← All Units← 返回单元列表 ← Course Hub← 课程主页
I B  M A T H  A A  H L
Unit A2 · SolutionsUnit A2 · 解析

Exponents & Logarithms — Solutions指数与对数 —— 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus 1.5, 1.7考纲 1.5, 1.7AA HL

PART I  ·  PAPER 1 SECTION A — SOLUTIONS第一部分  ·  第一卷 A 节 —— 解析No calculator · 20 marks不可使用计算器 · 20 分

Section A — Worked SolutionsA 节 —— 详细解析

Q1EASYPaper 1A1.5 Exponent Laws[4 marks]

Simplify (a) $\dfrac{2^{3} \cdot 2^{5}}{2^{4}}$, (b) $25^{3/2}$, (c) $\left(\dfrac{8}{27}\right)^{-2/3}$.化简 (a)、(b)、(c) 三式。

Answers:答案：  (a) $16$  ·  (b) $125$  ·  (c) $\dfrac{9}{4}$

(a) Product / Quotient laws A1

$\dfrac{2^{3} \cdot 2^{5}}{2^{4}} = 2^{3+5-4} = 2^{4} = 16$.

(b) Rational exponent A1

$25^{3/2} = (5^{2})^{3/2} = 5^{3} = 125$.

(c) Negative + rational + quotient M1·A1

Flip the fraction (negative exponent), then take cube root and square: $$ \left(\tfrac{8}{27}\right)^{-2/3} = \left(\tfrac{27}{8}\right)^{2/3} = \left(\tfrac{\sqrt[3]{27}}{\sqrt[3]{8}}\right)^{2} = \left(\tfrac{3}{2}\right)^{2} = \tfrac{9}{4}. $$
Order of operations on $a^{m/n}$. "Root first, power second" usually keeps numbers small. For $(8/27)^{-2/3}$, taking the cube root first gives $3/2$ — a tidy fraction — then squaring is trivial. The reverse ("square first, then cube-root") would force you through $(64/729)^{1/3} = \sqrt[3]{64}/\sqrt[3]{729} = 4/9 \ne 9/4$. Wait — that's not equal! Let me re-examine: actually $(8/27)^{-2/3}$ means $(8/27)^{-(2/3)} = ((8/27)^{-1})^{2/3} = (27/8)^{2/3}$ — and $(27/8)^{2/3} = (3/2)^{2} = 9/4$ matches. Or directly $(8/27)^{-2/3} = 1/(8/27)^{2/3} = 1/(8^{2/3}/27^{2/3}) = 1/(4/9) = 9/4$. Both routes agree.

(a) 积律 / 商律 A1

$\dfrac{2^{3} \cdot 2^{5}}{2^{4}} = 2^{3+5-4} = 2^{4} = 16$。

(b) 分数指数 A1

$25^{3/2} = (5^{2})^{3/2} = 5^{3} = 125$。

(c) 负 + 分数 + 商 M1·A1

负指数翻分子分母，再开立方根并平方： $$ \left(\tfrac{8}{27}\right)^{-2/3} = \left(\tfrac{27}{8}\right)^{2/3} = \left(\tfrac{\sqrt[3]{27}}{\sqrt[3]{8}}\right)^{2} = \left(\tfrac{3}{2}\right)^{2} = \tfrac{9}{4}. $$
$a^{m/n}$ 的运算顺序。"先开方再求幂"通常让数字更小。$(8/27)^{-2/3}$ 中先开立方根得 $3/2$，平方就是 $9/4$。若反过来"先求幂再开方"，要算 $\sqrt[3]{64/729} = 4/9$，再倒数得 $9/4$。两条路径都正确，但先开方更不容易出错。
Q2EASYPaper 1A1.5 Rational Exponents[5 marks]

Express each in the form $x^{p/q}$.将各式化为 $x^{p/q}$ 形式。

Answers:答案：  (a) $x^{7/6}$  ·  (b) $x^{5/6}$  ·  (c) $x^{3/2}$

(a) Radical → exponent → add M1·A1

$\sqrt[3]{x^{2}} \cdot \sqrt{x} = x^{2/3} \cdot x^{1/2} = x^{2/3 + 1/2} = x^{4/6 + 3/6} = x^{7/6}$.

(b) Subtract negative M1·A1

$\dfrac{x^{1/2}}{x^{-1/3}} = x^{1/2 - (-1/3)} = x^{1/2 + 1/3} = x^{3/6 + 2/6} = x^{5/6}$.

(c) Power-of-a-power A1

$(x^{4})^{3/8} = x^{4 \cdot 3/8} = x^{12/8} = x^{3/2}$.
Common denominator for adding exponents. Whenever you add or subtract rational exponents, convert to a common denominator once and write the sum in lowest terms. Half + third = three-sixths + two-sixths = five-sixths — always think "what's the LCM of the denominators."

(a) 根式转指数后相加 M1·A1

$\sqrt[3]{x^{2}} \cdot \sqrt{x} = x^{2/3} \cdot x^{1/2} = x^{2/3 + 1/2} = x^{4/6 + 3/6} = x^{7/6}$。

(b) 减去负指数 M1·A1

$\dfrac{x^{1/2}}{x^{-1/3}} = x^{1/2 - (-1/3)} = x^{1/2 + 1/3} = x^{3/6 + 2/6} = x^{5/6}$。

(c) 幂的幂 A1

$(x^{4})^{3/8} = x^{4 \cdot 3/8} = x^{12/8} = x^{3/2}$。
分数指数相加要通分。加减分数指数时统一通分一次，结果写最简。$\tfrac{1}{2} + \tfrac{1}{3} = \tfrac{3}{6} + \tfrac{2}{6} = \tfrac{5}{6}$ —— 先想"分母的最小公倍数"。
Q3MEDIUMPaper 1A1.7 Log Evaluation[5 marks]

Evaluate (a) $\log_{3}(81)$, (b) $\log_{10}(20) + \log_{10}(5)$, (c) $\ln(e^{4}) - \ln(\sqrt{e})$.求 (a)、(b)、(c) 三式。

Answers:答案：  (a) $4$  ·  (b) $2$  ·  (c) $\tfrac{7}{2}$

(a) Recognise the power A1

$81 = 3^{4}$, so $\log_{3}(81) = 4$.

(b) Product rule M1·A1

$\log_{10}(20) + \log_{10}(5) = \log_{10}(20 \cdot 5) = \log_{10}(100) = 2$.

(c) Power rule + inverse identity M1·A1

$\ln(e^{4}) = 4$, $\ln(\sqrt{e}) = \ln(e^{1/2}) = \tfrac{1}{2}$, so $\ln(e^{4}) - \ln(\sqrt{e}) = 4 - \tfrac{1}{2} = \tfrac{7}{2}$.
Combine before evaluating. Part (b)'s product-rule combination saves time over computing $\log_{10}(20) \approx 1.301$ and $\log_{10}(5) \approx 0.699$ separately. In Paper 1 (no calculator) you must combine — neither $\log_{10}(20)$ nor $\log_{10}(5)$ is recoverable on its own without a calculator.

(a) 识别幂次 A1

$81 = 3^{4}$，故 $\log_{3}(81) = 4$。

(b) 积律 M1·A1

$\log_{10}(20) + \log_{10}(5) = \log_{10}(20 \cdot 5) = \log_{10}(100) = 2$。

(c) 幂律 + 反函数恒等式 M1·A1

$\ln(e^{4}) = 4$，$\ln(\sqrt{e}) = \ln(e^{1/2}) = \tfrac{1}{2}$，故 $\ln(e^{4}) - \ln(\sqrt{e}) = 4 - \tfrac{1}{2} = \tfrac{7}{2}$。
先合并再求值。(b) 用积律一步合并比分别算 $\log_{10}(20) \approx 1.301$、$\log_{10}(5) \approx 0.699$ 快得多。在 Paper 1（不可用计算器）必须合并 —— 单独的 $\log_{10}(20)$、$\log_{10}(5)$ 没有计算器无法求出。
Q4HARDPaper 1A1.7 Log Laws — Expand & Express[6 marks]

$p = \log_{a}(2)$, $q = \log_{a}(3)$. Express (a) $\log_{a}(6)$, (b) $\log_{a}(8/9)$, (c) $\log_{a}(48)$.$p = \log_{a}(2)$、$q = \log_{a}(3)$。用 $p, q$ 表示 (a)、(b)、(c)。

Answers:答案：  (a) $p + q$  ·  (b) $3p - 2q$  ·  (c) $4p + q$

(a) Factor and apply product rule A1

$6 = 2 \cdot 3$, so $\log_{a}(6) = \log_{a}(2) + \log_{a}(3) = p + q$.

(b) Factor + product + power + quotient M1·A1

$\dfrac{8}{9} = \dfrac{2^{3}}{3^{2}}$, so $$ \log_{a}(8/9) \;=\; \log_{a}(2^{3}) - \log_{a}(3^{2}) \;=\; 3\log_{a}(2) - 2\log_{a}(3) \;=\; 3p - 2q. $$

(c) Factor first, then expand M1·A1·A1

$48 = 16 \cdot 3 = 2^{4} \cdot 3$, so $$ \log_{a}(48) \;=\; \log_{a}(2^{4}) + \log_{a}(3) \;=\; 4\log_{a}(2) + \log_{a}(3) \;=\; 4p + q. $$
Always factor first. The hardest step in these problems is recognising the prime factorisation of the argument. Build muscle memory: $6 = 2 \cdot 3$; $8 = 2^{3}$; $9 = 3^{2}$; $12 = 2^{2} \cdot 3$; $48 = 2^{4} \cdot 3$; $72 = 2^{3} \cdot 3^{2}$. Once factored, the log laws apply mechanically.

(a) 分解再用积律 A1

$6 = 2 \cdot 3$，故 $\log_{a}(6) = \log_{a}(2) + \log_{a}(3) = p + q$。

(b) 分解 + 积律 + 幂律 + 商律 M1·A1

$\dfrac{8}{9} = \dfrac{2^{3}}{3^{2}}$，故 $$ \log_{a}(8/9) \;=\; \log_{a}(2^{3}) - \log_{a}(3^{2}) \;=\; 3\log_{a}(2) - 2\log_{a}(3) \;=\; 3p - 2q. $$

(c) 先分解再展开 M1·A1·A1

$48 = 16 \cdot 3 = 2^{4} \cdot 3$，故 $$ \log_{a}(48) \;=\; \log_{a}(2^{4}) + \log_{a}(3) \;=\; 4\log_{a}(2) + \log_{a}(3) \;=\; 4p + q. $$
一律先分解。这类题最关键的一步是认出自变量的素因子分解。把常见分解烙进肌肉记忆：$6 = 2 \cdot 3$；$8 = 2^{3}$；$9 = 3^{2}$；$12 = 2^{2} \cdot 3$；$48 = 2^{4} \cdot 3$；$72 = 2^{3} \cdot 3^{2}$。分解到位后，运算律的应用就是机械的。
PART II  ·  PAPER 1 SECTION B — SOLUTIONS第二部分  ·  第一卷 B 节 —— 解析No calculator · 12 marks不可使用计算器 · 12 分

Section B — Worked SolutionsB 节 —— 详细解析

Q5HARDPaper 1B1.7 Quadratic in $a^{x}$[12 marks]

$4^{x} - 5 \cdot 2^{x} + 4 = 0$ — (a) why $u = 2^{x}$ converts to quadratic; (b) solve in $u$; (c) back to $x$; (d) sketch.$4^{x} - 5 \cdot 2^{x} + 4 = 0$ —— (a) 为何换元 $u = 2^{x}$ 化为二次；(b) 在 $u$ 下解；(c) 换回 $x$；(d) 画图。

Answers:答案：  (b) $u = 1$ or $u = 4$  ·  (c) $x = 0$ or $x = 2$  ·  (d) U-shaped with zeros at $0$ and $2$, minimum near $x = \log_{2}(5/2) \approx 1.32$ with $f_{\min} = -9/4$.U 形曲线，零点 $0$、$2$，极小值在 $x = \log_{2}(5/2) \approx 1.32$ 处，$f_{\min} = -9/4$。

(a) Substitution and domain M1·R1

Since $4^{x} = (2^{2})^{x} = (2^{x})^{2}$, setting $u = 2^{x}$ converts the equation into a quadratic in $u$. The function $x \mapsto 2^{x}$ has range $(0, \infty)$, so $u > 0$.

(b) Quadratic M1·M1·A1·A1

$$ u^{2} - 5u + 4 \;=\; 0 \;\Longrightarrow\; (u - 1)(u - 4) \;=\; 0 \;\Longrightarrow\; u = 1 \text{ or } u = 4. $$ Both roots satisfy $u > 0$ — both are admissible.

(c) Back to $x$ M1·A1·A1

$u = 2^{x} = 1 \Longrightarrow x = 0$. $u = 2^{x} = 4 = 2^{2} \Longrightarrow x = 2$. Both valid — no extraneous roots in this problem.

(d) Sketch A1·A1·A1

Let $f(x) = 4^{x} - 5 \cdot 2^{x} + 4$. Key features for the sketch:
  • $f(0) = 0$, $f(2) = 0$ (the two roots).
  • As $x \to -\infty$: $4^{x}, 2^{x} \to 0$, so $f(x) \to 4$ (horizontal asymptote on the left).
  • As $x \to +\infty$: $4^{x}$ dominates, $f(x) \to +\infty$.
  • Minimum: solve $f'(x) = 0$ via the $u$-form (substituting $u = 2^{x}$, $f = u^{2} - 5u + 4$, $df/du = 2u - 5 = 0$ at $u = 5/2$). So minimum is at $u = 5/2$, i.e. $x = \log_{2}(5/2) \approx 1.32$. Minimum value: $(5/2)^{2} - 5(5/2) + 4 = 25/4 - 25/2 + 4 = -9/4$.
The "hidden quadratic" pattern. Any equation of the form $a^{2x} + b \cdot a^{x} + c = 0$ collapses to a quadratic via $u = a^{x}$, with strict positivity $u > 0$ as the only admissibility constraint. Look for two exponential terms whose bases are $a$ and $a^{2}$ (or equivalent factorings like $9^{x} = (3^{x})^{2}$). The same substitution principle works when the "outer" function is anything monotone: $e^{2x}$, $9^{x}$, $\sin^{2}(x)$, etc.

(a) 换元与定义域 M1·R1

因 $4^{x} = (2^{2})^{x} = (2^{x})^{2}$，令 $u = 2^{x}$ 即把方程化为关于 $u$ 的二次方程。函数 $x \mapsto 2^{x}$ 值域为 $(0, \infty)$，故 $u > 0$。

(b) 二次方程 M1·M1·A1·A1

$$ u^{2} - 5u + 4 \;=\; 0 \;\Longrightarrow\; (u - 1)(u - 4) \;=\; 0, $$ 即 $u = 1$ 或 $u = 4$。 两根都满足 $u > 0$，均合法。

(c) 换回 $x$ M1·A1·A1

$u = 2^{x} = 1 \Longrightarrow x = 0$。 $u = 2^{x} = 4 = 2^{2} \Longrightarrow x = 2$。 两解均合法，本题无增根。

(d) 草图 A1·A1·A1

令 $f(x) = 4^{x} - 5 \cdot 2^{x} + 4$。关键特征：
  • $f(0) = 0$、$f(2) = 0$（两个零点）。
  • 当 $x \to -\infty$：$4^{x}, 2^{x} \to 0$，故 $f(x) \to 4$（左侧水平渐近线）。
  • 当 $x \to +\infty$：$4^{x}$ 占主导，$f(x) \to +\infty$。
  • 极小值：用 $u$ 形式求 $f'(x) = 0$（令 $u = 2^{x}$，$f = u^{2} - 5u + 4$，$df/du = 2u - 5 = 0$ 得 $u = 5/2$）。故极小值出现在 $u = 5/2$，即 $x = \log_{2}(5/2) \approx 1.32$。极小值 $(5/2)^{2} - 5(5/2) + 4 = 25/4 - 25/2 + 4 = -9/4$。
"隐藏二次"模式。任何 $a^{2x} + b \cdot a^{x} + c = 0$ 形式的方程，都可用 $u = a^{x}$ 化为二次方程，且唯一定义域约束是 $u > 0$。线索：方程中同时出现底数为 $a$ 与 $a^{2}$ 的指数项（或如 $9^{x} = (3^{x})^{2}$ 这类等价分解）。同一换元思想也适用于其它单调"外层"函数：$e^{2x}$、$9^{x}$、$\sin^{2}(x)$ 等。
PART III  ·  PAPER 2 — SOLUTIONS第三部分  ·  第二卷 —— 解析Calculator · 14 marks可使用计算器 · 14 分

Paper 2 — Worked Solutions第二卷 —— 详细解析

Q6MEDIUMPaper 21.7 Half-Life[7 marks]

Continuous decay $N(t) = N_{0}e^{kt}$. $N(30)/N_{0} = 0.4$. Find (a) $k$ to 4 dp, (b) half-life to nearest year, (c) time when $10\%$ remains.连续衰减 $N(t) = N_{0}e^{kt}$。$N(30)/N_{0} = 0.4$。求 (a) $k$（4 位小数）；(b) 半衰期（精确到年）；(c) 剩 $10\%$ 的时间。

Answers:答案：  (a) $k \approx -0.0305$  ·  (b) $\approx 23$ years年  ·  (c) $\approx 75$ years年

(a) Solve for $k$ M1·M1·A1

$$ e^{30k} \;=\; 0.4 \;\Longrightarrow\; 30k \;=\; \ln(0.4) \;\Longrightarrow\; k \;=\; \tfrac{\ln(0.4)}{30} \;\stackrel{\text{GDC}}{\approx}\; -0.0305. $$

(b) Half-life M1·A1

Half-life: $N(t)/N_{0} = 0.5$. Then $kt_{1/2} = \ln(0.5)$, so $$ t_{1/2} \;=\; \frac{\ln(0.5)}{k} \;=\; \frac{-\ln 2}{k} \;\stackrel{\text{GDC}}{\approx}\; \frac{-0.6931}{-0.0305} \;\approx\; 22.69 \;\to\; 23 \text{ years}. $$

(c) $10\%$ remaining M1·A1

$$ e^{kt} \;=\; 0.1 \;\Longrightarrow\; t \;=\; \tfrac{\ln(0.1)}{k} \;\stackrel{\text{GDC}}{\approx}\; \tfrac{-2.3026}{-0.0305} \;\approx\; 75.5 \;\to\; 75 \text{ years}. $$
Sanity check via half-lives. $10\% = 0.1 \approx (1/2)^{3.32}$, so the sample is about $3.32$ half-lives in. $3.32 \times 23 \approx 76$ — matches the $75$-year answer. Always cross-check exponential answers by counting half-lives.

(a) 解出 $k$ M1·M1·A1

$$ e^{30k} \;=\; 0.4 \;\Longrightarrow\; 30k \;=\; \ln(0.4) \;\Longrightarrow\; k \;=\; \tfrac{\ln(0.4)}{30} \;\stackrel{\text{GDC}}{\approx}\; -0.0305. $$

(b) 半衰期 M1·A1

半衰期对应 $N(t)/N_{0} = 0.5$。$kt_{1/2} = \ln(0.5)$，故 $$ t_{1/2} \;=\; \frac{\ln(0.5)}{k} \;=\; \frac{-\ln 2}{k} \;\stackrel{\text{GDC}}{\approx}\; \frac{-0.6931}{-0.0305} \;\approx\; 22.69 \;\to\; 23 \; \mathrm{yr}. $$

(c) 剩 $10\%$ M1·A1

$$ e^{kt} \;=\; 0.1 \;\Longrightarrow\; t \;=\; \tfrac{\ln(0.1)}{k} \;\stackrel{\text{GDC}}{\approx}\; \tfrac{-2.3026}{-0.0305} \;\approx\; 75.5 \;\to\; 75 \; \mathrm{yr}. $$
用半衰期检验。$10\% = 0.1 \approx (1/2)^{3.32}$，即样本约经过 $3.32$ 个半衰期。$3.32 \times 23 \approx 76$ —— 与 $75$ 年答案吻合。指数题做完后，习惯性用"几个半衰期"快速核对。
Q7HARDPaper 21.4 / 1.7 Compound Interest Comparison[7 marks]

€$10\,000$ at nominal $3.6\%$ p.a. Account X compounds monthly; Account Y continuously. Compare at $t = 10$, find gap-€$50$ time, explain Y > X.€$10\,000$ 名义年利率 $3.6\%$。账户 X 按月复利；账户 Y 连续复利。$t = 10$ 比较、求差 €$50$ 的时间、说明 Y > X。

Answers:答案：  (a) $X \approx$ €$14\,325.60$, $Y \approx$ €$14\,333.29$  ·  (b) $\approx 33$ years年  ·  (c) $e^{r} > (1 + r/n)^{n}$ for $n < \infty$.当 $n < \infty$ 时 $e^{r} > (1 + r/n)^{n}$。

(a) Values at $t = 10$ M1·A1·A1

$$ A_{X}(10) \;=\; 10\,000\,(1 + 0.036/12)^{120} \;=\; 10\,000\,(1.003)^{120} \;\stackrel{\text{GDC}}{\approx}\; \text{€}14\,325.60. $$ $$ A_{Y}(10) \;=\; 10\,000\,e^{0.036 \cdot 10} \;=\; 10\,000\,e^{0.36} \;\stackrel{\text{GDC}}{\approx}\; \text{€}14\,333.29. $$ Gap at $t = 10$: about €$7.69$.

(b) Time for gap to reach €$50$ M1·M1·A1

Solve $A_{Y}(t) - A_{X}(t) = 50$, i.e. $10\,000\,\bigl(e^{0.036 t} - (1.003)^{12 t}\bigr) = 50$. The expression grows monotonically since $e^{r} > (1 + r/n)^{n}$. Use the GDC's numerical solver (or scan $t$): at $t = 32$, gap $\approx$ €$48$; at $t = 33$, gap $\approx$ €$56$. So the gap first exceeds €$50$ around $t = 33$ years.

(c) Why $Y > X$ A1

For any nominal rate $r > 0$ and finite compounding frequency $n$, the inequality $\bigl(1 + \tfrac{r}{n}\bigr)^{n} < e^{r}$ holds (the LHS approaches the RHS as $n \to \infty$). Equivalently, the effective annual rate of continuous compounding $e^{r} - 1$ strictly exceeds the effective rate of any discrete scheme at the same nominal $r$.
Why the gap grows so slowly. The ratio $A_{Y}/A_{X} = (e^{r}/(1+r/n)^{n})^{t} = (1 + \epsilon)^{t}$ where $\epsilon = e^{r}/(1+r/n)^{n} - 1$ is tiny — for $r = 3.6\%$, $n = 12$, $\epsilon \approx 5 \times 10^{-5}$. Even after $10$ years the gap is single-digit euros. The lesson: continuous vs frequent-discrete compounding differs by a rounding error at these rates. Where compounding frequency matters more dramatically is at high rates (think credit-card APR ≈ $20\%$).

(a) $t = 10$ 时的金额 M1·A1·A1

$$ A_{X}(10) \;=\; 10\,000\,(1 + 0.036/12)^{120} \;=\; 10\,000\,(1.003)^{120} \;\stackrel{\text{GDC}}{\approx}\; \text{€}14\,325.60. $$ $$ A_{Y}(10) \;=\; 10\,000\,e^{0.036 \cdot 10} \;=\; 10\,000\,e^{0.36} \;\stackrel{\text{GDC}}{\approx}\; \text{€}14\,333.29. $$ $t = 10$ 时差约 €$7.69$。

(b) 差达到 €$50$ 的时间 M1·M1·A1

解 $A_{Y}(t) - A_{X}(t) = 50$，即 $10\,000\,\bigl(e^{0.036 t} - (1.003)^{12 t}\bigr) = 50$。由于 $e^{r} > (1 + r/n)^{n}$，该差关于 $t$ 单调增。用 GDC 数值求解（或扫值）：$t = 32$ 时差 $\approx$ €$48$；$t = 33$ 时差 $\approx$ €$56$。差首次超过 €$50$ 约在 $t = 33$ 年。

(c) 为何 $Y > X$ A1

对任意名义利率 $r > 0$ 与有限复利频率 $n$，恒有 $\bigl(1 + \tfrac{r}{n}\bigr)^{n} < e^{r}$（左侧在 $n \to \infty$ 时趋于右侧）。等价地，相同名义利率下，连续复利的有效年利率 $e^{r} - 1$ 严格大于任何离散方案的有效年利率。
差为何增长得慢。$A_{Y}/A_{X} = (e^{r}/(1+r/n)^{n})^{t} = (1 + \epsilon)^{t}$，其中 $\epsilon = e^{r}/(1+r/n)^{n} - 1$ 极小 —— $r = 3.6\%$、$n = 12$ 时 $\epsilon \approx 5 \times 10^{-5}$。即便 10 年差距也只是个位数欧元。结论：本题利率水平下，连续复利与高频离散复利的差异约等于舍入误差。复利频率对结果的影响在高利率（如信用卡 APR $\approx 20\%$）时才显著。
PART IV  ·  PAPER 3 — SOLUTIONS第四部分  ·  第三卷 —— 解析Calculator · HL extended exploration · 16 marks可使用计算器 · HL 长题探究 · 16 分

Paper 3 — Worked Solutions第三卷 —— 详细解析

Q8HARDPaper 31.7 Newton's Law of Cooling[16 marks]

$T(t) = 22 + 73 e^{-kt}$. $T(5) = 70$. Six parts: verify boundary conditions, find $k$, $T(15)$, time for $T = 40$, half-cooling formula and value.$T(t) = 22 + 73 e^{-kt}$。$T(5) = 70$。六小问：验证边界条件、求 $k$、$T(15)$、$T = 40$ 时间、半冷却时间公式与值。

Answers:答案：  (b) $k \approx 0.0839$  ·  (c) $\approx 43^\circ$C  ·  (d) $\approx 17$ min分钟  ·  (e) $t_{1/2} = \ln 2/k \approx 8.3$ min分钟

(a) Boundary conditions A1·A1

$T(0) = 22 + 73 \cdot e^{0} = 22 + 73 = 95 = T_{0}$. ✓ As $t \to \infty$, $e^{-kt} \to 0$ ($k > 0$), so $T(t) \to 22 = T_{\text{env}}$. ✓

(b) Find $k$ M1·M1·A1·A1

$T(5) = 70$ gives $22 + 73 e^{-5k} = 70$, so $e^{-5k} = \tfrac{48}{73}$. Take $\ln$: $$ -5k \;=\; \ln(48/73) \;\stackrel{\text{GDC}}{\approx}\; -0.4193 \;\Longrightarrow\; k \;\approx\; 0.0839. $$

(c) $T(15)$ M1·A1·A1

$$ T(15) \;=\; 22 + 73\,e^{-15(0.0839)} \;=\; 22 + 73\,e^{-1.258} \;\stackrel{\text{GDC}}{\approx}\; 22 + 73(0.284) \;\approx\; 22 + 20.76 \;\approx\; 43^\circ\text{C}. $$

(d) Time for $T = 40$ M1·M1·A1

$22 + 73 e^{-kt} = 40 \Longrightarrow e^{-kt} = \tfrac{18}{73}$. Take $\ln$: $$ -kt \;=\; \ln(18/73) \;\stackrel{\text{GDC}}{\approx}\; -1.401 \;\Longrightarrow\; t \;=\; \tfrac{1.401}{0.0839} \;\approx\; 16.7 \;\to\; 17 \text{ minutes}. $$

(e) Half-cooling time M1·A1·M1·A1

The temperature excess is $T(t) - T_{\text{env}} = 73 e^{-kt}$. Setting this equal to half the initial excess ($73/2$): $$ 73\,e^{-k t_{1/2}} \;=\; \tfrac{73}{2} \;\Longrightarrow\; e^{-k t_{1/2}} \;=\; \tfrac{1}{2} \;\Longrightarrow\; t_{1/2} \;=\; \frac{\ln 2}{k}. $$ With $k \approx 0.0839$: $t_{1/2} \approx \tfrac{0.6931}{0.0839} \approx 8.26 \to 8.3$ minutes.
Why "excess" not "temperature." A common slip: trying to "halve" $T(t)$ directly. But $T \to T_{\text{env}} = 22^\circ$C as $t \to \infty$, not zero — so $T$ itself doesn't halve over time. The right invariant is the excess over the environment, $T - T_{\text{env}}$, which does decay exponentially to zero. The half-cooling formula $t_{1/2} = \ln 2/k$ is then the direct analogue of nuclear half-life.

(a) 边界条件 A1·A1

$T(0) = 22 + 73 \cdot e^{0} = 22 + 73 = 95 = T_{0}$。 ✓ 当 $t \to \infty$，$e^{-kt} \to 0$（$k > 0$），故 $T(t) \to 22 = T_{\text{env}}$。 ✓

(b) 求 $k$ M1·M1·A1·A1

由 $T(5) = 70$ 得 $22 + 73 e^{-5k} = 70$，即 $e^{-5k} = \tfrac{48}{73}$。取 $\ln$： $$ -5k \;=\; \ln(48/73) \;\stackrel{\text{GDC}}{\approx}\; -0.4193 \;\Longrightarrow\; k \;\approx\; 0.0839. $$

(c) $T(15)$ M1·A1·A1

$$ T(15) \;=\; 22 + 73\,e^{-15(0.0839)} \;=\; 22 + 73\,e^{-1.258} \;\stackrel{\text{GDC}}{\approx}\; 22 + 73(0.284) \;\approx\; 22 + 20.76 \;\approx\; 43^\circ\text{C}. $$

(d) $T = 40$ 的时间 M1·M1·A1

$22 + 73 e^{-kt} = 40 \Longrightarrow e^{-kt} = \tfrac{18}{73}$。取 $\ln$： $$ -kt \;=\; \ln(18/73) \;\stackrel{\text{GDC}}{\approx}\; -1.401 \;\Longrightarrow\; t \;=\; \tfrac{1.401}{0.0839} \;\approx\; 16.7 \;\to\; 17 \; \mathrm{min}. $$

(e) 半冷却时间 M1·A1·M1·A1

温度差为 $T(t) - T_{\text{env}} = 73 e^{-kt}$。令其等于初始差之半（$73/2$）： $$ 73\,e^{-k t_{1/2}} \;=\; \tfrac{73}{2} \;\Longrightarrow\; e^{-k t_{1/2}} \;=\; \tfrac{1}{2} \;\Longrightarrow\; t_{1/2} \;=\; \frac{\ln 2}{k}. $$ 取 $k \approx 0.0839$：$t_{1/2} \approx \tfrac{0.6931}{0.0839} \approx 8.26 \to 8.3$ 分钟。
为何看"温差"而不是"温度本身"。常见失误：尝试直接将 $T(t)$ 减半。但当 $t \to \infty$ 时 $T \to T_{\text{env}} = 22^\circ$C，并非 $0$，故 $T$ 本身不会减半。正确的不变量是相对于环境的温差 $T - T_{\text{env}}$，它指数衰减至零。半冷却时间公式 $t_{1/2} = \ln 2/k$ 正是核衰变半衰期的直接类比。