📚 Exercise E.1: Exponential and Logarithmic Equations Practice | 练习 E.1：指数与对数方程练习
In IB Mathematics, Exercise E.1 provides essential practice in manipulating exponential and logarithmic expressions, solving equations, and applying these functions to real-world contexts. Mastery of this exercise ensures a solid foundation for calculus and modelling topics. This article breaks down the key concepts, offers step-by-step solutions to common problem types, and reinforces understanding through worked examples.
在 IB 数学中，练习 E.1 提供了对指数与对数表达式化简、方程求解以及实际应用考题的重要训练。掌握这组练习能为微积分与建模主题打下坚实基础。本文将分解核心概念，逐一详解典型问题，并通过例题加深理解。
1. What Exercise E.1 Actually Tests | 练习 E.1 究竟考查什么
Exercise E.1 sits inside the functions and algebra strand of IB Mathematics, for both Analysis and Approaches (AA) and Applications and Interpretation (AI). Its purpose is to build fluency with the two function families that describe repeated multiplication: exponential functions and their inverses, logarithms. Long before you meet calculus, you must be able to rewrite, combine and solve these expressions reliably.
练习 E.1 位于 IB 数学（无论是分析与方法 AA 还是应用与解释 AI）的函数与代数板块。它的目的是让你熟练处理描述"重复相乘"的两类函数：指数函数及其逆函数——对数。在你接触微积分之前，就必须能够可靠地改写、合并并求解这些表达式。
The exercise normally asks you to do four things: simplify an expression using the laws of logarithms; solve an equation where the unknown sits in an exponent; solve an equation where the unknown sits inside a logarithm; and interpret a growth, decay or scale model. Nearly every later topic - differentiating eˣ and ln x, integrating 1/x, geometric sequences, differential equations - depends on these four skills.
该练习通常要求你完成四件事：用对数法则化简表达式；求解未知数位于指数上的方程；求解未知数位于对数内部的方程；解释增长、衰减或标度模型。之后几乎每个主题——eˣ 与 ln x 的求导、1/x 的积分、等比数列、微分方程——都依赖这四项技能。
- Recognise the standard form f(x) = a·bˣ and read off the parameters a and b.
识别标准形式 f(x) = a·bˣ 并读出参数 a 与 b。 - Convert freely between y = logₐ(x) and aʸ = x.
在 y = logₐ(x) 与 aʸ = x 之间自由转换。 - Apply the product, quotient and power laws in both directions.
正向与反向运用积、商、幂三条法则。 - Check the domain of every logarithm you produce, and reject invalid roots.
检验你所得到的每个对数的定义域，并舍去不合法的根。
| Skill | 技能 | Typical command term | 常见指令词 | Where it reappears | 后续出现位置 |
| Manipulating logs | 化简对数 | Write as a single logarithm | 写成单个对数 | Integration, series | 积分、数列 |
| Solving exponential equations | 解指数方程 | Solve, giving your answer to 3 s.f. | 求解并保留三位有效数字 | Calculus Paper 1 and 2 | 微积分卷一、卷二 |
| Solving log equations | 解对数方程 | Find x, justifying your answer | 求 x 并说明理由 | Functions, proof | 函数、证明 |
| Modelling growth and decay | 增长与衰减建模 | Find the time taken for... | 求所需时间 | Paper 3 investigations | 卷三探究题 |
2. Exponential Functions Refresher | 指数函数复习
An exponential function takes the form f(x) = a·bˣ d b ≠ 1. The base b determines growth (b > 1) or decay (0 < b < 1). For instance, f(x) = 2ˣ grows rapidly as x increases, whereas g(x) = (1/2)ˣ decays toward zero.
指数函数形如 f(x) = a·bˣ，其中 a ≠ 0，b > 0 且 b ≠ 1。底数 b 决定增长 (b > 1) 或衰减 (0 < b < 1)。例如，f(x) = 2ˣ 随 x 增大而快速上升，而 g(x) = (1/2)ˣ 则逐渐趋近于零。
mptote at y = 0: the function approaches zero but never reaches it. Third, the function is one-to-one, which is exactly why it has an inverse - the logarithm.
有三个结构特征在每个题目中都很关键。第一，f(0) = a，故 y 轴截距即初始值。第二，图像有水平渐近线 y = 0：函数趋近于零但永不到达。第三，函数是一一对应的，这正解释了它为何存在逆函数——对数。
The natural exponential function with base e ≈ 2.71828 is ubiquitous: f(x) = eˣ to calculus, and it is the base used for continuous growth models in finance and physics.
以 e ≈ 2.71828 为底的自然指数函数处处可见：f(x) = eˣ。其导数等于自身，因此在微积分中地位核心，它也是金融与物理中连续增长模型所用的底数。
f(x) = a·bˣ, a ≠ 0, b > 0, b ≠ 1 ⇒ asymptote: y = 0
3. Logarithmic Functions as Inverses | 对数函数作为逆函数
The logarithm logₐ(x) answers a single question: "To what power must a be raised to get x?" Formally, y = logₐ(x) ⇔ aʸ on logarithm has base 10, written log(x); the natural logarithm has base e, written ln(x).
对数 logₐ(x) 回答一个问题："a 需要升到几次幂才能得到 x？"严格定义为 y = logₐ(x) ⇔ aʸ = x。例如 log₂(8) = 3，因为 2³ = 8。常用对数以 10 为底，写作 log(x)；自然对数以 e 为底，写作 ln(x)。
ain of a logarithmic function is x > 0, and its range is all real numbers. Graphs of y = bˣ ions often ask you to sketch both curves on the same axes and label the asymptote of each.
对数函数的定义域为 x > 0，值域为全体实数。y = bˣ 与 y = log_b(x) 的图像关于直线 y = x 对称。这种对称性值得记忆，因为考题常要求你在同一坐标系中画出两条曲线并标注各自的渐近线。
- logₐ(1) = 0 for every valid base a.
对任意合法底数 a，都有 logₐ(1) = 0。 - logₐ(a) = 1.
logₐ(a) = 1。 - logₐ(aˣ) = x and alogₐ x = x - the two cancellation identities.
logₐ(aˣ) = x 与 alogₐ x = x——两条"抵消"恒等式。 - The argument must be strictly positive: log of zero or of a negative number is undefined in real numbers.
真数必须严格为正：零或负数的对数在实数范围内无定义。
4. Laws of Logarithms | 对数运算法则
e laws, used in both directions, solve almost every manipulation question in Exercise E.1.
积法则：logₐ(xy) = logₐ(x) + logₐ(y)。商法则：logₐ(x/y) = logₐ(x) - logₐ(y)。幂法则：logₐ(xⁿ) = n·logₐ(x)。这三条法则，正向与反向使用，几乎能解决练习 E.1 中所有化简题。
Reading the laws backwards is the harder skill and the one examiners reward. Going from logₐ(x) + logₐ(y) to logₐ(xy) is called "writing as a single logarithm", and it is the gateway to solving logarithmic equations.
反向读法更难，也更容易得分。从 logₐ(x) + logₐ(y) 得到 logₐ(xy) 被称为"写成单个对数"，这是解对数方程的必经之门。
xy) = logₐx + logₐy logₐ(x/y) = logₐx − logₐy logₐ(xⁿ) = n·logₐx
A frequent trap: logₐ(x + y) cannot be simplified. The product rule applies to a product inside the bracket, never to a sum. Likewise, (logₐ x)² is not 2·logₐ x, and logₐ(x)/logₐ(y) is not logₐ(x − y).
一个常见陷阱：logₐ(x + y) 无法化简。积法则只作用于括号内的乘积，绝不作用于和。同理，(logₐ x)² 不等于 2·logₐ x，而 logₐ(x)/logₐ(y) 也不等于 logₐ(x − y)。
5. Solving Exponential Equations | 解指数方程
There are three standard strategies. Choose the one that matches the structure of the equation, because the marks are awarded for the method shown, not only for the final number.
解指数方程有三种标准策略。请根据方程结构选择合适方法，因为分数是给所展示的方法的，而不只是给最终数字。
- Same base method | 同底法: if both sides can be written as powers of the same base, equate the exponents.
若两边都能写成同底幂，则令指数相等。 - Logarithm method | 取对数法: if the bases differ and cannot be matched, take log or ln of both sides.
若底数不同且无法统一，对两边取 log 或 ln。 - Substitution method | 换元法: if the equation is quadratic in bˣ, let u = bˣ and solve for u first.
若方程关于 bˣ 是二次型，先令 u = bˣ 解出 u。
Example 1 - same base. Solve 32x − 1 = 27. Since 27 = 3³, we get 2x − 1 = 3, hence x = 2.
例 1——同底。解 32x − 1 = 27。因 27 = 3³，得 2x − 1 = 3，故 x = 2。
Example 2 - different bases. Solve 3x + 1 = 52x (ln 3 − 2ln 5) = −ln 3 and x = ln 3 / (2ln 5 − ln 3) ≈ 0.518 to three significant figures.
例 2——异底。解 3x + 1 = 52x。取自然对数：(x + 1)ln 3 = 2x·ln 5，即 x·ln 3 + ln 3 = 2x·ln 5，得 x(ln 3 − 2ln 5) = −ln 3，故 x = ln 3 / (2ln 5 − ln 3) ≈ 0.518（三位有效数字）。
Example 3 - hidden quadratic. Solve 4ˣ − 5·2ˣ + 4 = 0. Notice 4ˣ = (2ˣ)². Let u = 2ˣ, so u² − 5u + 4 = 0 and (u − 1)(u − 4) = 0. Hence 2ˣ = 1 giving x = 0, or 2ˣ = 4 giving x = 2. Both roots are positive real solutions because 2ˣ is always defined.
例 3——隐藏的二次型。解 4ˣ − 5·2ˣ + 4 = 0。注意 4ˣ = (2ˣ)²。令 u = 2ˣ，则 u² − 5u + 4 = 0，(u − 1)(u − 4) = 0。故 2ˣ = 1 得 x = 0，或 2ˣ = 4 得 x = 2
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