📚 Exercise G.1: Laws of Exponents and Logarithms | 指数与对数法则练习
Exponents and logarithms form the backbone of many advanced mathematical topics in the IB Mathematics curriculum. Exercise G.1 is designed to build your confidence in applying the fundamental rules of indices and logs, both algebraically and in real-world contexts. Mastering these laws will prepare you for solving exponential equations, modelling growth and decay, and tackling calculus involving natural logarithms.
指数与对数是IB数学课程中许多高级主题的基础。练习G.1旨在帮助同学们建立对指数和对数基本法则的代数推导与实际问题应用上的信心。熟练这些法则将为后续求解指数方程、建立增长与衰减模型以及处理涉及自然对数的微积分打下坚实的基础。
1. Review of Exponent Laws | 指数法则复习
ents: aᵐ ÷ aⁿ = aᵐ⁻ⁿ. A power raised to another power multiplies the exponents: (aᵐ)ⁿ = aᵐˣ extend to rational exponents, and they are the foundation upon which logarithms are built.
回想一下，对于任何非零实数 a 和整数 m, n, 同底数幂的乘法满足 aᵐ × aⁿ = aᵐ⁺ⁿ。除法时指数相减：aᵐ ÷ aⁿ = aᵐ⁻ⁿ。幂的乘方则将指数相乘：(aᵐ)ⁿ = aᵐˣⁿ。这些规则同样适用于有理指数，它们构成了对数运算的基础。
ple, simplify 2³ × 2⁴. Using aᵐ × aⁿ = aᵐ⁺ⁿ, we get 2³⁺⁴ = 2⁷ = 128. Similarly, (x²)³ = x²ˣ er many expressions requiring the fluent use of these laws before moving on to logarithms.
例如，化简 2³ × 2⁴，利用 aᵐ × aⁿ = aᵐ⁺ⁿ，得到 2³⁺⁴ = 2⁷ = 128。类似地，(x²)³ = x²ˣ³ = x⁶。在练习 G.1 中，你会遇到许多需要熟练运用这些法则进行化简的式子，然后才深入学习对数。
2. Zero and Negative Exponents | 零指数与负指数
istent with the subtraction of exponents, e.g. a³ ÷ a³ = a⁰ = 1, and a² ÷ a⁵ = a⁻³ = 1/a³.
任何非零底数的零次幂等于 1：a⁰ = 1 (a ≠ 0)。负指数则表示正指数幂的倒数：a⁻ⁿ = 1/(aⁿ)。这些定义与指数的减法运算是一致的，比如 a³ ÷ a³ = a⁰ = 1，而 a² ÷ a⁵ = a⁻³ = 1/a³。
ep in mind that a negative exponent does not make the value negative; it produces a fraction.
练习 G.1 通常会要求同学们将含有零指数或负指数的式子改写成不含这些指数的形式。例如，5⁻² = 1/5² = 1/25，而 (3x)⁰ = 1（x ≠ 0）。请记住，负指数并不代表这个数是负数，它产生的是一个分数。
3. Rational Exponents and Roots | 有理指数与根式
An exponent expressed as a fraction m/n combines a power and a root: am/n = (ⁿ√a)ᵐ = ⁿ√aᵐ. In particular, a1/n = ⁿ√a, and a1/2 = √a. This connection allows us to convert between radical notation and exponential form seamlessly, which is crucial in calculus and advanced algebra.
用分数 m/n 表示的指数同时包含了乘方和开方：am/n = (ⁿ√a)ᵐ = ⁿ√aᵐ。特别地，a1/n = ⁿ√a，a1/2 = √a。这种联系使我们能够在根式记法同指数形式之间自如转换，这在微积分和高等代数中尤为重要。
In Exercise G.1, you might be asked to evaluate 272/3. Rewrite as (³√27)² = 3² = 9. Or simplify √x⁴ as x4/2 to the domain is essential, because even roots of negative numbers are not real unless we enter the complex plane.
在练习 G.1 中，你可能会被要求计算 272/3。改写为 (³√27)² = 3² = 9。或者将 √x⁴ 化简为 x4/2 = x²，前提是 x ≥ 0。关注定义域非常重要，因为负数的偶次方根不是实数，除非我们进入复数域。
4. Introduction to Logarithms | 对数入门
A logarithm answers the question: 'To what exponent must we raise the base to obtain a given number?' If bˣ and y > 0. Common bases are 10 (common logarithm, often written simply as log) and e (natural logarithm, ln). Logarithms are the inverse operations of exponentiation.
对数所回答的问题是：“底数要乘方到哪个指数才能得到给定的数？” 如果 bˣ = y，那么 log_b(y) = x，其中 b > 0, b ≠ 1, y > 0。常见的底数有 10（常用对数，常简记为 log）和 e（自然对数，记作 ln）。对数运算是指数运算的逆运算。
. In Exercise G.1, you will translate between exponential and logarithmic forms, a skill that underpins the solution of many IB equations.
例如，2³ = 8，因此 log₂(8) = 3。同样，10² = 100，可得 log₁₀(100) = 2。在练习 G.1 中，你将反复在这两种形式间转换，这一技能是求解许多 IB 方程的基础。
5. Logarithm Laws | 对数运算法则
(M). These laws are valid for M, N > 0, and they enable the simplification of complex logarithmic expressions into sums, differences, and multiples of simpler logs.
由指数法则可以推出三条核心对数法则。积的法则：log_b(MN) = log_b(M) + log_b(N)。商的法则：log_b(M/N) = log_b(M) - log_b(N)。幂的法则：log_b(Mᵖ) = p·log_b(M)。这些法则对于 M, N > 0 成立，并且能够将复杂的对数式化简为简单对数的和、差与倍数。
not to misapply the product law to sums; log_b(M + N) is not equal to log_b(M) + log_b(N).
在练习 G.1 中你会综合运用这些法则。例如，利用法则展开 log₂(8x²/y)：log₂(8) + log₂(x²) - log₂(y) = 3 + 2log₂(x) - log₂(y)。注意不要将积的法则误用到加法上；log_b(M + N) 并不等于 log_b(M) + log_b(N)。
6. Change of Base Formula | 换底公式
arithm. A common choice is c = 10 or c = e, giving log_b(a) = log(a)/log(b) = ln(a)/ln(b).
大多数计算器只能计算以 10 或以 e 为底的对数。换底公式 log_b(a) = log_c(a) / log_c(b)（其中 c 为任意不等于 1 的正底数）使得我们可以计算任意底数的对数。通常取 c = 10 或 c = e，得到 log_b(a) = log(a)/log(b) = ln(a)/ln(b)。
o essential when solving exponential equations with unfamiliar bases, as it allows us to take logs of both sides and isolate the variable.
练习 G.1 中有类似求解 log₃(20) 的题目。使用以 10 为底：log₃(20) = log(20)/log(3) ≈ 1.3010/0.4771 ≈ 2.727。当求解带有陌生底数的指数方程时，换底公式同样不可或缺，因为它允许对方程两边取对数并分离出变量。
7. Solving Exponential Equations | 求解指数方程
s in the exponent, we often use logarithms to bring it down. For an equation of the form bˣ = k, take log (or ln) of both sides: log(bˣ log(b). If both sides can be expressed with the same base, we can simply equate exponents.
当变量出现在指数上时，我们通常利用对数将其“拿下来”。对于形如 bˣ = k 的方程，两边同时取对数（或自然对数）：log(bˣ) = log(k) → x·log(b) = log(k)，于是 x = log(k)/log(b)。如果两边能够写成同底数幂，那么只需令指数相等即可快速求解。
Example from Exercise G.1: Solve 5²ˣ⁺¹ = 125. Notice that 125 = 5³, so 5²ˣ⁺¹ = 5³ ⇒ 2x + 1 = 3 ⇒ x = 1. Alternatively, with unequal bases, e.g. 3ˣ = 7, we use x = ln(7)/ln(3) ≈ 1.771. Always check your answer by substituting back.
来自练习 G.1 的例子：解方程 5²ˣ⁺¹ = 125。注意到 125 = 5³，于是 5²ˣ⁺¹ = 5³ ⇒ 2x + 1 = 3 ⇒ x = 1。另一种情况是底数不同，例如 3ˣ = 7，则使用 x = ln(7)/ln(3) ≈ 1.771。务必代回原式验证答案。
8. Solving Logarithmic Equations | 求解对数方程
Logarithmic equations often involve one or more logarithmic terms. The key is to condense logs using the product and quotient laws into a single log, then rewrite in exponential form. Alternatively, if both sides have a log with the same base, equate the arguments. Always check that proposed solutions do not produce a log of a non-positive number.
对数方程通常包含一个或多个对数项。关键步骤是利用积和商的法则将对数合并成单个对数，然后改写为指数形式求解。另一种情形是，如果方程两边都是同底数的对数，则直接使真数相等。一定要检验得到的解是否导致对数中的真数为非正数，否则为增根。
, solve log₂(x) + log₂(x - 2) = 3. Combine: log₂[x(x - 2)] = 3. Rewrite: x(x - 2) = 2³ = 8 ⇒ x² - 2x - 8 = 0 ⇒ undefined, the only valid solution is x = 4. Exercises in G.1 reinforce this domain check.
例如，求解 log₂(x) + log₂(x - 2) = 3。合并：log₂[x(x - 2)] = 3。改写：x(x - 2) = 2³ = 8 ⇒ x² - 2x - 8 = 0 ⇒ (x - 4)(x + 2) = 0。得 x = 4 或 x = -2。因为 log₂(-2) 无定义，所以唯一有效解是 x = 4。G.1 中的练习充分强化了这种定义域的检查。
9. Natural Logarithms and e | 自然对数与 e
st be comfortable with the properties ln(e) = 1, ln(1) = 0, and the inverse relationship eln x = x for x > 0. Derivative rules later also rely heavily on ln.
自然对数 ln 以 e ≈ 2.71828 为底。数 e 在连续增长、微积分和复数中自然出现。在 IB 数学中，你必须熟知 ln(e) = 1, ln(1) = 0，以及对于 x > 0 的逆关系 eln x = x。后续导数法则也大量依赖 ln。
In Exercise G.1, you might be asked to solve e²ˣ = 5. Take natural log both sides: ln(e²ˣ) = ln(5) → 2x = ln(5) → x = ln(5)/2. Whenever you see an equation with ex, ln is the natural tool. The change of base formula also works with ln, so it is the preferred logarithm for many theoretical purposes.
在练习 G.1 中，你或许会碰到解 e²ˣ = 5。两边取自然对数：ln(e²ˣ) = ln(5) → 2x = ln(5) → x = ln(5)/2。看到含有 eˣ 的方程，ln 就是天然工具。换底公式在 ln 下同样成立，因此 ln 在理论推导中往往是首选的对数。
10. Applications in Compound Interest | 复利中的应用
ter t years with principal P, annual rate r, compounded n times per year is A = P(1 + r/n)nt. When interest is compounded continuously, the formula becomes A = Pert. Logarithms help solve for t or r.
指数函数和对数函数可以用来为金融场景建模，例如复利。本金 P 在年利率 r、每年计息 n 次的情况下，t 年后的积累金额为 A = P(1 + r/n)nt。当连续复利时，公式变为 A = Pert。对数可用来求解时间 t 或利率 r。
Typical Exercise G.1 application: How long will it take for 1000 to grow to 1500 at 5% compounded monthly? Set up 1500 = 1000(1 + 0.05/12)12t → 1.5 = (1 + 0.05/12)12t (1.5) / [12 ln(1 + 0.05/12)] ≈ 8.14 years. Logarithms make solving for exponents feasible.
练习 G.1 中的典型应用题：1000元以年利率5%每月复利，要多久才能增长到1500元？列方程：1500 = 1000(1 + 0.05/12)12t → 1.5 = (1 + 0.05/12)12t。取 ln：ln(1.5) = 12t · ln(1 + 0.05/12)。于是 t = ln(1.5) / [12 ln(1 + 0.05/12)] ≈ 8.14 年。对数让求解指数成为可能。
11. Common Mistakes to Avoid | 常见错误避免
One frequent error is confusing the negative exponent with a negative number. For instance, 2⁻³ is 0.125, not -8. Another pitfall is mishandling log properties: log(a + b) is not log a + log b, and log(a - b) is not log a / log b. Additionally, forgetting to check the domain of logarithmic functions can lead to extraneous solutions. Always ensure the argument is positive.
一个常见错误是混淆负指数与负数。例如，2⁻³ 等于 0.125，而不是 -8。另一个陷阱是对数性质的错误运用：log(a + b) 不等于 log a + log b，且 log(a - b) 也不等于 log a / log b。此外，忘记检验对数函数的定义域会导致增根。务必确保真数为正。
but log(p·M) ≠ p·log(M). Keep these distinctions in mind as you work through Exercise G.1.
在复利的情景中，同学们有时会错误解读计息频率，或忘记将年利率转化为周期利率。在代数变形时，使用幂法则也要小心：log(Mᵖ) = p·log(M) 是成立的，但 log(p·M) ≠ p·log(M)。在完成练习 G.1 的过程中，请牢记这些区别。
12. Practice Problems and Solutions | 练习题与解答
Let's work through a selection of problems mirroring those in Exercise G.1. Attempt each before reading the solution to strengthen your skills.
让我们来做一些与练习 G.1 中类似的题目，每道题先尝试自行解答再看解析，以此增强技能。
Problem 1: Simplify (3⁻² · 91/2) / 27⁻¹.
问题1： 化简 (3⁻² · 91/2) / 27⁻¹。
Solution: Write all bases as 3. 91/2 = (3²)1/2 ; 27⁻¹ = (3³)⁻¹ = 3⁻³. Numerator: 3⁻² · 3¹ = 3⁻¹. Division: 3⁻¹ / 3⁻³ = 3⁻¹⁻⁽⁻³⁾ = 3² = 9.
解答：将所有底数化为 3。91/2 = (3²)1/2 = 3¹；27⁻¹ = (3³)⁻¹ = 3⁻³。分子：3⁻² · 3¹ = 3⁻¹。除法：3⁻¹ / 3⁻³ = 3⁻¹⁻⁽⁻³⁾ = 3² = 9。
Problem 2: Express 2 log x - 3 log y + 1/2 log z as a single logarithm.
问题2： 将 2 log x - 3 log y + 1/2 log z 写成单个对数。
Solution: Use power rule first: log(x²) - log(y³) + log(z1/2). Combine: log( (x² · √z) / y³ ).
解答：先用幂法则：log(x²) - log(y³) + log(z1/2)。合并：log( (x² · √z) / y³ )。
Problem 3: Solve for x: 2x+2 - 2x = 24.
问题3： 求解 x：2x+2 - 2x = 24。
Solution: Factor 2x: 2x (2² - 1) = 24 → 2x (4 - 1) = 24 → 3·2x = 24 → 2x = 8 → x = 3.
解答：提取因子 2x：2x (2² - 1) = 24 → 2x (4 - 1) = 24 → 3·2x = 24 → 2x = 8 → x = 3。
Problem 4: If ln(p) = 2 and ln(q) = 5, evaluate ln(√e³p²/q).
问题4： 已知 ln(p) = 2，ln(q) = 5，求 ln(√e³p²/q)。
Solution: Simplify inside: √e³p²/q = (e³p²/q)1/2 ) = (1/2) [ln(e³) + ln(p²) - ln(q)] = (1/2)[3 + 2·2 - 5] = (1/2)[3 + 4 - 5] = (1/2)·2 = 1.
解答：化简内部：√e³p²/q = (e³p²/q)1/2。·) = (1/2) [ln(e³) + ln(p²) - ln(q)] = (1/2)[3 + 2·2 - 5] = (1/2)[3 + 4 - 5] = (1/2)·2 = 1。
These examples capture the essence of Exercise G.1. Consistent practice will make these manipulations second nature.
以上例子抓住了练习 G.1 的核心。持之以恒的练习将使这些运算变得得心应手。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图