诸如5²ˣ = 5ˣ⁺³这类方程的首要策略。
at a⁰ = 1 for any non-zero a, and a¹ = a. In IB problems you often see expressions like 2²ˣ⁻¹ rewritten as 2²ˣ / 2 or 1/2 × 4ˣ. Fluency with these conversions reduces mistakes when solving for the unknown exponent.
另一条基本规则是：对于任意非零数a，a⁰ = 1，且a¹ = a。在IB问题中，你常会遇到将2²ˣ⁻¹改写为2²ˣ / 2 或 1/2 × 4ˣ的情形。熟练进行这类转换能减少求解未知指数时的错误。
2. Equating Exponents: Same Base Method | 同底数幂比较法
ress both sides of an equation as a power of the same positive base b, then the equation bᴬ = bᴮ implies A = B. For example, to solve 3²ˣ equently in the early parts of Exercise 7G.2, where numbers are chosen to work out neatly.
若能将方程两边表示为相同正底数b的幂，则方程bᴬ = bᴮ意味着A = B。例如，解3²ˣ⁻¹ = 81时，注意到81 = 3⁴，因此2x − 1 = 4，解得x = 2.5。此法常出现在练习7G.2的前半部分，题目中的数字会设计成恰好可化简的形式。
Watch out for compound bases: equations like 4ˣ⁺² = 8²ˣ you to express 4 and 8 as powers of 2. Since 4 = 2² and 8 = 2³, the equation becomes (2²)ˣ⁺² = (2³)²ˣ⁻¹ → 2²ˣ⁺⁴ = 2⁶ˣ = 6x − 3, so x = 7/4. IB markers expect you to show the base conversion steps explicitly.
注意底数为合数的情况：对于方程4ˣ⁺² = 8²ˣ⁻¹，需将4和8表示为2的幂。4 = 2²，8 = 2³，于是方程化为(2²)ˣ⁺² = (2³)²ˣ⁻¹ → 2²ˣ⁺⁴ = 2⁶ˣ⁻³。令指数相等得2x + 4 = 6x − 3，因此x = 7/4。IB阅卷人期望你将底数转换的步骤明确写出。
3. Introduction to Logarithms for Equation Solving | 用对数解方程入门
ced to the same base, logarithms become the tool of choice. Taking logs of both sides of aʸ a. Any base can be used, but IB students typically work with log base 10 (lg) or natural log (ln), as these are directly available on GDC calculators.
当方程两边无法化为同底数时，对数便成为首选工具。对aʸ = b两端取对数得y·log a = log b，因此y = log b / log a。可取任意底数，但IB学生通常使用以10为底的对数（lg）或自然对数（ln），因为图形计算器（GDC）直接提供这些功能。
A classic Exercise 7G.2 problem is 2ˣ g₂7. Always check whether the question asks for an exact value or a decimal approximation.
练习7G.2中的典型题目如2ˣ = 7。取自然对数：x·ln 2 = ln 7，得x = ln 7 / ln 2 ≈ 2.807。若要求精确值，IB通常需要保留对数形式，如x = log₂7。务必检查题目要求的是精确值还是近似小数。
4. Applying the Change-of-Base Formula | 换底公式的应用
= log₂x / log₂4 = log₂x / 2. The equation becomes (3/2)log₂x = 3, so log₂x = 2 and x = 4.
换底公式 logₐb = logₓb / logₓa 在比较对数或化简涉及不同底数的表达式时特别有用。在练习7G.2中，你可能会遇到log₂x + log₄x = 3这样的方程。将log₄x化为以2为底：log₄x = log₂x / log₂4 = log₂x / 2。方程变为(3/2)log₂x = 3，因此log₂x = 2，得x = 4。
Calculators use base 10 or e, so any log expression you input must be converted. For instance, to evaluate log₅0.2, type ln 0.2 / ln 5, or use the GDC's logBASE function. Practice this skill until it becomes automatic; IB exam questions on logarithmic equations notoriously test the change-of-base in multi-step contexts.
计算器采用以10或e为底，因此输入的任何对数表达式都须经过转换。例如，计算log₅0.2时，可输入ln 0.2 / ln 5，或使用GDC的logBASE功能。反复练习直至熟练；IB对数方程考题尤爱在多步骤情境下考查换底公式。
5. Solving Exponential Equations with Quadratics | 解含二次型的指数方程
ntial equations transform into quadratic equations via substitution. For example, solve e²ˣ − 5eˣ + 6 = 0. Let y = eˣ, then y² − 5y + 6 = 0, so y = 2 or y = 3. Back-substituting gives eˣ = 2 ⇒ x = ln 2, and eˣ = 3 ⇒ x = ln 3. Always verify that the solution for y is positive, as eˣ > 0 for all real x.
某些指数方程可通过代换化为二次方程。例如，解e²ˣ − 5eˣ + 6 = 0。设y = eˣ，则y² − 5y + 6 = 0，得y = 2 或 y = 3。回代得eˣ = 2 ⇒ x = ln 2，以及eˣ = 3 ⇒ x = ln 3。务必验证y的解为正，因为对所有实数x，eˣ > 0。
This method extends to bases other than e. Consider 3·2²ˣ − 7·2ˣ + 2 = 0. Set y = 2ˣ giving 3y² − 7y + 2 = 0, factorised to (3y − 1)(y − 2) = 0. Thus y = 1/3 or y = 2. From 2ˣ = 1/3 we obtain x = log₂(1/3) = −log₂3; from 2ˣ are valid. The idea of disguised quadratics is a favourite in Exercise 7G.2 and IB exams.
此法同样适用于e以外的底数。例如3·2²ˣ − 7·2ˣ + 2 = 0。设y = 2ˣ，得3y² − 7y + 2 = 0，因式分解为(3y − 1)(y − 2) = 0。于是y = 1/3或y = 2。从2ˣ = 1/3得x = log₂(1/3) = −log₂3；从2ˣ = 2得x = 1。由于1/3为正，两个解均成立。隐藏二次型的概念是练习7G.2和IB考试中的常见考点。
6. Solving Logarithmic Equations Using Properties | 利用对数性质解对数方程
Core log properties— = logₐM + logₐN; quotient rule: logₐ(M/N) = logₐM − logₐN; power rule: logₐ(Mᵖ) = p·logₐM— 2)(x−1) = 2³ = 8. Solve the quadratic x² + x − 2 = 8 → x² + x − 10 = 0, yielding x = (−1 ± √41)/2. Check domain: original logs require x+2 > 0 and x−1 > 0, so x > 1. Only the positive root (−1 + √41)/2 ≈ 2.70 passes the check.
核心对数性质——积的性质：logₐ(MN) = logₐM + logₐN；商的性质：logₐ(M/N) = logₐM − logₐN；幂的性质：logₐ(Mᵖ) = p·logₐM——是压缩或展开对数式的关键。练习7G.2中典型的指令是解log₂(x+2) + log₂(x−1) = 3。将左边合并：log₂[(x+2)(x−1)] = 3，于是(x+2)(x−1) = 2³ = 8。解二次方程 x² + x − 2 = 8 → x² + x − 10 = 0，得x = (−1 ± √41)/2。检验定义域：原对数要求x+2 > 0且x−1 > 0，故x > 1。只有正根(−1 + √41)/2 ≈ 2.70符合要求。
Domain restrictions are often overlooked by students. Whenever you solve a log equation, write the condition that the argument of each logarithm must be strictly positive before you even begin manipulating. In IB markschemes, failing to reject an extraneous solution can cost a mark.
定义域限制常被学生忽略。解对数方程时，应在开始变形之前，先将每个对数的真数必须严格为正的条件写出。在IB评分标准中，未能排除增根可能导致丢分。
7. Equations with Logarithms on Both Sides | 两端均含对数的方程
0 → (x−3)(x+1) = 0. Check domain: for ln(x²−3) we need x²−3>0, and for ln(2x) we need 2x>0 ⇒ x>0. x = 3 satisfies both, while x = −1 fails the 2x>0 condition, so it is rejected.
当方程形如logₐf(x) = logₐg(x)时，只要底数a相同，便可消去对数，因为对数函数是一一映射的。例如，ln(x² − 3) = ln(2x)。消去ln得x² − 3 = 2x → x² − 2x − 3 = 0 → (x−3)(x+1) = 0。检验定义域：对于ln(x²−3)需x²−3>0，对于ln(2x)需2x>0 ⇒ x>0。x = 3满足两者，而x = −1不满足2x>0的条件，故舍去。
ert everything to exponential form. Always consider whether using the definition logₐx = y ⇔ aʸ = x simplifies the problem more directly.
有时对数方程以不同底数出现。在练习7G.2中，一种好的策略是利用换底公式统一底数，或将所有项转化为指数形式。始终想一想，使用定义 logₐx = y ⇔ aʸ = x 是否能够更直接地简化问题。
8. Handling the Natural Exponential and Logarithm | 处理自然指数与自然对数
Equations involving eˣ and ln x are the most common transcendental equations in IB. The key relationship is ln(eˣ) = x for all real x, and eˡⁿˣ = x for x > 0. These inverse properties let you 'undo' operations. For example, solve 2eˣ⁻¹ = 10. First isolate eˣ⁻¹ = 5, then take ln: x − 1 = ln 5, so x = 1 + ln 5.
涉及eˣ和ln x的方程是IB中最常见的超越方程。关键关系是：对所有实数x有ln(eˣ) = x，且对x > 0有eˡⁿˣ = x。这些反函数性质能让你“撤销”运算。例如，解2eˣ⁻¹ = 10。先分离得eˣ⁻¹ = 5，然后取自然对数：x − 1 = ln 5，故x = 1 + ln 5。
When the equation contains both eˣ and ln, you often need to exponentiate or log both sides carefully. For instance, e²ˡⁿˣ = 16. Use e²ˡⁿˣ = (eˡⁿˣ is the only valid answer. The IB loves to hide these domain issues in compound functions.
当方程同时包含eˣ和ln时，常常需要谨慎地两边取指数或取对数。例如，e²ˡⁿˣ = 16。利用e²ˡⁿˣ = (eˡⁿˣ)² = x²，得x² = 16 → x = ±4。定义域检验：ln x要求x > 0，因此仅x = 4为有效解。IB喜欢将这类定义域问题隐藏在复合函数中。
9. Mixed Exponential-Logarithmic Word Problems | 指数-对数混合文字题
oblems, such as modelling temperature decay T = 20 + 80e⁻ᵏᵗ, or radioactive decay m = m₀e⁻ˡ 05) ≈ 32.19. IB expects you to interpret the answer in context (e.g., after 32.2 minutes).
练习7G.2常包含应用题，例如模拟温度衰减T = 20 + 80e⁻ᵏᵗ，或放射性衰变m = m₀e⁻ˡᵗ。任务是求常数或解出t。例如，某物质按m = 50e⁻⁰·⁰⁵ᵗ衰变，求m = 10时的t。设50e⁻⁰·⁰⁵ᵗ = 10 → e⁻⁰·⁰⁵ᵗ = 0.2 → −0.05t = ln 0.2 → t = ln 0.2 / (−0.05) ≈ 32.19。IB期望你结合上下文解释答案（如32.2分钟后）。
quation when solving for t. Taking logs is unavoidable. Practise rewriting statements like 'the population doubles every 5 years' as P = P₀·2ᵗ/⁵ before you solve.
其他问题涉及翻倍时间或半衰期。复利公式A = P(1 + r/n)ⁿᵗ 在求解t时转化为指数方程，取对数不可避免。在求解之前，练习将诸如“每5年人口翻倍”的表述改写为P = P₀·2ᵗ/⁵。
10. Common Mistakes and How to Avoid Them | 常见错误与规避方法
start. Mistake 3: Misapplying power rule: logₐ(x²) = 2logₐx holds for x>0; for x<0 you need 2logₐ|x|.
错误1：忘记logₐ(M + N) 不等于 logₐM + logₐN。学生常错误地拆分和式。正确的做法是，只有当真数内部为乘积时才能合并对数。错误2：在未检验定义域时消去对数，导致出现增根。务必一开始就声明并验证定义域。错误3：滥用幂的性质：logₐ(x²) = 2logₐx 仅在x>0时成立；当x<0时需要写成2logₐ|x|。
ding errors. Mistake 5: Ignoring the possibility that a quadratic substitution (like y = eˣ) might give a negative root which is invalid. In Exercise 7G.2, always state 'eˣ > 0, so y must be positive' and discard negative y.
错误4：在多步计算中过早取近似值。保留精确的对数表达式直至最后一步，以避免累积舍入误差。错误5：忽略二次代换（如y = eˣ）可能产生无效的负根。在练习7G.2中，始终声明“eˣ > 0，因此y必须为正”，并舍去负y值。
11. Exercise 7G.2 Sample Problems with Worked Steps | 练习7G.2例题与分步详解
Let us work through three typical questions that mirror the exercise. (i) Solve 5ˣ⁺¹ = 2ˣ (ln4 + ln5) / (ln5 − ln2) = −ln20 / ln(5/2). This is the exact form; approximate to −2.88.
让我们演练三道体现该练习风格的标准题。（i）解5ˣ⁺¹ = 2ˣ⁻²。取自然对数： 2 − ln5 → x(ln5 − ln2) = −(2 ln2 + ln5) → x = −(ln4 + ln5) / (ln5 − ln2) = −ln20 / ln(5/2)。此为精确形式；近似值为−2.88。
→ 1+6 = 3x − x → 2x = 7 → x = 3.5. Domain requires x+1>0 and x−2>0 → x>2. x=3.5 is valid.
（ii）解log₃(x+1) − log₃(x−2) = 1。合并：log₃[(x+1)/(x−2)] = 1 → (x+1)/(x−2) = 3¹ = 3。交叉相乘：x+1 = 3(x−2) = 3x − 6 → 1+6 = 3x − x → 2x = 7 → x = 3.5。定义域要求x+1>0且x−2>0 ⇒ x>2。x=3.5有效。
(iii) Solve 2²ˣ − 3·2ˣ⁺² + 32 = 0. Rewrite the middle term: 3·2ˣ⁺² = 3·2ˣ·2² = 12·2ˣ. Let y = 2ˣ, then 2²ˣ = (2ˣ)² = y². Equation becomes y² − 12y + 32 = 0 → (y−4)(y−8) = 0. So y = 4 or y = 8. Then 2ˣ = 4 ⇒ x = 2; 2ˣ = 8 ⇒ x = 3. Both are valid.
（iii）解2²ˣ − 3·2ˣ⁺² + 32 = 0。重写中间项：3·2ˣ⁺² = 3·2ˣ·2² = 12·2ˣ。设y = 2ˣ，则2²ˣ = (2ˣ)² = y²。方程变为y² − 12y + 32 = 0 → (y−4)(y−8) = 0。于是y = 4或y = 8。则2ˣ = 4 ⇒ x = 2；2ˣ = 8 ⇒ x = 3。两者均有效。
12. GDC Tips and Paper 1 Skills | 图形计算器技巧与Paper 1技能
nd the intersection. However, you must still be able to confirm domain and show reasoning.
虽然Paper 1禁止使用计算器，但其考查的是相同的代数技能。Paper 1中的对数方程常用2、3、5、10等标准底数，且数字设计成可得到精确整数或简单分数解。在Paper 2中，你可以使用GDC绘制Y1 = 左侧，Y2 = 右侧，然后求交点。但是，你仍需能够确认定义域并展示推理过程。
Use the GDC' linearise, but first isolate the exponential term if possible. For example, to solve 3e⁰·²ˣ − 7 = 5, rewrite as 3e⁰·²ˣ = 12 → e⁰·²ˣ h the window when using graphical methods; know the typical ranges for exponential growth.
对于不易线性化的指数方程，可使用GDC的方程求解器，但尽可能先分离指数项。例如，解3e⁰·²ˣ − 7 = 5，改写为3e⁰·²ˣ = 12 → e⁰·²ˣ = 4。然后可手动取自然对数或用原方程使用求解器。使用图形法时务必注意窗口设置；了解指数增长的一般取值范围。
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