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）。
Integer Exponents整数指数 SL 1.5
▸ 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）。
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 $\bigl(4x^{0}y^{-2}\bigr)^{-1} \cdot \bigl(2x^{-3}y\bigr)^{2}$ for $x, y \ne 0$.化简 $\bigl(4x^{0}y^{-2}\bigr)^{-1} \cdot \bigl(2x^{-3}y\bigr)^{2}$（$x, y \ne 0$）。
Clear the zero exponent first. $x^{0} = 1$ regardless of what $x$ is, so it drops out of the first bracket entirely:
先消去零指数。无论 $x$ 取何值，$x^{0} = 1$，第一个括号里的 $x^{0}$ 直接消失：
$$ \bigl(4x^{0}y^{-2}\bigr)^{-1} \;=\; \bigl(4y^{-2}\bigr)^{-1} \;=\; 4^{-1}y^{2} \;=\; \frac{y^{2}}{4}. $$Expand the second bracket. Apply $(ab)^{n} = a^{n}b^{n}$:
展开第二个括号。用 $(ab)^{n} = a^{n}b^{n}$：
$$ \bigl(2x^{-3}y\bigr)^{2} \;=\; 2^{2}x^{-6}y^{2} \;=\; 4x^{-6}y^{2}. $$Multiply the two results.
相乘两个结果。
$$ \frac{y^{2}}{4} \cdot 4x^{-6}y^{2} \;=\; x^{-6}y^{4} \;=\; \frac{y^{4}}{x^{6}}. $$Why this is a good test of the zero law. A common slip is to treat $x^{0}$ as if it vanishes the whole term (giving $0$) rather than reducing it to the multiplicative identity $1$ — exactly the distinction the "going deeper" derivation above proves from the product rule.
为何这题特别检验零指数律。常见错误是把 $x^{0}$ 当作让整项消失为 $0$，而不是化为乘法单位元 $1$ —— 这正是上面"深入"推导从乘积律证明的那个区别。
Rational Exponents & Radicals分数指数与根式 SL 1.5
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）。
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：
$$ x^{1/2}\,x^{1/3} \;=\; x^{1/2 + 1/3} \;=\; x^{3/6 + 2/6} \;=\; x^{5/6} \;=\; \sqrt[6]{x^{5}}. $$Solve $x^{2/3} = 4$ for $x > 0$.求解 $x^{2/3} = 4$（$x > 0$）。
Raise both sides to the reciprocal power. To undo the exponent $2/3$, raise both sides to $3/2$:
两边取倒数次幂。要消去指数 $2/3$，两边同时取 $3/2$ 次幂：
$$ \bigl(x^{2/3}\bigr)^{3/2} \;=\; 4^{3/2} \quad \Longrightarrow \quad x \;=\; 4^{3/2}. $$Evaluate. Read $4^{3/2}$ as "square root, then cube":
计算。将 $4^{3/2}$ 读作"先开平方再立方"：
$$ 4^{3/2} \;=\; \bigl(\sqrt{4}\bigr)^{3} \;=\; 2^{3} \;=\; 8. $$Check. $8^{2/3} = \bigl(\sqrt[3]{8}\bigr)^{2} = 2^{2} = 4$. ✓
检验。$8^{2/3} = \bigl(\sqrt[3]{8}\bigr)^{2} = 2^{2} = 4$。 ✓
Logarithm Definition & Laws对数的定义与运算律 SL 1.7
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）。
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。
(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：
$$ \ln\!\left(\tfrac{x^{2}\,y}{z}\right) \;=\; \ln(x^{2}) + \ln(y) - \ln(z) \;=\; 2\ln(x) + \ln(y) - \ln(z). $$Solve $\log_{x}(9) = 2$ for $x$.求解 $\log_{x}(9) = 2$。
Rewrite in exponential form. Here the base $x$ is the unknown, not the argument — a question style that reads differently from Example A2.3 but uses the same definition:
改写为指数形式。此题未知量是底数 $x$，不是真数 —— 题型与例题 A2.3 不同，但用的是同一个定义：
$$ \log_{x}(9) = 2 \quad \Longleftrightarrow \quad x^{2} = 9. $$Solve the resulting equation. Algebraically $x = \pm 3$, but the logarithm base must satisfy $x > 0$ and $x \ne 1$ (the standing restriction from the A2.3 definition box). Discard $x = -3$:
求解方程。代数上 $x = \pm 3$，但对数的底数必须满足 $x > 0$ 且 $x \ne 1$（A2.3 定义框中的固定限制）。舍去 $x = -3$：
$$ x \;=\; 3. $$Check. $\log_{3}(9) = 2$ since $3^{2} = 9$. ✓ This base-domain restriction is easy to forget precisely because it never shows up when the argument is the unknown — only when the base is.
检验。$\log_{3}(9) = 2$，因 $3^{2} = 9$。 ✓ 这条底数限制很容易被忽略，正因为当未知量是真数时它根本不会出现 —— 只有未知量是底数时才要检验。
Change of Base换底公式 SL 1.7
▸ 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 $c$：
$$ c \;=\; \frac{\log_{b}(x)}{\log_{b}(a)} \quad \Longrightarrow \quad \log_{a}(x) \;=\; \frac{\log_{b}(x)}{\log_{b}(a)}. \qquad \square $$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。 ✓
Which is larger: $\log_{3}(50)$ or $\log_{2}(17)$?$\log_{3}(50)$ 与 $\log_{2}(17)$ 哪个更大？
Change both to the same base. Using $\ln$:
换到同一底。用 $\ln$：
$$ \log_{3}(50) = \frac{\ln 50}{\ln 3}, \qquad \log_{2}(17) = \frac{\ln 17}{\ln 2}. $$Estimate each. $\ln 50 \approx 3.912$, $\ln 3 \approx 1.099$, so $\log_{3}(50) \approx 3.56$. Meanwhile $\ln 17 \approx 2.833$, $\ln 2 \approx 0.693$, so $\log_{2}(17) \approx 4.09$.
逐一估算。$\ln 50 \approx 3.912$，$\ln 3 \approx 1.099$，故 $\log_{3}(50) \approx 3.56$；而 $\ln 17 \approx 2.833$，$\ln 2 \approx 0.693$，故 $\log_{2}(17) \approx 4.09$。
Conclusion. $\log_{2}(17) > \log_{3}(50)$. Sanity check without a calculator: $3^{3} = 27 < 50 < 81 = 3^{4}$ (so the first is between 3 and 4), while $2^{4} = 16 < 17 < 32 = 2^{5}$ but $17$ is barely past $16$ (so the second is just above 4) — consistent with the estimates.
结论。$\log_{2}(17) > \log_{3}(50)$。不用计算器的验证：$3^{3} = 27 < 50 < 81 = 3^{4}$（第一个在 3、4 之间），而 $2^{4} = 16 < 17 < 32 = 2^{5}$ 且 $17$ 刚过 $16$（第二个略高于 4）——与估算一致。
Solving Exponential & Logarithmic Equations指数方程与对数方程的求解 SL 1.7
- 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 — $x = a^{c}$.
- 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 —— $x = a^{c}$。
extraneous root）。
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 $x \approx 1.3219$。
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 $x = 5$。
▸ 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。
Exponential Growth & Decay Models指数增长与衰减模型 SL 1.5 / 1.7
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。
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。 ✓
$5000 is invested at a nominal annual rate of $4\%$, compounded quarterly. (a) Find the balance after $10$ years. (b) Compare it to continuous compounding at the same nominal rate.$5000 以名义年利率 $4\%$ 按季度复利投资。(a) 求 $10$ 年后的余额。(b) 与相同名义利率下的连续复利比较。
(a) Discrete model. Here $P = 5000$, $r = 0.04$, $n = 4$ (quarterly), $t = 10$:
(a) 离散模型。$P = 5000$、$r = 0.04$、$n = 4$（按季）、$t = 10$：
$$ A \;=\; 5000\left(1 + \frac{0.04}{4}\right)^{4 \times 10} \;=\; 5000(1.01)^{40} \;\approx\; 5000 \times 1.4889 \;\approx\; \$7444.73. $$(b) Continuous model at the same nominal rate. Using $A(t) = Pe^{rt}$:
(b) 相同名义利率下的连续模型。用 $A(t) = Pe^{rt}$：
$$ A \;=\; 5000\,e^{0.04 \times 10} \;=\; 5000\,e^{0.4} \;\approx\; 5000 \times 1.4918 \;\approx\; \$7459.12. $$Compare. Continuous compounding earns only $\$14.39$ more than quarterly over $10$ years — quarterly compounding is already very close to the $n \to \infty$ limit from the concept box above. Compounding daily ($n = 365$) would close almost all of that remaining gap.
比较。$10$ 年下来，连续复利只比季度复利多赚 $\$14.39$ —— 说明季度复利已非常接近上方概念框中 $n \to \infty$ 的极限。若改为按日计息（$n = 365$），差距会被进一步抹平。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- 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。
- 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。
Flashcards闪卡
Unit A2 — Practice Quiz单元 A2——练习测验
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。
- State and apply all seven integer exponent laws (product, quotient, power, distribution, zero, negative)写出并应用全部七条整数指数律（积、商、幂、分配、零、负）
- Convert between $a^{m/n}$ and $\sqrt[n]{a^{m}}$ fluently熟练在 $a^{m/n}$ 与 $\sqrt[n]{a^{m}}$ 之间互换
- Justify $a^{0} = 1$ and $a^{-n} = 1/a^{n}$ from the product rule (extension by consistency)能用积律一致性论证 $a^{0} = 1$ 与 $a^{-n} = 1/a^{n}$ 的合理性
- State the log/exponential equivalence: $\log_{a}(b) = c \Longleftrightarrow a^{c} = b$写出对数与指数的等价关系：$\log_{a}(b) = c \Longleftrightarrow a^{c} = b$
- Apply the three log laws (product, quotient, power) in either direction at speed能秒速双向应用三条对数运算律（积、商、幂）
- Reproduce the proof of $\log(xy) = \log x + \log y$ from the exponent product rule能用指数积律推导 $\log(xy) = \log x + \log y$
- State and apply change of base; use $\ln$ or $\log_{10}$ as the universal helper熟练换底公式；以 $\ln$ 或 $\log_{10}$ 作万能工具
- Solve exponential equations $a^{x} = b$ by taking $\log$ of both sides解 $a^{x} = b$ 类指数方程（两边取对数）
- Solve logarithmic equations and check the domain to discard extraneous roots解对数方程并通过定义域检验剔除增根
- HL Recognise and solve "hidden quadratic" equations in $a^{x}$ by substitution能识别并用换元解 $a^{x}$ 的"隐藏二次"方程
- Set up continuous growth/decay models $N(t) = N_{0}e^{kt}$ from given conditions由条件建立连续增长/衰减模型 $N(t) = N_{0}e^{kt}$
- Compute half-life $t_{1/2} = \ln 2/|k|$ and doubling time $t_{2} = \ln 2/k$熟练计算半衰期 $t_{1/2} = \ln 2/|k|$ 与倍增时间 $t_{2} = \ln 2/k$
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。
How to use this guide本指南使用说明
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：
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。
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。