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Exercise 4G: Exponential and Logarithmic Equations | 练习4G:指数与对数方程

2026-08-02 21:47 · 本文部分内容由AI辅助创作

📚 Exercise 4G: Exponential and Logarithmic Equations | 练习4G:指数与对数方程

In IB Mathematics, solving equations involving exponents and logarithms is a core skill that appears across both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Exercise 4G provides focused practice on manipulating exponential expressions, applying logarithm laws, and solving a variety of equations that model real-world growth and decay. Mastering these techniques is essential for success in the IB exam, where such equations are frequently combined with functions, graphs, and problem-solving contexts.

在IB数学中,求解涉及指数和对数的方程是一项核心技能,无论是在分析与方法(AA)还是应用与解释(AI)课程中都会出现。练习4G提供了针对性的训练,帮助你熟练变形指数表达式、运用对数运算法则,并求解各种模拟现实世界中增长与衰减的方程。掌握这些技巧对于在IB考试中取得成功至关重要,因为这类方程常常与函数、图像以及实际问题情境结合在一起考查。


1. Understanding Exponential Functions | 理解指数函数

An exponential function is of the form f(x) = a · bx, where a ≠ 0, b > 0, and b ≠ 1. The base b determines the growth (b > 1) or decay (0 < b < 1). In IB, the natural exponential function ex is especially important, and you must be comfortable moving between different bases.

指数函数的形式为 f(x) = a · bx,其中 a ≠ 0,b > 0 且 b ≠ 1。底数 b 决定了函数的增长(b > 1)或衰减(0 < b < 1)。在IB课程中,自然指数函数 ex 尤为重要,你必须能够灵活地在不同底数之间进行转换。


2. Graphs of Exponential Functions | 指数函数的图像

have a horizontal asymptote at y = 0 (when there is no vertical shift). The graph of y = ex is increasing and concave up, while y = e−x is decreasing. Sketching these graphs quickly will help you visualise solutions and check the reasonableness of your algebraic work.

所有指数函数的图像都经过点 (0, a),并且存在一条水平渐近线 y = 0(在没有垂直平移的情况下)。y = ex 的图像单调递增且凹向上,而 y = e−x 则单调递减。快速绘制这些图像能帮助你直观地看到解的情形,并检验代数求解结果的合理性。


3. Logarithmic Functions as Inverses | 作为反函数的对数函数

The logarithmic function y = logb x (b > 0, b ≠ 1, x > 0) is the inverse of the exponential function y = bx. This means logb (bx) = x and blogb x nverse relationship is key to solving equations where the unknown appears in the exponent.

对数函数 y = logb x(b > 0,b ≠ 1,x > 0)是指数函数 y = bx 的反函数。这意味着 logb (bx) = x 且 blogb x = x。理解这一反函数关系对于求解未知数出现在指数位置的方程至关重要。


4. Logarithm Laws and Properties | 对数运算法则与性质

You must memorise and apply the three core logarithm laws: logb(MN) = logb M + logb N, logb(M/N) = logb M − logb N, and logb(Mk) = k·logb M. Also remember that logb 1 = 0 and logb to condense or expand logarithmic expressions and to bring variables down from exponents.

你必须牢记并运用三条核心对数法则:logb(MN) = logb M + logb N,logb(M/N) = logb M − logb N,以及 logb(Mk) = k·logb M。同时要记住 logb 1 = 0 和 logb b = 1。这些法则用于压缩或展开对数表达式,并能把变量从指数位置上“拉下来”。


5. Change of Base Formula | 换底公式

When your GDC only has natural log (ln) or common log (log₁₀), the change of base formula is vital: logb a = (logc a) / (logc ₂ 5 = ln 5 / ln 2. This allows you to solve equations with any base using your calculator.

当你的图形计算器只有自然对数(ln)或常用对数(log₁₀)时,换底公式就非常重要:logb a = (logc a) / (logc b),其中 c 为任意不等于1的正数。在IB中,我们通常使用以10为底或以e为底。例如,log₂ 5 = ln 5 / ln 2。这样你就能用计算器求解任何底数的方程。

log₂ 8 = ln 8 / ln 2 = 3

log₂ 8 = ln 8 / ln 2 = 3


6. Solving Exponential Equations | 解指数方程

There are two main strategies for exponential equations. If both sides can be expressed with the same base, equate the exponents. For example, 32x = 27 can be written as 32x so 2x = 3, giving x = 1.5. When this is not possible, take the logarithm of both sides: 2x = 5 ⇒ ln(2x) = ln 5 ⇒ x ln 2 = ln 5 ⇒ x = (ln 5)/(ln 2).

求解指数方程主要有两种策略。如果两边可以写成同底数的幂,可以直接让指数相等。例如,32x = 27 可化为 32x = 3³,因此 2x = 3,解得 x = 1.5。当无法化为同底数时,对两边同时取对数:2x = 5 ⇒ ln(2x) = ln 5 ⇒ x ln 2 = ln 5 ⇒ x = (ln 5)/(ln 2)。


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7. Solving Logarithmic Equations | 解对数方程

ial form. For instance, solve log₂(x + 1) + log₂(x − 1) = 3. Combine: log₂[(x+1)(x−1)] = 3 ⇒ log₂(x² − 1) = 3. Rewrite as 2³ = x² − 1 ⇒ x² = 9 ⇒ 3. Always check the domain: x + 1 > 0 and x − 1 > 0, so x = 3 is the only valid solution.

对数方程常常需要先用法则合并对数,再转化为指数形式。例如,求解 log₂(x + 1) + log₂(x − 1) = 3。合并得 log₂[(x+1)(x−1)] = 3 ⇒ log₂(x² − 1) = 3。改写成指数形式为 2³ = x² − 1 ⇒ x² = 9 ⇒ x = 3。务必检查定义域:x + 1 > 0 且 x − 1 > 0,因此 x = 3 是唯一有效解。


8. Equations with Different Bases | 不同底数的方程

When an equation involves exponential terms with different bases that cannot be made identical, isolating one term and applying logarithms is usually best. For example, 3x+1 = 42x. Take ln: (x+1) ln 3 = 2x ln 4. Expand and rearrange: x ln 3 + ln 3 = 2x ln 4 ⇒ x(ln 3 − 2 ln 4) = −ln 3 ⇒ is type of algebraic manipulation is frequently tested in IB Paper 1 without a calculator.

当方程中出现不同底数的指数项且无法化为同底数时,通常最好的方法是分离一项后取对数。例如,3x+1 = 42x。取自然对数得:(x+1) ln 3 = 2x ln 4。展开并整理:x ln 3 + ln 3 = 2x ln 4 ⇒ x(ln 3 − 2 ln 4) = −ln 3 ⇒ x = (−ln 3) / (ln 3 − ln 16)。这类代数变形常在IB试卷1(无计算器部分)中考查。


9. Applications: Growth and Decay | 应用:增长与衰减

e population growth, radioactive decay, and compound interest. A typical model is A = A₀ ekt. You may need to find k given two data points, or determine the time when a quantity reaches a certain value. Always interpret the question carefully and state your answer in context with correct units.

指数方程和对数方程用于模拟现实中的情形,如人口增长、放射性衰变和复利计算。典型的模型是 A = A₀ ekt。你可能需要根据两个数据点求出 k,或确定数量达到某一特定值的时间。务必仔细审题,并在情境中用正确的单位表达答案。

A(t) = A₀ ekt (growth when k > 0, decay when k < 0)

A(t) = A₀ ekt(k > 0 为增长,k < 0 为衰减)


10. Using Technology: GDC Tips | 使用科技:图形计算器技巧

Your graphing calculator can solve exponential and logarithmic equations directly using the equation solver or by graphing. To solve 2x = x + 3, graph y = 2x and y = x + 3, then find the intersection. For numerical solving, use the "solve" s allowed in Papers 2 and 3, but you must show correct setup before relying on technology.

你的图形计算器可以利用方程求解功能或通过作图直接求解指数方程和对数方程。要解 2x = x + 3,可以画出 y = 2x 和 y = x + 3,然后求交点。对于数值求解,使用“solve”功能。记住,在试卷2和3中允许使用图形计算器,但你必须在依赖技术之前给出正确的方程设定。


11. Common Mistakes to Avoid | 常见错误及避免方法

ays verify your algebraic steps and check your final answers in the original equation. Below is a table of frequent errors.

学生经常忘记 log(x) 仅在 x 为正数时才有定义,从而引入增解。另一个陷阱是错误地使用法则,例如误以为 log(u + v) = log u + log v。务必验证每一步代数变形,并在原方程中检查最终答案。下面是一张常见错误表。

Common Mistake / 常见错误Correction / 正确做法
log(x+2)+log(x−1)=2 ⇒ (x+2)+(x−1)=10²Must multiply inside one log: log[(x+2)(x−1)] = 2 ⇒ (x+2)(x−1)=100
Squaring both sides of a log equation without checking domain.After solving, reject values that make any argument ≤ 0.
Using log(a − b) = log a − log b.Only log(a/b) = log a − log b, not subtraction inside.

12. Summary and Exam Tips | 总结与考试技巧

Exercise 4G builds the fluency you need to handle exponential and logarithmic equations confidently. In the exam, show clear working: state the logarithm law you are using, indicate when you are exponentiating both sides, and always include a domain check for log equations. With practice, you will not only solve these equations accurately but also apply them to challenging modelling problems.

练习4G帮助你培养流利地处理指数方程和对数方程的能力。在考试中,要展示清晰的解题过程:说明你正在使用哪条对数法则,明确何时对两边取指数,并且始终包含对数方程的定义域检查。通过练习,你不但能准确地求解这些方程,还能将它们应用到富有挑战性的建模问题中。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

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