IB Mathematics: Analysis & Approaches HL · 鼎睿学苑

Unit A2: Exponents
& Logarithms单元 A2：指数与对数

The second sub-unit of Topic 1 (Number & Algebra). Master the algebraic laws of exponents (integer, rational, real), the logarithm as the inverse of exponentiation, the change-of-base formula, solving exponential and logarithmic equations, and the standard growth/decay models that appear on every Paper 2.Topic 1（数与代数）的第二个子单元。掌握指数（exponent）的整数、分数、实数次幂运算律，对数作为指数运算的逆运算（inverse），换底公式（change-of-base formula），指数方程与对数方程的求解，以及每场 Paper 2 都会出现的标准指数增长与衰减模型（growth/decay model）。

IB AA HL · Topic 1.5, 1.7 Papers 1 · 2 6 Concepts6 个核心概念
Read me first先读这里

How to use this guide本指南使用说明

This guide is built to serve two students at once:本指南同时为两类学生设计：

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If you're cramming如果你在临阵磨枪

Read only the dashed-gold "Cram-Mode Cheat" box at the top of each section, plus the formula boxes. Skim one worked example per section. Skip the ▸ expandable details. Take the practice quiz. That's a 25-minute pass.

只看每节顶端的金色虚线 "Cram-Mode Cheat" 速记框和公式框。每节挑一道 worked example（例题）扫一眼，跳过 ▸ 折叠的"深入"部分，做一次练习测验。约 25 分钟过完一遍。

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If you're going for a 7如果你目标是 7 分

Read straight through. Open every ▸ Going deeper details block — that's where derivations and the "why $a^0 = 1$" justification live. Practice fluency with logarithm laws until you can manipulate $\log(xy)$, $\log(x/y)$, $\log(x^n)$ in either direction at speed. Re-do the quiz with no hints.

从头读到尾。打开每个 ▸ Going deeper 折叠块——推导和"为什么 $a^0 = 1$"的解释都在里面。把对数运算律练到滚瓜烂熟：$\log(xy)$、$\log(x/y)$、$\log(x^n)$ 双向都能秒切。再做一遍测验不看提示。

HL flagHL 标记说明 A2 content is shared between SL and HL — exponents and logarithms appear on both syllabi. The few HL-specific extensions (e.g. solving equations that reduce to a quadratic in $a^x$) are flagged inline with HL.A2 内容在 SL 与 HL 大纲中重合 —— 指数与对数两边都考。少数 HL 专属拓展（例如可化为 $a^x$ 的二次方程）会用 HL 标出。
Topic A2.1 · ExponentsTopic A2.1 · 指数

Integer Exponents整数指数 SL 1.5

Seven exponent laws cover every Paper 1 manipulation. For $a, b \ne 0$ and integers $m, n$: $$ a^{m}\,a^{n} = a^{m+n}, \quad \frac{a^{m}}{a^{n}} = a^{m-n}, \quad (a^{m})^{n} = a^{mn}, \quad (ab)^{n} = a^{n}b^{n}, $$ $$ \left(\tfrac{a}{b}\right)^{n} = \tfrac{a^{n}}{b^{n}}, \qquad a^{0} = 1, \qquad a^{-n} = \tfrac{1}{a^{n}}. $$ These hold for any integer $n$ (positive, negative, zero) and extend to rational and real $n$ once we redefine "exponentiation" via roots and limits.
七条指数运算律覆盖了 Paper 1 上所有指数化简题。对 $a, b \ne 0$、整数 $m, n$： $$ a^{m}\,a^{n} = a^{m+n}, \quad \frac{a^{m}}{a^{n}} = a^{m-n}, \quad (a^{m})^{n} = a^{mn}, \quad (ab)^{n} = a^{n}b^{n}, $$ $$ \left(\tfrac{a}{b}\right)^{n} = \tfrac{a^{n}}{b^{n}}, \qquad a^{0} = 1, \qquad a^{-n} = \tfrac{1}{a^{n}}. $$ 这些对任意整数 $n$（正、负、零）均成立；通过根式与极限的延拓，可推广到分数指数与实数指数。
Definition For a positive integer $n$, $a^{n}$ is defined as repeated multiplication: $a^{n} = \underbrace{a \cdot a \cdots a}_{n\text{ factors}}$. The definition is extended to $n = 0$ by setting $a^{0} = 1$ (for $a \ne 0$), and to negative integers $n = -k$ by $a^{-k} = 1/a^{k}$. These extensions are forced by the requirement that the product rule $a^{m}a^{n} = a^{m+n}$ remain valid across all integer exponents.
定义 对正整数 $n$，$a^{n}$ 定义为反复相乘：$a^{n} = \underbrace{a \cdot a \cdots a}_{n\text{ 个因子}}$。再令 $a^{0} = 1$（$a \ne 0$），$a^{-k} = 1/a^{k}$（负整数）将定义推广。这些推广不是凭空规定，而是为了让乘积律 $a^{m}a^{n} = a^{m+n}$ 在所有整数指数上仍成立。
▸ Going deeper — why $a^{0} = 1$ and $a^{-n} = 1/a^{n}$▸ 深入 —— 为什么 $a^{0} = 1$ 与 $a^{-n} = 1/a^{n}$

Start from the product rule $a^{m}a^{n} = a^{m+n}$, which is obvious for positive integer exponents (just count factors). Apply it with $n = 0$:

从对正整数显然成立的乘积律 $a^{m}a^{n} = a^{m+n}$ 出发（数因子即可证明）。取 $n = 0$：

$$ a^{m} \cdot a^{0} \;=\; a^{m+0} \;=\; a^{m}. $$

Dividing both sides by $a^{m}$ (which requires $a \ne 0$) gives $a^{0} = 1$. Then apply the rule with $m + n = 0$, i.e. $n = -m$:

两边除以 $a^{m}$（要求 $a \ne 0$）得 $a^{0} = 1$。再取 $m + n = 0$，即 $n = -m$：

$$ a^{m} \cdot a^{-m} \;=\; a^{0} \;=\; 1 \quad \Longrightarrow \quad a^{-m} \;=\; \tfrac{1}{a^{m}}. $$

So both definitions are forced — they're the unique extensions of the integer-power definition that preserve the product rule. This is the standard "extension by consistency" argument that runs through all of A2 (and reappears in A4 for $i^{0}$, $i^{-1}$).

所以这两个定义是被迫的 —— 它们是保持乘积律成立的唯一推广。这就是贯穿整个 A2 的"按一致性延拓"思路（A4 中关于 $i^{0}$、$i^{-1}$ 的处理也是同一个套路）。

Worked Example A2.1 — Simplify with multiple lawsA2.1 例题 —— 综合运用指数律

Simplify $\dfrac{(2x^{3})^{4} \cdot x^{-2}}{x^{5}}$ for $x \ne 0$.化简 $\dfrac{(2x^{3})^{4} \cdot x^{-2}}{x^{5}}$（$x \ne 0$）。

Set up. Apply $(ab)^{n} = a^{n}b^{n}$ to the numerator first:

列式。先对分子使用 $(ab)^{n} = a^{n}b^{n}$：

$$ (2x^{3})^{4} \;=\; 2^{4}\,(x^{3})^{4} \;=\; 16\,x^{12}. $$

Combine $x$-powers.

合并 $x$ 的幂次。

$$ \frac{16\,x^{12} \cdot x^{-2}}{x^{5}} \;=\; \frac{16\,x^{12 + (-2)}}{x^{5}} \;=\; \frac{16\,x^{10}}{x^{5}} \;=\; 16\,x^{10 - 5} \;=\; 16\,x^{5}. $$

Answer. $\boxed{16\,x^{5}}$.

答案。$\boxed{16\,x^{5}}$。

Simplify $\dfrac{a^{5} \cdot a^{-3}}{a^{-4}}$ for $a \ne 0$.化简 $\dfrac{a^{5} \cdot a^{-3}}{a^{-4}}$（$a \ne 0$）。
A2.1 · Q1
$a^{-2}$
$a^{4}$
$a^{6}$
$a^{12}$
Numerator: $a^{5+(-3)} = a^{2}$. Then $a^{2} / a^{-4} = a^{2-(-4)} = a^{6}$.分子：$a^{5+(-3)} = a^{2}$。再 $a^{2} / a^{-4} = a^{2-(-4)} = a^{6}$。
Trap: subtracting a negative exponent adds to the exponent. $a^{2}/a^{-4} = a^{2-(-4)} = a^{6}$.陷阱：除以负指数等价于加该指数的绝对值。$a^{2}/a^{-4} = a^{2-(-4)} = a^{6}$。
Topic A2.2 · Roots & Rational ExponentsTopic A2.2 · 根式与分数指数

Rational Exponents & Radicals分数指数与根式 SL 1.5

Rational exponents are roots. For $a > 0$ and positive integer $n$: $$ a^{1/n} \;=\; \sqrt[n]{a}, \qquad a^{m/n} \;=\; \sqrt[n]{a^{m}} \;=\; \bigl(\sqrt[n]{a}\bigr)^{m}. $$ Combine with $a^{-k} = 1/a^{k}$ for negative rationals. All seven integer laws extend unchanged. Restriction: $a > 0$ avoids ambiguity (e.g. $(-1)^{1/2}$ is not real).
分数指数就是根式（radical）。对 $a > 0$、正整数 $n$： $$ a^{1/n} \;=\; \sqrt[n]{a}, \qquad a^{m/n} \;=\; \sqrt[n]{a^{m}} \;=\; \bigl(\sqrt[n]{a}\bigr)^{m}. $$ 负分数指数再配上 $a^{-k} = 1/a^{k}$。所有七条整数指数律原样适用。限制：取 $a > 0$ 以避免歧义（例如 $(-1)^{1/2}$ 在实数中无定义）。
Why $a^{1/n} = \sqrt[n]{a}$ The defining property of $\sqrt[n]{a}$ is "the unique positive number whose $n$-th power is $a$." For the exponent definition to extend consistently — preserving $(a^{m})^{n} = a^{mn}$ — we need $a^{1/n}$ to satisfy $(a^{1/n})^{n} = a^{(1/n) \cdot n} = a^{1} = a$. Only one positive number does that: $\sqrt[n]{a}$.
为何 $a^{1/n} = \sqrt[n]{a}$ $\sqrt[n]{a}$ 的定义性质是"$n$ 次幂等于 $a$ 的唯一正数"。要让指数律 $(a^{m})^{n} = a^{mn}$ 在分数情形仍成立，必须 $(a^{1/n})^{n} = a^{(1/n) \cdot n} = a^{1} = a$。能满足这一条件的唯一正数就是 $\sqrt[n]{a}$。
Worked Example A2.2 — Rational exponentsA2.2 例题 —— 分数指数

Evaluate $8^{2/3}$, $16^{-3/4}$, and simplify $x^{1/2} \cdot x^{1/3}$.求 $8^{2/3}$、$16^{-3/4}$，并化简 $x^{1/2} \cdot x^{1/3}$。

(a) $8^{2/3}$. Read as "cube root, then square" (or square then cube root — both yield the same value):

(a) $8^{2/3}$。读作"先开三次方再平方"（或反向都可）：

$$ 8^{2/3} \;=\; \bigl(\sqrt[3]{8}\bigr)^{2} \;=\; 2^{2} \;=\; 4. $$

(b) $16^{-3/4}$. Combine the negative sign and the rational exponent:

(b) $16^{-3/4}$。把负号和分数指数合并：

$$ 16^{-3/4} \;=\; \frac{1}{16^{3/4}} \;=\; \frac{1}{(\sqrt[4]{16})^{3}} \;=\; \frac{1}{2^{3}} \;=\; \frac{1}{8}. $$

(c) $x^{1/2} \cdot x^{1/3}$. Add the exponents with a common denominator:

(c) $x^{1/2} \cdot x^{1/3}$。通分加幂次：

$$ x^{1/2}\,x^{1/3} \;=\; x^{1/2 + 1/3} \;=\; x^{3/6 + 2/6} \;=\; x^{5/6} \;=\; \sqrt[6]{x^{5}}. $$
Simplify $27^{-2/3}$.化简 $27^{-2/3}$。
A2.2 · Q1
$-9$
$\tfrac{1}{9}$
$9$
$-\tfrac{1}{9}$
$27^{-2/3} = 1/27^{2/3} = 1/(\sqrt[3]{27})^{2} = 1/3^{2} = 1/9$.$27^{-2/3} = 1/27^{2/3} = 1/(\sqrt[3]{27})^{2} = 1/3^{2} = 1/9$。
A negative exponent gives a reciprocal, not a negative number. $27^{-2/3} = 1/9$.负指数表示倒数，不是负数。$27^{-2/3} = 1/9$。
Topic A2.3 · LogarithmsTopic A2.3 · 对数

Logarithm Definition & Laws对数的定义与运算律 SL 1.7

A logarithm is "what exponent did I raise the base to?" For $a > 0$, $a \ne 1$, $b > 0$: $$ \log_{a}(b) = c \quad \Longleftrightarrow \quad a^{c} = b. $$ Three laws (mirroring the exponent product/quotient/power laws): $$ \log_{a}(xy) = \log_{a}x + \log_{a}y, \quad \log_{a}\!\left(\tfrac{x}{y}\right) = \log_{a}x - \log_{a}y, \quad \log_{a}(x^{n}) = n\log_{a}x. $$ Plus the identity slots: $\log_{a}(1) = 0$, $\log_{a}(a) = 1$. Two reserved bases: $\log = \log_{10}$ (common), $\ln = \log_{e}$ (natural).
对数（logarithm）就是"底数被取到了几次方？"对 $a > 0$、$a \ne 1$、$b > 0$： $$ \log_{a}(b) = c \quad \Longleftrightarrow \quad a^{c} = b. $$ 三条运算律（对应指数的积、商、幂律）： $$ \log_{a}(xy) = \log_{a}x + \log_{a}y, \quad \log_{a}\!\left(\tfrac{x}{y}\right) = \log_{a}x - \log_{a}y, \quad \log_{a}(x^{n}) = n\log_{a}x. $$ 两个恒等位置：$\log_{a}(1) = 0$、$\log_{a}(a) = 1$。两种保留底数：$\log = \log_{10}$（常用对数）、$\ln = \log_{e}$（自然对数）。
Definition For a positive base $a$ with $a \ne 1$, the logarithm base $a$ of a positive real $b$ is the unique real $c$ such that $a^{c} = b$. Symbolically, $c = \log_{a}(b)$ iff $a^{c} = b$. The logarithm is the inverse function of the exponential function $x \mapsto a^{x}$; the two cancel: $$ a^{\log_{a}(b)} = b, \qquad \log_{a}(a^{x}) = x. $$
定义 对正底数 $a$（$a \ne 1$），正实数 $b$ 关于底 $a$ 的对数是使 $a^{c} = b$ 成立的唯一实数 $c$。符号上，$c = \log_{a}(b) \Longleftrightarrow a^{c} = b$。对数是指数函数 $x \mapsto a^{x}$ 的反函数（inverse function），两者相互抵消： $$ a^{\log_{a}(b)} = b, \qquad \log_{a}(a^{x}) = x. $$
▸ Going deeper — proof of the product rule▸ 深入 —— 乘积律的证明

Set $\log_{a}(x) = p$ and $\log_{a}(y) = q$. By definition, $a^{p} = x$ and $a^{q} = y$. Multiply:

设 $\log_{a}(x) = p$、$\log_{a}(y) = q$。由定义，$a^{p} = x$、$a^{q} = y$。相乘：

$$ xy \;=\; a^{p} \cdot a^{q} \;=\; a^{p+q} \quad \text{(exponent product rule)}. $$

Read off the logarithm of the LHS: $\log_{a}(xy) = p + q = \log_{a}(x) + \log_{a}(y)$. $\square$

左边取对数：$\log_{a}(xy) = p + q = \log_{a}(x) + \log_{a}(y)$。 $\square$

The quotient rule comes from the same algebra with $x/y = a^{p}/a^{q} = a^{p-q}$. The power rule comes from $x^{n} = (a^{p})^{n} = a^{pn}$.

商律由 $x/y = a^{p}/a^{q} = a^{p-q}$ 同理可得；幂律由 $x^{n} = (a^{p})^{n} = a^{pn}$ 得到。

Worked Example A2.3 — Compute and combineA2.3 例题 —— 计算与合并

(a) Evaluate $\log_{2}(32)$. (b) Simplify $\log_{10}(50) + \log_{10}(2)$. (c) Expand $\ln\!\left(\dfrac{x^{2}\,y}{z}\right)$ into a sum of single-variable logs.(a) 求 $\log_{2}(32)$。(b) 化简 $\log_{10}(50) + \log_{10}(2)$。(c) 将 $\ln\!\left(\dfrac{x^{2}\,y}{z}\right)$ 展开为单变量对数之和。

(a) Ask: $2^{?} = 32$? Since $2^{5} = 32$, $\log_{2}(32) = 5$.

(a) 反问：$2^{?} = 32$？因 $2^{5} = 32$，故 $\log_{2}(32) = 5$。

(b) Combine using the product rule:

(b) 用积律合并：

$$ \log_{10}(50) + \log_{10}(2) \;=\; \log_{10}(50 \cdot 2) \;=\; \log_{10}(100) \;=\; 2. $$

(c) Split, then power-down:

(c) 先拆，再降幂：

$$ \ln\!\left(\tfrac{x^{2}\,y}{z}\right) \;=\; \ln(x^{2}) + \ln(y) - \ln(z) \;=\; 2\ln(x) + \ln(y) - \ln(z). $$
If $\log_{a}(2) = p$ and $\log_{a}(3) = q$, express $\log_{a}(72)$ in terms of $p, q$.已知 $\log_{a}(2) = p$、$\log_{a}(3) = q$，将 $\log_{a}(72)$ 用 $p, q$ 表示。
A2.3 · Q1
$3p + 2q$
$2p + 3q$
$p + q$
$6pq$
$72 = 8 \cdot 9 = 2^{3} \cdot 3^{2}$, so $\log_{a}(72) = 3\log_{a}(2) + 2\log_{a}(3) = 3p + 2q$.$72 = 8 \cdot 9 = 2^{3} \cdot 3^{2}$，故 $\log_{a}(72) = 3\log_{a}(2) + 2\log_{a}(3) = 3p + 2q$。
Factor $72 = 2^{3} \cdot 3^{2}$, then use power + product rules: $\log_{a}(72) = 3p + 2q$.先分解 $72 = 2^{3} \cdot 3^{2}$，再用幂律和积律：$\log_{a}(72) = 3p + 2q$。
Topic A2.4 · Change of BaseTopic A2.4 · 换底

Change of Base换底公式 SL 1.7

Convert any logarithm to any other base in one step. $$ \log_{a}(x) \;=\; \frac{\log_{b}(x)}{\log_{b}(a)}, \qquad a, b > 0,\; a, b \ne 1,\; x > 0. $$ On the GDC, take $b = 10$ or $b = e$ to compute any $\log_{a}(x)$ that doesn't fall out by inspection.
任意对数都可一步换到其它底。 $$ \log_{a}(x) \;=\; \frac{\log_{b}(x)}{\log_{b}(a)}, \qquad a, b > 0,\; a, b \ne 1,\; x > 0. $$ 用 GDC 时取 $b = 10$ 或 $b = e$，就能算出任何 $\log_{a}(x)$（一眼看不出来的也行）。
▸ Going deeper — derivation of change of base▸ 深入 —— 换底公式的推导

Set $c = \log_{a}(x)$, so $a^{c} = x$. Take $\log_{b}$ of both sides:

设 $c = \log_{a}(x)$，故 $a^{c} = x$。两边取 $\log_{b}$：

$$ \log_{b}(a^{c}) \;=\; \log_{b}(x) \quad \Longrightarrow \quad c\,\log_{b}(a) \;=\; \log_{b}(x). $$

Solving for $c$:

解出 $c$：

$$ c \;=\; \frac{\log_{b}(x)}{\log_{b}(a)} \quad \Longrightarrow \quad \log_{a}(x) \;=\; \frac{\log_{b}(x)}{\log_{b}(a)}. \qquad \square $$
Worked Example A2.4 — GDC-friendly computationA2.4 例题 —— 用 GDC 计算

Compute $\log_{2}(7)$ to 4 decimal places using the change-of-base formula and natural logarithms.用换底公式与自然对数（natural log），将 $\log_{2}(7)$ 保留 4 位小数。

Set up. Apply with $b = e$:

列式。取 $b = e$ 代入：

$$ \log_{2}(7) \;=\; \frac{\ln(7)}{\ln(2)}. $$

Compute. $\ln(7) \approx 1.94591$, $\ln(2) \approx 0.69315$, so

计算。$\ln(7) \approx 1.94591$，$\ln(2) \approx 0.69315$，故

$$ \log_{2}(7) \;\approx\; \frac{1.94591}{0.69315} \;\approx\; 2.8074. $$

Sanity check. $2^{2} = 4 < 7 < 8 = 2^{3}$, so $\log_{2}(7)$ should lie between $2$ and $3$. ✓

自我验证。$2^{2} = 4 < 7 < 8 = 2^{3}$，故 $\log_{2}(7)$ 应在 $2$ 与 $3$ 之间。 ✓

Using change of base, $\log_{5}(125) = ?$用换底公式，$\log_{5}(125) = ?$
A2.4 · Q1
$\tfrac{1}{3}$
$2$
$3$
$\log_{10}(625)$
$125 = 5^{3}$, so $\log_{5}(125) = 3$ directly. Change-of-base check: $\log(125)/\log(5) = 3\log(5)/\log(5) = 3$. ✓$125 = 5^{3}$，故 $\log_{5}(125) = 3$。换底验证：$\log(125)/\log(5) = 3\log(5)/\log(5) = 3$。 ✓
$125 = 5^{3}$, so $\log_{5}(125) = 3$.$125 = 5^{3}$，故 $\log_{5}(125) = 3$。
Topic A2.5 · Exponential & Log EquationsTopic A2.5 · 指对数方程

Solving Exponential & Logarithmic Equations指数方程与对数方程的求解 SL 1.7

Two reciprocal techniques.
  • Exponential equation ($a^{x} = b$): take $\log$ of both sides — typically $\log_{10}$ or $\ln$. Then $x = \log_{a}(b) = \log(b)/\log(a)$.
  • Logarithmic equation ($\log_{a}(x) = c$): exponentiate both sides — $x = a^{c}$.
Always check domain. Logs require positive arguments — discard solutions that make any $\log(\cdot)$ argument $\le 0$.
两种互逆技巧。
  • 指数方程 $a^{x} = b$：两边取对数（一般取 $\log_{10}$ 或 $\ln$），得 $x = \log_{a}(b) = \log(b)/\log(a)$。
  • 对数方程 $\log_{a}(x) = c$：两边取指数 —— $x = a^{c}$。
务必检验定义域。对数要求自变量为正 —— 解出后凡使任何 $\log(\cdot)$ 自变量非正的解都要舍去（extraneous root）。
Worked Example A2.5a — Exponential equationA2.5a 例题 —— 指数方程

Solve $2^{x+1} = 5$ for $x$ exactly and to 4 decimal places.求 $2^{x+1} = 5$，精确解与 4 位小数近似各给一份。

Take $\log$ of both sides.

两边取对数。

$$ (x+1)\log_{10}(2) \;=\; \log_{10}(5). $$

Solve for $x$.

解 $x$。

$$ x \;=\; \frac{\log_{10}(5)}{\log_{10}(2)} - 1 \;=\; \log_{2}(5) - 1. $$

GDC value. $\log_{2}(5) \approx 2.3219$, so $x \approx 1.3219$.

GDC 数值。$\log_{2}(5) \approx 2.3219$，故 $x \approx 1.3219$。

Worked Example A2.5b — Logarithmic equation with domain checkA2.5b 例题 —— 含定义域检验的对数方程

Solve $\log_{10}(x) + \log_{10}(x - 3) = 1$.求解 $\log_{10}(x) + \log_{10}(x - 3) = 1$。

Combine via product rule.

用积律合并。

$$ \log_{10}\bigl(x(x-3)\bigr) \;=\; 1 \quad \Longrightarrow \quad x(x-3) \;=\; 10^{1} \;=\; 10. $$

Solve the quadratic.

解二次方程。

$$ x^{2} - 3x - 10 \;=\; 0 \quad \Longrightarrow \quad (x - 5)(x + 2) \;=\; 0 \quad \Longrightarrow \quad x = 5 \text{ or } x = -2. $$

Domain check. The original equation requires $x > 0$ AND $x - 3 > 0$, i.e. $x > 3$. The candidate $x = -2$ fails (negative), so reject. Only $x = 5$ is valid.

定义域检验。原方程要求 $x > 0$ 且 $x - 3 > 0$，即 $x > 3$。候选 $x = -2$ 为负数，舍去（extraneous）。唯一有效解为 $x = 5$。

Pitfall — extraneous roots Combining logs via $\log a + \log b = \log(ab)$ enlarges the implicit domain. The combined equation $\log(ab) = c$ only requires $ab > 0$ — which allows both $a, b$ positive AND both $a, b$ negative. The original equation needed each individually positive. So always check candidate solutions against the original equation's domain.
陷阱 —— 增根 用 $\log a + \log b = \log(ab)$ 合并对数时，定义域被扩大：合并后的方程只要 $ab > 0$，允许 $a, b$ 同正或同负；但原方程要求两者各自为正。所以候选解必须代回原方程的定义域里核对。
▸ Going deeper — quadratic in $a^{x}$ HL▸ 深入 —— 化为 $a^{x}$ 的二次方程 HL

Equations like $4^{x} - 5 \cdot 2^{x} + 4 = 0$ look exponential but reduce to a quadratic. Substitute $u = 2^{x}$ (so $4^{x} = (2^{2})^{x} = (2^{x})^{2} = u^{2}$):

$4^{x} - 5 \cdot 2^{x} + 4 = 0$ 这类方程看似指数方程，其实可化为二次。令 $u = 2^{x}$（注意 $4^{x} = (2^{2})^{x} = (2^{x})^{2} = u^{2}$）：

$$ u^{2} - 5u + 4 = 0 \;\Longrightarrow\; (u - 1)(u - 4) = 0 \;\Longrightarrow\; u = 1 \text{ or } u = 4. $$

Convert back: $2^{x} = 1 \Rightarrow x = 0$; $2^{x} = 4 \Rightarrow x = 2$. Both valid (no domain restriction on exponentials). This "hidden quadratic" pattern appears regularly on HL Paper 1 — spot it by looking for $a^{2x}$ and $a^{x}$ side-by-side.

代回：$2^{x} = 1 \Rightarrow x = 0$；$2^{x} = 4 \Rightarrow x = 2$。两解都合法（指数函数无定义域限制）。这种"隐藏二次"模式在 HL Paper 1 上常见 —— 看到 $a^{2x}$ 与 $a^{x}$ 同时出现就尝试此换元。

Solve $5^{x} = 12$ for $x$ in terms of $\ln$.求 $5^{x} = 12$ 用 $\ln$ 表达的解。
A2.5 · Q1
$x = \dfrac{\ln 12}{\ln 5}$
$x = \ln 12 - \ln 5$
$x = \ln\!\left(\tfrac{12}{5}\right)$
$x = 12 \ln 5$
Take $\ln$ of both sides: $x \ln 5 = \ln 12 \Rightarrow x = \ln 12 / \ln 5 = \log_{5}(12)$.两边取 $\ln$：$x \ln 5 = \ln 12 \Rightarrow x = \ln 12 / \ln 5 = \log_{5}(12)$。
Take $\ln$ of both sides: $x \ln 5 = \ln 12$, so $x = \ln 12 / \ln 5$. The "subtraction" answer would only be right if the equation were $5 \cdot x = 12$.两边取 $\ln$：$x \ln 5 = \ln 12$，故 $x = \ln 12 / \ln 5$。"相减"答案只在 $5 \cdot x = 12$ 时才对。
Topic A2.6 · Growth & DecayTopic A2.6 · 增长与衰减

Exponential Growth & Decay Models指数增长与衰减模型 SL 1.5 / 1.7

Two standard models. $$ \text{Continuous: } N(t) = N_{0}\,e^{kt}, \qquad \text{Discrete: } A(t) = P\,(1 + r/n)^{nt}. $$ For continuous: $k > 0$ growth, $k < 0$ decay. Half-life (decay): $t_{1/2} = \dfrac{\ln 2}{|k|}$. Doubling time (growth): $t_{2} = \dfrac{\ln 2}{k}$. For discrete compound interest: $P$ principal, $r$ annual rate, $n$ compoundings per year, $t$ years.
两个标准模型。 $$ \text{连续：} N(t) = N_{0}\,e^{kt}, \qquad \text{离散：} A(t) = P\,(1 + r/n)^{nt}. $$ 连续模型：$k > 0$ 增长，$k < 0$ 衰减。半衰期（half-life，衰减）：$t_{1/2} = \dfrac{\ln 2}{|k|}$。倍增时间（doubling time，增长）：$t_{2} = \dfrac{\ln 2}{k}$。 离散复利：$P$ 本金，$r$ 年利率，$n$ 每年计息次数，$t$ 年数。
Worked Example A2.6 — Carbon-14 datingA2.6 例题 —— 碳-14 测年

Carbon-14 has a half-life of $5730$ years. (a) Find the decay constant $k$. (b) A sample contains $35\%$ of its original C-14. Estimate its age to the nearest year.碳-14（Carbon-14）半衰期为 $5730$ 年。(a) 求衰减常数 $k$。(b) 某样本含原 C-14 的 $35\%$，估计样本年龄（精确到年）。

(a) Decay constant. At $t = t_{1/2} = 5730$, $N(t)/N_{0} = 1/2$. From the model $N(t) = N_{0}e^{kt}$:

(a) 衰减常数。$t = t_{1/2} = 5730$ 时 $N(t)/N_{0} = 1/2$。代入 $N(t) = N_{0}e^{kt}$：

$$ e^{5730\,k} \;=\; \tfrac{1}{2} \;\Longrightarrow\; 5730\,k \;=\; \ln(1/2) \;=\; -\ln 2 \;\Longrightarrow\; k \;=\; -\tfrac{\ln 2}{5730} \;\approx\; -1.210 \times 10^{-4}\text{ yr}^{-1}. $$

(b) Age estimate. Set $N(t)/N_{0} = 0.35$:

(b) 估算年龄。令 $N(t)/N_{0} = 0.35$：

$$ e^{kt} \;=\; 0.35 \;\Longrightarrow\; t \;=\; \frac{\ln(0.35)}{k} \;=\; \frac{\ln(0.35)}{-\ln(2)/5730} \;=\; -\frac{5730 \ln(0.35)}{\ln 2} \;\approx\; 8678\text{ years}. $$

Sanity check. $0.35$ is between $0.25$ and $0.5$, so the age should be between $5730$ (one half-life) and $11460$ (two half-lives). $8678$ falls in that range. ✓

自我验证。$0.35$ 介于 $0.25$ 与 $0.5$ 之间，故年龄应在 $5730$（一个半衰期）与 $11460$（两个半衰期）之间。$8678$ 落在该范围。 ✓

A population grows continuously at $5\%$ per year. The doubling time, to the nearest year, is:某人口按连续 $5\%$ 年增长率增长。倍增时间（精确到年）为：
A2.6 · Q1
$5$ years
$10$ years
$14$ years
$20$ years
$t_{2} = \ln(2)/k = \ln(2)/0.05 \approx 0.6931/0.05 \approx 13.86 \to 14$ years.$t_{2} = \ln(2)/k = \ln(2)/0.05 \approx 0.6931/0.05 \approx 13.86 \to 14$ 年。
Doubling time formula: $t_{2} = \ln(2)/k$. Here $k = 0.05$, so $t_{2} \approx 13.86 \to 14$ years. (The "rule of 70" approximates this: $70/5 = 14$.)倍增时间公式：$t_{2} = \ln(2)/k$。$k = 0.05$，故 $t_{2} \approx 13.86 \to 14$ 年。（"70 法则"给出近似：$70/5 = 14$。）
Exam Tactics考试技巧

Exam Strategy & Common Pitfalls考试策略与常见陷阱

Paper 1 — No calculatorPaper 1 —— 不可用计算器
  • Recognise tidy numbers. $\log_{2}(32) = 5$, $\log_{3}(81) = 4$, $\ln(e^{5}) = 5$ — these are gimme marks. Always check for clean integer arguments before reaching for change of base.
  • 识别"齐整"的数值。$\log_{2}(32) = 5$、$\log_{3}(81) = 4$、$\ln(e^{5}) = 5$ —— 这些是送分。换底公式之前，先看自变量能否凑成幂。
  • Use log laws to combine before evaluating. $\log(2) + \log(50) = \log(100) = 2$ is faster than computing two separate logs and adding.
  • 先用对数律合并再求值。$\log(2) + \log(50) = \log(100) = 2$ 比分别算两个对数再相加更快。
  • Domain awareness. Every log equation gets a domain check at the end. Mark schemes deduct for missing it.
  • 注意定义域。对数方程做完后必须检验定义域。漏检会被扣分。
Paper 2 — Calculator permittedPaper 2 —— 可使用计算器
  • Use $\ln$ as the universal helper. The GDC computes $\ln$ instantly; combined with change of base, that handles every logarithm in any base.
  • 把 $\ln$ 当万能工具。GDC 上 $\ln$ 是秒算，再配合换底公式，任何底数的对数都能算。
  • Set up before evaluating. Even when the final number comes from the calculator, write the formula and the substituted values. Method marks are awarded for the set-up, not the answer.
  • 列式先于求值。即便最终用计算器算数值，也要写出公式与代入的值。方法分给的是列式，不是答案。
  • Half-life / doubling-time formulas. Memorise both $t_{1/2} = \ln 2/|k|$ and $t_{2} = \ln 2/k$. These shave 30 seconds off every real-world problem.
  • 记住半衰期和倍增公式。$t_{1/2} = \ln 2/|k|$ 与 $t_{2} = \ln 2/k$ 都要背 —— 在实际题里能省 30 秒。
Active Recall主动回忆

Flashcards闪卡

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Product rule for exponents?指数的乘积律？
$$a^{m}\,a^{n} = a^{m+n}$$
Negative exponent?负指数？
$$a^{-n} = \tfrac{1}{a^{n}}$$
Rational exponent meaning?分数指数的含义？
$$a^{m/n} = \sqrt[n]{a^{m}}$$
$\log_{a}(b) = c \Longleftrightarrow$ ?$\log_{a}(b) = c \Longleftrightarrow$ ?
$$a^{c} = b$$
Log product rule?对数的积律？
$$\log(xy) = \log x + \log y$$
Log quotient rule?对数的商律？
$$\log\!\left(\tfrac{x}{y}\right) = \log x - \log y$$
Log power rule?对数的幂律？
$$\log(x^{n}) = n\log x$$
Change of base formula?换底公式？
$$\log_{a}(x) = \tfrac{\log_{b}(x)}{\log_{b}(a)}$$
Inverse identities for $\log_{a}$ and $a^{x}$?$\log_{a}$ 与 $a^{x}$ 的反函数恒等式？
$$a^{\log_{a}b} = b, \;\; \log_{a}(a^{x}) = x$$
Continuous growth model?连续增长模型？
$$N(t) = N_{0}\,e^{kt}$$
Half-life from decay constant?由衰减常数求半衰期？
$$t_{1/2} = \tfrac{\ln 2}{|k|}$$
Compound-interest formula?复利公式？
$$A = P\,\bigl(1 + \tfrac{r}{n}\bigr)^{nt}$$
Mixed Practice综合练习

Unit A2 — Practice Quiz单元 A2——练习测验

Simplify $\left(\dfrac{x^{4}\,y^{-2}}{x^{-1}\,y^{3}}\right)^{2}$.化简 $\left(\dfrac{x^{4}\,y^{-2}}{x^{-1}\,y^{3}}\right)^{2}$。
Q1
$x^{5}\,y^{-5}$
$x^{8}\,y^{-10}$
$x^{10}\,y^{-10}$
$x^{6}\,y^{-2}$
Inside the bracket: $x^{4-(-1)} = x^{5}$, $y^{-2-3} = y^{-5}$. Square both: $x^{10}\,y^{-10}$.括号内：$x^{4-(-1)} = x^{5}$、$y^{-2-3} = y^{-5}$。整体平方：$x^{10}\,y^{-10}$。
Simplify the inside first: $x^{5}\,y^{-5}$, then square: $x^{10}\,y^{-10}$.先化简括号内：$x^{5}\,y^{-5}$，再平方：$x^{10}\,y^{-10}$。
If $\log_{2}(x) + \log_{2}(x - 6) = 4$, find $x$.已知 $\log_{2}(x) + \log_{2}(x - 6) = 4$，求 $x$。
Q2
$x = -2$ or $x = 8$
$x = 8$
$x = 4$
$x = 2$
$\log_{2}(x(x-6)) = 4 \Rightarrow x(x-6) = 16 \Rightarrow x^{2} - 6x - 16 = 0 \Rightarrow (x-8)(x+2) = 0$. Domain requires $x > 6$, so reject $x = -2$. Only $x = 8$.$\log_{2}(x(x-6)) = 4 \Rightarrow x(x-6) = 16 \Rightarrow x^{2} - 6x - 16 = 0 \Rightarrow (x-8)(x+2) = 0$。定义域要求 $x > 6$，舍 $x = -2$。仅 $x = 8$。
The quadratic gives $x = 8$ or $x = -2$, but domain $x > 6$ rejects $-2$. Answer: $x = 8$.二次方程给出 $x = 8$ 或 $x = -2$，但定义域 $x > 6$ 舍掉 $-2$。答案：$x = 8$。
A radioactive substance has half-life $25$ years. Approximately what fraction remains after $80$ years?某放射性物质半衰期为 $25$ 年。$80$ 年后约剩多少比例？
Q3
$\tfrac{1}{4}$
$\tfrac{1}{8}$
$0.20$
$0.11$
After $n$ half-lives, fraction $= (1/2)^{n}$. Here $n = 80/25 = 3.2$, so fraction $= (1/2)^{3.2} \approx 0.109 \approx 0.11$.经过 $n$ 个半衰期后，剩余比例 $= (1/2)^{n}$。$n = 80/25 = 3.2$，故 $(1/2)^{3.2} \approx 0.109 \approx 0.11$。
After $n$ half-lives, fraction $= (1/2)^{n}$. With $n = 3.2$: $(1/2)^{3.2} \approx 0.109$.经过 $n$ 个半衰期后比例为 $(1/2)^{n}$。$n = 3.2$ 时 $(1/2)^{3.2} \approx 0.109$。
If $\log_{a}(b) = 3$, what is $\log_{b}(a)$?已知 $\log_{a}(b) = 3$，求 $\log_{b}(a)$。
Q4
$\tfrac{1}{3}$
$-3$
$3$
$\log(1/3)$
By change of base, $\log_{b}(a) = 1/\log_{a}(b) = 1/3$. Equivalently, $a^{3} = b \Rightarrow b^{1/3} = a$, so $\log_{b}(a) = 1/3$.由换底公式，$\log_{b}(a) = 1/\log_{a}(b) = 1/3$。等价地，$a^{3} = b \Rightarrow b^{1/3} = a$，故 $\log_{b}(a) = 1/3$。
Inverse log identity: $\log_{b}(a) = 1/\log_{a}(b) = 1/3$.对数倒底恒等式：$\log_{b}(a) = 1/\log_{a}(b) = 1/3$。
Solve $3^{2x+1} = 2 \cdot 3^{x} + 1$. HL求解 $3^{2x+1} = 2 \cdot 3^{x} + 1$。 HL
Q5
$x = 0$ only
$x = 0$ only (the other root is rejected)
$x = 0$ or $x = \log_{3}(-1/3)$
$x = 1$
$3^{2x+1} = 3 \cdot (3^{x})^{2}$. Let $u = 3^{x}$ (so $u > 0$): $3u^{2} - 2u - 1 = 0 \Rightarrow (3u + 1)(u - 1) = 0 \Rightarrow u = 1$ or $u = -1/3$. Reject $u = -1/3$ (negative). So $3^{x} = 1 \Rightarrow x = 0$.$3^{2x+1} = 3 \cdot (3^{x})^{2}$。令 $u = 3^{x}$（$u > 0$）：$3u^{2} - 2u - 1 = 0 \Rightarrow (3u + 1)(u - 1) = 0 \Rightarrow u = 1$ 或 $u = -1/3$。$u = -1/3$ 负，舍去。故 $3^{x} = 1 \Rightarrow x = 0$。
Substitute $u = 3^{x}$ (note $u > 0$). The quadratic gives $u = 1$ or $u = -1/3$; reject the negative root, so $3^{x} = 1 \Rightarrow x = 0$.换元 $u = 3^{x}$（$u > 0$）。二次给 $u = 1$ 或 $u = -1/3$；舍去负根，故 $3^{x} = 1 \Rightarrow x = 0$。
Self-check自查

Readiness Checklist备考清单

Tick each one when you can do it cold — without notes, without the formula box, on your first attempt.

每一条都要"裸做"做对（不看笔记、不看公式框、一次过）才打勾。

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IB Paper-Style PracticeIB 试卷风格练习

A2 Practice + Solutions are still on the roadmap (sibling of A1/A3/A4). When they ship they'll live under Practice Questions/Unit_A2_*.html. For now, the A1 and A3 practice sets give you scaffolded warmup on adjacent Topic 1 material.

A2 配套的 Practice + Solutions 仍在排期（与 A1/A3/A4 并列）。上线后将位于 Practice Questions/Unit_A2_*.html。在此之前，A1 与 A3 的练习题可作为 Topic 1 相邻内容的过渡热身。

A1 Practice →A1 练习题 → A3 Practice →A3 练习题 →
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IB Math Analysis & Approaches HL — Unit A2: Exponents & Logarithms · First Assessment 2029IB 数学分析与方法 HL——单元 A2：指数与对数 · 首次考核 2029