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Exponential and Logarithmic Functions · Solutions指数函数与对数函数 · 答案

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC styles练习题配套答案 · 按分逐步讲解 · SAT / AP 衔接 / 安大略 / 不列颠哥伦比亚省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 答案SAT MCQ + ON/BC short answer · 22 marksSAT 风格选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Exp Parent & Asymptote指数母函数与渐近线 · HSF-IF.C.7e [3 marks][3 分]

$y$-intercept and horizontal asymptote of $f(x) = 4 \cdot 2^{x}$.求 $f(x) = 4 \cdot 2^{x}$ 的 $y$ 轴截距与水平渐近线。

Answer:答案：  (B)  $y$-intercept $(0, 4)$; asymptote $y = 0$(B)  $y$ 轴截距 $(0, 4)$；渐近线 $y = 0$

(a) $y$-intercept by substituting $x = 0$代入 $x = 0$ 求 $y$ 轴截距 M1·A1

$f(0) = 4 \cdot 2^{0} = 4 \cdot 1 = 4$, so the $y$-intercept is $(0, 4)$. The coefficient out in front of $b^{x}$ is always the $y$-intercept of $a \cdot b^{x}$, because $b^{0} = 1$ for every base $b > 0$.$f(0) = 4 \cdot 2^{0} = 4 \cdot 1 = 4$，所以 $y$ 轴截距为 $(0, 4)$。$a \cdot b^{x}$ 中 $b^{x}$ 前面的系数恒为该函数的 $y$ 轴截距，因为对任意底数 $b > 0$ 都有 $b^{0} = 1$。

(b) Horizontal asymptote from the parent shape由母函数形状判断水平渐近线 A1

As $x \to -\infty$, $2^{x} \to 0^{+}$, so $4 \cdot 2^{x} \to 0$ as well. The horizontal asymptote is $y = 0$. (The vertical scaling by $4$ stretches the graph but does not move the asymptote: $4 \cdot 0 = 0$.) Matches option (B).当 $x \to -\infty$ 时，$2^{x} \to 0^{+}$，故 $4 \cdot 2^{x} \to 0$。水平渐近线为 $y = 0$。（纵向 $4$ 倍拉伸只改变曲线高度，不移动渐近线：$4 \cdot 0 = 0$。）对应选项 (B)。
Why the wrong choices fail.错误选项分析：
  • (A) $(0, 1)$; $y = 0$, forgets the leading coefficient $4$; this would be the intercept of the bare parent $2^{x}$.：忽略了前系数 $4$；这是裸母函数 $2^{x}$ 的截距。
  • (C) $(0, 4)$; $y = 4$, confuses the $y$-intercept with the asymptote. For $a \cdot b^{x}$ (no vertical shift) the asymptote is always $y = 0$; the constant $4$ multiplies, it does not translate.：混淆了 $y$ 轴截距与渐近线。无纵向平移的 $a \cdot b^{x}$ 渐近线恒为 $y = 0$；常数 $4$ 是乘法因子，不产生平移。
  • (D) $(0, 2)$; $x = 0$, reads "$2$" off the base instead of evaluating $f(0)$, and names the wrong axis: exponentials have a horizontal asymptote, not a vertical one.：直接从底数读出 "$2$"，未计算 $f(0)$；且方向写错：指数函数的渐近线是水平渐近线而非竖直渐近线。
$a \cdot b^{x}$: the $a$ is the $y$-intercept; the asymptote stays at $y = 0$ until you add a constant.$a \cdot b^{x}$：$a$ 即为 $y$ 轴截距；在加入常数前渐近线恒为 $y = 0$。 The only thing that moves the horizontal asymptote of an exponential is an added constant ($+ k$ at the end). Multiplying out in front by $a$ stretches the graph vertically but leaves the asymptote at $y = 0$, because $0$ times anything is still $0$. Lock this in early; downstream questions on $g(x) = 3 e^{2x} - 5$ (Q8) and continuous compounding (Q9) all chain off the same shape.能改变指数函数水平渐近线的唯一因素是末尾加上的常数（$+ k$）。前面乘以 $a$ 只纵向拉伸图像，不移动渐近线，因为 $0$ 乘以任何数仍为 $0$。务必早早记牢；后面的 $g(x) = 3 e^{2x} - 5$（Q8）与连续复利（Q9）都基于同一形状。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Same-Base Exp Eq同底指数方程 · HSF-LE.A.4 [3 marks][3 分]

Solve $9^{x + 1} = 27$ exactly.精确求解 $9^{x + 1} = 27$。

Answer:答案：  (A)  $x = 1/2$

(a) Rewrite both sides on the common base $3$两边化为相同底数 $3$ M1·A1

$9 = 3^{2}$ and $27 = 3^{3}$, so$9 = 3^{2}$，$27 = 3^{3}$，因此 $$ 9^{x + 1} \;=\; (3^{2})^{x + 1} \;=\; 3^{2(x + 1)} \;=\; 3^{2 x + 2}, \qquad 27 \;=\; 3^{3}. $$

(b) Equate exponents and solve指数相等并求解 A1

Since $3^{u}$ is one-to-one, $3^{2 x + 2} = 3^{3}$ forces $2 x + 2 = 3$, hence $2 x = 1$ and $x = \tfrac{1}{2}$. Matches option (A). (Sanity-check: $9^{1.5} = (3^{2})^{1.5} = 3^{3} = 27$. $\checkmark$)由于 $3^{u}$ 是一一映射，$3^{2 x + 2} = 3^{3}$ 给出 $2 x + 2 = 3$，从而 $2 x = 1$，$x = \tfrac{1}{2}$。对应选项 (A)。（验证：$9^{1.5} = (3^{2})^{1.5} = 3^{3} = 27$。$\checkmark$）
Why the wrong choices fail.错误选项分析：
  • (B) $x = 2$, likely the result of equating $9^{x + 1} = 27 \Rightarrow x + 1 = 3$ without rebasing $9$. This ignores that $9$ and $27$ are different powers of $3$.：可能直接令 $9^{x + 1} = 27 \Rightarrow x + 1 = 3$，未将 $9$ 换为同底，忽略了 $9$ 与 $27$ 是 $3$ 的不同次幂。
  • (C) $x = 3/2$, sign or chain slip: $2(x + 1) = 3 \Rightarrow x + 1 = 3/2 \Rightarrow x = 1/2$, but stopping at $3/2$ skips the final subtraction of $1$.：链式或符号失误：$2(x + 1) = 3 \Rightarrow x + 1 = 3/2 \Rightarrow x = 1/2$，但停在 $3/2$ 漏掉了最后减去 $1$。
  • (D) $x = -1/2$, sign-flip in the final step.：末步符号写反。
Same-base attack: rebuild both sides on the smallest prime base before equating exponents.同底法：先把两边都化为最小质数底数，再令指数相等。 $9 = 3^{2}$, $27 = 3^{3}$, $81 = 3^{4}$; $4 = 2^{2}$, $8 = 2^{3}$, $16 = 2^{4}$, $32 = 2^{5}$. The one-to-one property $b^{u} = b^{v} \Rightarrow u = v$ only fires once both sides share a base; never apply it before that. Same-base is preferred on a no-calculator paper because the answer is exact; reach for "take the log" (Q5) only when no common base exists.$9 = 3^{2}$、$27 = 3^{3}$、$81 = 3^{4}$；$4 = 2^{2}$、$8 = 2^{3}$、$16 = 2^{4}$、$32 = 2^{5}$。一一映射性质 $b^{u} = b^{v} \Rightarrow u = v$ 仅在两边底数相同后才能使用，绝不可提前应用。无计算器卷面优先用同底法，因为答案精确；当不存在公共底数时再使用"两边取对数"（见 Q5）。
Q3EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Log as Inverse对数即反函数 · HSF-BF.B.5 (+) [3 marks][3 分]

Equivalent of $\log_{5}(x) = 3$.$\log_{5}(x) = 3$ 的等价形式。

Answer:答案：  (B)  $5^{3} = x$

(a) Apply the definition of logarithm利用对数定义 M1·A1·A1

By definition, $\log_{b}(x) = y \iff b^{y} = x$ (for $b > 0$, $b \ne 1$, $x > 0$). With $b = 5$ and $y = 3$ this becomes $5^{3} = x$, i.e. $x = 125$. Matches option (B).由定义 $\log_{b}(x) = y \iff b^{y} = x$（$b > 0$，$b \ne 1$，$x > 0$）。取 $b = 5$、$y = 3$ 得 $5^{3} = x$，即 $x = 125$。对应选项 (B)。
Why the wrong choices fail.错误选项分析：
  • (A) $3^{5} = x$, swaps the base and the exponent. The base of the log $\log_{b}$ becomes the base of the exponential, not its exponent.：底数与指数交换。$\log_{b}$ 的底数应作指数式的底数，而非指数。
  • (C) $5 \cdot 3 = x$, treats $\log$ as multiplication. There is no "$\log$ times exponent" identity; this confuses the log with $\log_{b}(b) \cdot y$.：把对数当成乘法。不存在"$\log$ 乘以指数"的恒等式；这与 $\log_{b}(b) \cdot y$ 相混。
  • (D) $x^{3} = 5$, base and the unknown swapped, then exponent moved. Two errors compounded.：底数与未知数交换，又调换了指数。两处错误叠加。
$\log_{b}(\,\cdot\,)$ and $b^{\,\cdot\,}$ are inverse functions; the log answers "what exponent?"$\log_{b}(\,\cdot\,)$ 与 $b^{\,\cdot\,}$ 互为反函数；对数回答的是"指数是多少？" Read $\log_{b}(x) = y$ aloud as "the exponent you put on $b$ to get $x$ is $y$." The mnemonic base-to-the-answer-equals-argument ($b^{y} = x$) flips one definition into the other in one move. The conversion is the single most-used identity in this unit, it powers Q4 evaluations, Q5 different-base solving, Q7 log-equation extraction, and Q8(c) inverse construction.把 $\log_{b}(x) = y$ 读作"使 $b$ 升到某次幂得到 $x$ 的指数为 $y$"。记忆口诀底数^答案 = 真数（$b^{y} = x$）能一步互换两种定义形式。该转换是本单元最常用的恒等式，贯穿 Q4 求值、Q5 异底求解、Q7 对数方程剥离与 Q8(c) 反函数构造。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Laws of Logs对数运算律 · MHF4U Exp & Log指数与对数 [6 marks][6 分]

(a) Evaluate $\log_{2}(32) - \log_{2}(4)$. (b) Condense $2 \log_{3}(x) + \log_{3}(5) - \log_{3}(y)$. (c) Expand $\log_{10}\!\left(\tfrac{100 \, x^{3}}{\sqrt{y}}\right)$.(a) 求 $\log_{2}(32) - \log_{2}(4)$ 的值。(b) 合并 $2 \log_{3}(x) + \log_{3}(5) - \log_{3}(y)$。(c) 展开 $\log_{10}\!\left(\tfrac{100 \, x^{3}}{\sqrt{y}}\right)$。

Answer:答案：  (a) $3$  ·  (b) $\log_{3}\!\left(\dfrac{5 x^{2}}{y}\right)$  ·  (c) $2 + 3 \log_{10}(x) - \tfrac{1}{2} \log_{10}(y)$

(a) Quotient law, then evaluate用除法律再求值 M1·A1

Apply the quotient law $\log_{b}(M) - \log_{b}(N) = \log_{b}(M/N)$:应用除法律 $\log_{b}(M) - \log_{b}(N) = \log_{b}(M/N)$： $$ \log_{2}(32) - \log_{2}(4) \;=\; \log_{2}\!\left(\tfrac{32}{4}\right) \;=\; \log_{2}(8) \;=\; 3, $$ since $2^{3} = 8$. (Alternative direct read: $\log_{2}(32) = 5$, $\log_{2}(4) = 2$, $5 - 2 = 3$. $\checkmark$)因为 $2^{3} = 8$。（也可直接计算：$\log_{2}(32) = 5$，$\log_{2}(4) = 2$，$5 - 2 = 3$。$\checkmark$）

(b) Power, then product, then quotient先幂律、再乘法律、再除法律 M1·A1

Power law on the first term: $2 \log_{3}(x) = \log_{3}(x^{2})$.
Product law: $\log_{3}(x^{2}) + \log_{3}(5) = \log_{3}(5 x^{2})$.
Quotient law: $\log_{3}(5 x^{2}) - \log_{3}(y) = \log_{3}\!\left(\dfrac{5 x^{2}}{y}\right)$.先对首项用幂律：$2 \log_{3}(x) = \log_{3}(x^{2})$。
乘法律：$\log_{3}(x^{2}) + \log_{3}(5) = \log_{3}(5 x^{2})$。
除法律：$\log_{3}(5 x^{2}) - \log_{3}(y) = \log_{3}\!\left(\dfrac{5 x^{2}}{y}\right)$。

(c) Expand using all three laws综合运用三条运算律展开 M1·A1

$$ \log_{10}\!\left(\tfrac{100 \, x^{3}}{\sqrt{y}}\right) \;=\; \log_{10}(100) + \log_{10}(x^{3}) - \log_{10}\!\left(y^{1/2}\right) \;=\; 2 + 3 \log_{10}(x) - \tfrac{1}{2} \log_{10}(y). $$ ($\log_{10}(100) = 2$ since $10^{2} = 100$; the $\sqrt{y} = y^{1/2}$ rewrite is the move that unlocks the power law.)（$\log_{10}(100) = 2$，因为 $10^{2} = 100$；把 $\sqrt{y}$ 写成 $y^{1/2}$ 是激活幂律的关键步骤。）
The three log laws are the three exponent laws read backwards, and $\log(a + b) \ne \log a + \log b$ is the perennial trap.三条对数律即三条指数律的"反向读"，而 $\log(a + b) \ne \log a + \log b$ 是经典陷阱。 Product, quotient, and power laws all come from $b^{m} \cdot b^{n} = b^{m + n}$, $b^{m}/b^{n} = b^{m - n}$, $(b^{m})^{p} = b^{m p}$ flipped through the inverse $\log_{b}$. The single most-graded error in MHF4U: writing $\log(x + y) = \log x + \log y$. This is FALSE, the product law uses multiplication inside the log ($\log(xy)$), never addition. Habit move: never split a log across a $+$ or $-$ inside the argument; only across multiplication, division, or exponents.乘法律、除法律、幂律都由 $b^{m} \cdot b^{n} = b^{m + n}$、$b^{m}/b^{n} = b^{m - n}$、$(b^{m})^{p} = b^{m p}$ 通过反函数 $\log_{b}$ 翻转得来。MHF4U 中最常被扣分的错误：写出 $\log(x + y) = \log x + \log y$。这是错的，乘法律针对的是真数内部的乘法（$\log(xy)$），而非加法。习惯动作：真数内若出现 $+$ 或 $-$，绝不可拆开对数；只有乘法、除法、幂才能拆。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Different-Base Exp异底指数方程 · BC PC12 exp eqsBC PC12 指数方程 [7 marks][7 分]

(a) $3^{x} = 20$. (b) $5 \cdot 2^{x} = 90$. (c) $7^{x - 1} = 4^{x}$ with change-of-base.(a) $3^{x} = 20$。(b) $5 \cdot 2^{x} = 90$。(c) $7^{x - 1} = 4^{x}$，用换底公式。

Answer:答案：  (a) $x = \log_{3}(20) = \dfrac{\ln 20}{\ln 3} \approx 2.727$  ·  (b) $x = \log_{2}(18) = \dfrac{\ln 18}{\ln 2} \approx 4.170$  ·  (c) $x = \dfrac{\ln 7}{\ln 7 - \ln 4} = \dfrac{\ln 7}{\ln(7/4)} \approx 3.477$

(a) Take the natural log of both sides两边取自然对数 M1·A1·A1

Apply $\ln$ to both sides and use the power law:两边取 $\ln$ 并用幂律： $$ \ln(3^{x}) \;=\; \ln(20) \;\Longrightarrow\; x \ln 3 \;=\; \ln 20 \;\Longrightarrow\; x \;=\; \frac{\ln 20}{\ln 3}. $$ Numerically: $\ln 20 \approx 2.9957$, $\ln 3 \approx 1.0986$, so $x \approx 2.7268 \approx 2.727$. Equivalent exact form: $x = \log_{3}(20)$ (by change-of-base).数值：$\ln 20 \approx 2.9957$，$\ln 3 \approx 1.0986$，故 $x \approx 2.7268 \approx 2.727$。等价精确形式：$x = \log_{3}(20)$（由换底公式）。

(b) Isolate the exponential first, then take the log先孤立指数，再取对数 M1·A1

Divide by $5$: $2^{x} = 18$. Take $\ln$:两边除以 $5$：$2^{x} = 18$。取 $\ln$： $$ x \ln 2 \;=\; \ln 18 \;\Longrightarrow\; x \;=\; \frac{\ln 18}{\ln 2} \;=\; \log_{2}(18). $$ Numerically: $\ln 18 \approx 2.8904$, $\ln 2 \approx 0.6931$, so $x \approx 4.1699 \approx 4.170$.数值：$\ln 18 \approx 2.8904$，$\ln 2 \approx 0.6931$，故 $x \approx 4.1699 \approx 4.170$。

(c) Take $\ln$ of both sides; change-of-base falls out两边取 $\ln$，换底公式自然出现 M1·A1

$$ \ln(7^{x - 1}) \;=\; \ln(4^{x}) \;\Longrightarrow\; (x - 1) \ln 7 \;=\; x \ln 4. $$ Distribute and collect $x$-terms on one side:展开并把 $x$ 项移到同侧： $$ x \ln 7 - \ln 7 \;=\; x \ln 4 \;\Longrightarrow\; x (\ln 7 - \ln 4) \;=\; \ln 7 \;\Longrightarrow\; x \;=\; \frac{\ln 7}{\ln 7 - \ln 4} \;=\; \frac{\ln 7}{\ln(7/4)}. $$ Numerically: $\ln 7 \approx 1.9459$, $\ln(7/4) = \ln 1.75 \approx 0.5596$, so $x \approx 3.4773 \approx 3.477$. The change-of-base step is the move from $\log_{b}(a)$ to $\ln a / \ln b$, used implicitly to write the final form as a ratio of natural logs.数值：$\ln 7 \approx 1.9459$，$\ln(7/4) = \ln 1.75 \approx 0.5596$，故 $x \approx 3.4773 \approx 3.477$。换底公式即从 $\log_{b}(a)$ 写成 $\ln a / \ln b$ 的步骤，最终形式正是两个自然对数之比。
$\ln$ vs $\log_{10}$: it does not matter, pick one and stay.$\ln$ 还是 $\log_{10}$ 都行：选一个一直用即可。 Either log works on every step because change-of-base $\log_{b}(a) = \log_{c}(a)/\log_{c}(b)$ tells you the choice of base cancels in any ratio you build. Calculators have both $\ln$ and $\log$ keys, but $\ln$ is the cleaner default for AP Calculus continuity (derivative $\tfrac{d}{dx} \ln x = 1/x$ has no extra constant). Habit move: take $\ln$, never mix $\ln$ and $\log_{10}$ in the same chain. Common slip on part (c): cancelling $\ln$ across $\ln 7 - \ln 4$ to get $\ln 3$. WRONG, $\ln 7 - \ln 4 = \ln(7/4) \ne \ln 3$. The quotient law subtracts logs, it does not subtract arguments.每一步两种对数都可用，因为换底公式 $\log_{b}(a) = \log_{c}(a)/\log_{c}(b)$ 表明在任何比值中底数都会约掉。计算器同时有 $\ln$ 与 $\log$ 键，但 $\ln$ 更适配 AP Calculus（$\tfrac{d}{dx} \ln x = 1/x$ 无额外常数）。习惯动作：取 $\ln$ 即可，不要在同一题中混用 $\ln$ 与 $\log_{10}$。(c) 部分常见错误：把 $\ln 7 - \ln 4$ 当作 $\ln 3$。错的：$\ln 7 - \ln 4 = \ln(7/4) \ne \ln 3$。除法律是对数相减，不是真数相减。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 答案AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Change-of-Base换底公式 · HSF-BF.B.5 (+) [8 marks][8 分]

$L = \log_{4}(50)$. (a) Change-of-base form via $\ln$ + 4-dp value. (b) Prove $\log_{b}(x^{p}) = p \log_{b}(x)$ from definition. (c) Evaluate $\log_{4}(50) - \log_{4}(2) - \log_{4}(25)$.$L = \log_{4}(50)$。(a) 用 $\ln$ 写出换底形式并取四位小数。(b) 由定义证明 $\log_{b}(x^{p}) = p \log_{b}(x)$。(c) 计算 $\log_{4}(50) - \log_{4}(2) - \log_{4}(25)$。

Answer:答案：  (a) $L = \dfrac{\ln 50}{\ln 4} \approx 2.8219$  ·  (b) proof below见下证明  ·  (c) $0$

(a) Change-of-base via natural log用自然对数换底 M1·A1·A1

Apply change-of-base $\log_{b}(a) = \dfrac{\ln a}{\ln b}$ with $b = 4$, $a = 50$:取 $b = 4$、$a = 50$，应用换底公式 $\log_{b}(a) = \dfrac{\ln a}{\ln b}$： $$ L \;=\; \log_{4}(50) \;=\; \frac{\ln 50}{\ln 4}. $$ Numerically: $\ln 50 \approx 3.9120$, $\ln 4 \approx 1.3863$, so $L \approx 2.82193 \approx 2.8219$. (Cross-check: $4^{2.8219} = e^{2.8219 \ln 4} = e^{3.9120} \approx 50.00$. $\checkmark$)数值：$\ln 50 \approx 3.9120$，$\ln 4 \approx 1.3863$，故 $L \approx 2.82193 \approx 2.8219$。（验证：$4^{2.8219} = e^{2.8219 \ln 4} = e^{3.9120} \approx 50.00$。$\checkmark$）

(b) Proof of the power law幂律证明 M1·A1·A1

Let $y = \log_{b}(x^{p})$. By definition of logarithm, $b^{y} = x^{p}$.
Take both sides to the power $1/p$ (valid since $b > 0$ and $x > 0$): $b^{y/p} = x$.
By definition of logarithm again, $\log_{b}(x) = y/p$, i.e. $y = p \log_{b}(x)$. QED.
Alternative chain: let $u = \log_{b}(x)$, so $b^{u} = x$. Then $x^{p} = (b^{u})^{p} = b^{u p}$, and applying $\log_{b}$ to both sides gives $\log_{b}(x^{p}) = u p = p \log_{b}(x)$.设 $y = \log_{b}(x^{p})$。由对数定义，$b^{y} = x^{p}$。
两边同取 $1/p$ 次方（因 $b > 0$，$x > 0$，合法）：$b^{y/p} = x$。
再用对数定义：$\log_{b}(x) = y/p$，即 $y = p \log_{b}(x)$。证毕。
另一证法：令 $u = \log_{b}(x)$，则 $b^{u} = x$。于是 $x^{p} = (b^{u})^{p} = b^{u p}$，两边取 $\log_{b}$ 得 $\log_{b}(x^{p}) = u p = p \log_{b}(x)$。 AG

(c) Combine using product and quotient laws用乘法律与除法律合并 M1·A1

$$ \log_{4}(50) - \log_{4}(2) - \log_{4}(25) \;=\; \log_{4}\!\left(\frac{50}{2 \cdot 25}\right) \;=\; \log_{4}(1) \;=\; 0. $$ ($\log_{b}(1) = 0$ for every base $b$, since $b^{0} = 1$.)（对任意底数 $b$ 都有 $\log_{b}(1) = 0$，因为 $b^{0} = 1$。）
Change-of-base mechanics: pick any base $c$ you can compute, divide.换底机制：任选你能计算的底数 $c$，然后相除。 The full identity is $\log_{b}(a) = \dfrac{\log_{c}(a)}{\log_{c}(b)}$ for any $c > 0$, $c \ne 1$. Calculators give you two ready-made $c$'s: $e$ (the $\ln$ key) and $10$ (the $\log$ key). For exact-form provincial answers, leave as $\ln a / \ln b$; for AP free-response, the same answer can be typed into a calculator without intermediate rounding. Common slip: writing $\log_{4}(50) = \dfrac{\ln 4}{\ln 50}$, inverse of the correct ratio. The mnemonic top-is-argument, bottom-is-base matches the definition $b^{\text{answer}} = a$.完整恒等式：$\log_{b}(a) = \dfrac{\log_{c}(a)}{\log_{c}(b)}$，其中 $c > 0$，$c \ne 1$。计算器给你两个现成的 $c$：$e$（$\ln$ 键）与 $10$（$\log$ 键）。省考要精确形式时保留为 $\ln a / \ln b$；AP 简答题可直接代入计算器避免中途舍入。常见错误：把 $\log_{4}(50)$ 写成 $\dfrac{\ln 4}{\ln 50}$，分子分母颠倒。记忆口诀真数在上，底数在下，与定义 $b^{\text{答案}} = a$ 一致。
Q7HARD难 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Log Equations + Extraneous对数方程 + 增根 · MHF4U Exp & Log指数与对数 [9 marks][9 分]

(a) $\log_{2}(x) + \log_{2}(x - 2) = 3$. (b) $\log_{5}(2 x + 3) - \log_{5}(x - 1) = 1$. (c) Why is checking the domain required, not optional?(a) $\log_{2}(x) + \log_{2}(x - 2) = 3$。(b) $\log_{5}(2 x + 3) - \log_{5}(x - 1) = 1$。(c) 为何验证定义域是必需的、不可省略？

Answer:答案：  (a) $x = 4$  ·  (b) $x = \tfrac{8}{3}$  ·  (c) the equation $\log_{b}(M) = \log_{b}(N)$ is only equivalent to $M = N$ where both logs are defined, i.e. where $M, N > 0$; combining first can manufacture solutions outside the original domain.$\log_{b}(M) = \log_{b}(N)$ 与 $M = N$ 等价仅在两边对数都有定义（即 $M, N > 0$）的范围内成立；先合并可能制造出原始定义域之外的解。

(a) Combine, exponentiate, check domain合并、化为指数、验证定义域 M1·A1·A1·A1

Domain: $x > 0$ AND $x - 2 > 0$, so $x > 2$.
Apply the product law: $\log_{2}\bigl(x(x - 2)\bigr) = 3$. Convert to exponential form: $x(x - 2) = 2^{3} = 8$. Expand and rearrange:定义域：$x > 0$ 且 $x - 2 > 0$，即 $x > 2$。
用乘法律：$\log_{2}\bigl(x(x - 2)\bigr) = 3$。化为指数形式：$x(x - 2) = 2^{3} = 8$。展开重排： $$ x^{2} - 2 x - 8 \;=\; 0 \;\Longrightarrow\; (x - 4)(x + 2) \;=\; 0 \;\Longrightarrow\; x = 4 \;\text{or}\; x = -2. $$ Domain check:定义域检验： $x = 4 > 2$ $\checkmark$; $x = -2 \not> 2$, discard as extraneous (it makes $\log_{2}(x)$ undefined). Solution set: $\{4\}$.$x = 4 > 2$ $\checkmark$；$x = -2 \not> 2$，作为增根舍去（它使 $\log_{2}(x)$ 无定义）。解集：$\{4\}$。

(b) Combine, exponentiate, check domain合并、化为指数、验证定义域 M1·A1·A1

Domain: $2 x + 3 > 0$ AND $x - 1 > 0$, so $x > 1$.
Quotient law: $\log_{5}\!\left(\dfrac{2 x + 3}{x - 1}\right) = 1$. Exponentiate: $\dfrac{2 x + 3}{x - 1} = 5^{1} = 5$.
Cross-multiply (valid since $x > 1 \Rightarrow x - 1 > 0$):定义域：$2 x + 3 > 0$ 且 $x - 1 > 0$，即 $x > 1$。
除法律：$\log_{5}\!\left(\dfrac{2 x + 3}{x - 1}\right) = 1$。化为指数：$\dfrac{2 x + 3}{x - 1} = 5^{1} = 5$。
交叉相乘（因 $x > 1 \Rightarrow x - 1 > 0$，合法）： $$ 2 x + 3 \;=\; 5 (x - 1) \;=\; 5 x - 5 \;\Longrightarrow\; 8 \;=\; 3 x \;\Longrightarrow\; x \;=\; \tfrac{8}{3}. $$ Domain check:定义域检验： $x = \tfrac{8}{3} \approx 2.67 > 1$ $\checkmark$. Both arguments positive: $2(\tfrac{8}{3}) + 3 = \tfrac{25}{3} > 0$, $\tfrac{8}{3} - 1 = \tfrac{5}{3} > 0$. Solution: $x = \tfrac{8}{3}$.$x = \tfrac{8}{3} \approx 2.67 > 1$ $\checkmark$。两个真数均为正：$2(\tfrac{8}{3}) + 3 = \tfrac{25}{3} > 0$，$\tfrac{8}{3} - 1 = \tfrac{5}{3} > 0$。解：$x = \tfrac{8}{3}$。

(c) Why the domain check is mandatory为何定义域检验是必需的 R1·R1

The product/quotient laws $\log_{b}(M) + \log_{b}(N) = \log_{b}(MN)$ and $\log_{b}(M) - \log_{b}(N) = \log_{b}(M/N)$ are only valid on the intersection of the domains $M > 0$ AND $N > 0$. Combining first widens the algebra to allow solutions where the combined expression $\log_{b}(MN)$ is defined but the originals are not (e.g. $MN > 0$ from two negatives). Those candidates satisfy the combined equation but make the original undefined, hence "extraneous." Checking is required, not a courtesy.乘法律与除法律 $\log_{b}(M) + \log_{b}(N) = \log_{b}(MN)$ 与 $\log_{b}(M) - \log_{b}(N) = \log_{b}(M/N)$ 仅在 $M > 0$ 且 $N > 0$ 的定义域交集上成立。先合并会拓宽代数解集，使得合并后的 $\log_{b}(MN)$ 有定义但原对数无定义（例如两负数相乘也使 $MN > 0$）。这些候选满足合并后的方程，却使原方程无定义，因而是"增根"。验证是必要步骤，而非礼节性动作。
The domain restriction $\log_{b}(\text{arg}) > 0$, argument must be strictly positive, is non-negotiable.定义域要求 $\log_{b}(\text{真数}) > 0$（真数必须严格大于零）不可妥协。 Every log law preserves the equation only on the original domain. Habit move: write the domain inequalities first, before combining; then at the end intersect candidate solutions with that domain. In part (a) the quadratic gave two roots, $4$ and $-2$, and $-2$ secretly satisfies $\log_{2}\bigl((-2)(-2 - 2)\bigr) = \log_{2}(8) = 3$ (the combined form), but the originals $\log_{2}(-2)$ and $\log_{2}(-4)$ are both undefined over $\mathbb{R}$. Provincial markers split marks for stating the domain (R1) and for explicitly discarding the extraneous root (A1).每条对数律仅在原始定义域上保留方程等价性。习惯动作：合并前先写出定义域不等式；最后把候选解与该定义域取交集。(a) 中二次方程给出两根 $4$ 与 $-2$，$-2$ 暗中满足 $\log_{2}\bigl((-2)(-2 - 2)\bigr) = \log_{2}(8) = 3$（合并后形式），但原始 $\log_{2}(-2)$ 与 $\log_{2}(-4)$ 在 $\mathbb{R}$ 上都无定义。省考阅卷会单独给写出定义域 (R1) 与明确舍弃增根 (A1) 的分。
Q8HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 Base $e$ & Natural Log底数 $e$ 与自然对数 · BC PC12 / HSF-LE.A.4 [10 marks][10 分]

$g(x) = 3 e^{2 x} - 5$. (a) HA and $y$-intercept. (b) Solve $g(x) = 16$. (c) $g^{-1}(x)$ + domain/range. (d) BC PC12 Big Idea on inverses.$g(x) = 3 e^{2 x} - 5$。(a) 水平渐近线与 $y$ 轴截距。(b) 解 $g(x) = 16$。(c) $g^{-1}(x)$ 及其定义域 / 值域。(d) BC PC12 反函数大概念。

Answer:答案：  (a) HA $y = -5$, $y$-int $(0, -2)$渐近线 $y = -5$，$y$ 轴截距 $(0, -2)$  ·  (b) $x = \tfrac{1}{2} \ln 7 \approx 0.973$  ·  (c) $g^{-1}(x) = \tfrac{1}{2} \ln\!\left(\tfrac{x + 5}{3}\right)$, $D_{g^{-1}} = (-5, \infty)$, $R_{g^{-1}} = \mathbb{R}$  ·  (d) see (d) below见下方 (d)

(a) Asymptote and intercept渐近线与截距 A1·A1

As $x \to -\infty$, $e^{2 x} \to 0^{+}$, so $g(x) \to 3 (0) - 5 = -5$. Horizontal asymptote: $y = -5$.
$y$-intercept: $g(0) = 3 e^{0} - 5 = 3 - 5 = -2$, so $(0, -2)$. (The $-5$ outside the exponential shifts the parent's asymptote $y = 0$ down to $y = -5$, Q1 insight applied with a vertical shift.)当 $x \to -\infty$ 时，$e^{2 x} \to 0^{+}$，故 $g(x) \to 3 (0) - 5 = -5$。水平渐近线：$y = -5$。
$y$ 轴截距：$g(0) = 3 e^{0} - 5 = 3 - 5 = -2$，即 $(0, -2)$。（指数外面的 $-5$ 把母函数的渐近线 $y = 0$ 平移至 $y = -5$，即 Q1 的结论加上纵向平移。）

(b) Solve $g(x) = 16$解 $g(x) = 16$ M1·A1·A1

$$ 3 e^{2 x} - 5 \;=\; 16 \;\Longrightarrow\; 3 e^{2 x} \;=\; 21 \;\Longrightarrow\; e^{2 x} \;=\; 7. $$ Take $\ln$: $2 x = \ln 7$, so $x = \tfrac{1}{2} \ln 7$. Numerically: $\ln 7 \approx 1.9459$, $x \approx 0.97296 \approx 0.973$.取 $\ln$：$2 x = \ln 7$，即 $x = \tfrac{1}{2} \ln 7$。数值：$\ln 7 \approx 1.9459$，$x \approx 0.97296 \approx 0.973$。

(c) Inverse function with domain/range反函数及其定义域 / 值域 M1·A1·A1

Write $y = 3 e^{2 x} - 5$ and swap $x \leftrightarrow y$: $x = 3 e^{2 y} - 5$. Solve for $y$:设 $y = 3 e^{2 x} - 5$，交换 $x \leftrightarrow y$：$x = 3 e^{2 y} - 5$。解 $y$： $$ x + 5 \;=\; 3 e^{2 y} \;\Longrightarrow\; \frac{x + 5}{3} \;=\; e^{2 y} \;\Longrightarrow\; 2 y \;=\; \ln\!\left(\frac{x + 5}{3}\right) \;\Longrightarrow\; y \;=\; \tfrac{1}{2} \ln\!\left(\frac{x + 5}{3}\right). $$ So $g^{-1}(x) = \tfrac{1}{2} \ln\!\left(\dfrac{x + 5}{3}\right)$. The domain of $g^{-1}$ equals the range of $g$: since $g(x) > -5$ for all real $x$ and $g(x) \to \infty$ as $x \to \infty$, range of $g$ is $(-5, \infty)$. Hence $D_{g^{-1}} = (-5, \infty)$, $R_{g^{-1}} = \mathbb{R}$ (the domain of $g$). Sanity-check using (b): $g^{-1}(16) = \tfrac{1}{2} \ln(21/3) = \tfrac{1}{2} \ln 7$. $\checkmark$于是 $g^{-1}(x) = \tfrac{1}{2} \ln\!\left(\dfrac{x + 5}{3}\right)$。$g^{-1}$ 的定义域等于 $g$ 的值域：对所有实数 $x$ 都有 $g(x) > -5$，且当 $x \to \infty$ 时 $g(x) \to \infty$，故 $g$ 的值域为 $(-5, \infty)$。因此 $D_{g^{-1}} = (-5, \infty)$，$R_{g^{-1}} = \mathbb{R}$（即 $g$ 的定义域）。用 (b) 验证：$g^{-1}(16) = \tfrac{1}{2} \ln(21/3) = \tfrac{1}{2} \ln 7$。$\checkmark$

(d) BC PC12 inverse Big IdeaBC PC12 反函数大概念 R1·R1

BC PC12 Big Idea:BC PC12 大概念： "Using inverses is the foundation of solving equations." Solving $g(x) = 16$ is precisely an instance: $g$ is a composition of multiplication-by-$3$, addition-of-$-5$, and the exponential $e^{2 x}$. To isolate $x$ we undo each operation in reverse, add $5$, divide by $3$, apply the inverse of $e^{2 x}$ (which is $\tfrac{1}{2} \ln$). The step $e^{2 x} = 7 \Rightarrow 2 x = \ln 7$ is the application of the inverse function, the natural log is what undoes $e^{\,\cdot\,}$."使用反函数是解方程的基础。"解 $g(x) = 16$ 正是该概念的实例：$g$ 是"乘以 $3$"、"加 $-5$"和指数 $e^{2 x}$ 的复合。要孤立 $x$，需按相反顺序逐步逆运算：加 $5$、除以 $3$、再应用 $e^{2 x}$ 的反函数（即 $\tfrac{1}{2} \ln$）。$e^{2 x} = 7 \Rightarrow 2 x = \ln 7$ 这一步就是反函数的应用，自然对数正是抵消 $e^{\,\cdot\,}$ 的工具。
Base $e$ is the natural base for inverse-of-exponential because $\ln$ undoes $e^{\,\cdot\,}$ cleanly.底数 $e$ 是指数反函数的自然底，因为 $\ln$ 干净地抵消 $e^{\,\cdot\,}$。 Continuous-compounding form $P e^{r t}$ (Q9) and natural decay $M_{0} e^{-k t}$ both invert to $t = \tfrac{1}{r} \ln(\,\cdot\,)$ in one step. Other bases work, you could solve $3 \cdot 10^{x} - 5 = 16$ with $\log_{10}$, but $e$ matches the AP Calculus derivative $\tfrac{d}{dx} e^{x} = e^{x}$ and $\tfrac{d}{dx} \ln x = 1/x$, with no extra constants. Habit move on inverses: peel from the outside in, applying the inverse of each operation in reverse order. The last operation to be applied to $x$ is the first to be undone.连续复利形式 $P e^{r t}$（Q9）与自然衰减 $M_{0} e^{-k t}$ 都可一步反演为 $t = \tfrac{1}{r} \ln(\,\cdot\,)$。其他底数也行（如用 $\log_{10}$ 解 $3 \cdot 10^{x} - 5 = 16$），但 $e$ 与 AP Calculus 的求导公式 $\tfrac{d}{dx} e^{x} = e^{x}$ 与 $\tfrac{d}{dx} \ln x = 1/x$ 相符，无额外常数。反函数的习惯动作：由外向内剥，按相反顺序对每个运算应用反函数。最后施加于 $x$ 的运算最先被消去。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §6 Continuous Compounding连续复利 · BC PC12 base $e$BC PC12 底数 $e$ / HSF-LE.B.5 [9 marks][9 分]

$P(t) = P_{0} e^{0.045 t}$. (a) Balance on \$2{,}000 after 10 yr. (b) Doubling time $T$. (c) Interpret $P_{0}$ and $0.045$ via HSF-LE.B.5. (d) Continuous vs. annually-compounded after 10 yr, by how much.$P(t) = P_{0} e^{0.045 t}$。(a) \$2{,}000 经 10 年后的余额。(b) 倍增时间 $T$。(c) 用 HSF-LE.B.5 解释 $P_{0}$ 与 $0.045$。(d) 连续复利与年复利在 10 年后的余额差。

Answer:答案：  (a) \$3{,}136.62  ·  (b) $T = \tfrac{\ln 2}{0.045} \approx 15.40$ yr年  ·  (c) $P_{0}$ = initial principal; $0.045$ = continuous nominal rate$P_{0}$ = 初始本金；$0.045$ = 连续名义利率  ·  (d) continuous wins by \$30.68连续复利多 \$30.68

(a) Balance after 10 years10 年后的余额 M1·A1

$P(10) = 2000 \cdot e^{0.045 \cdot 10} = 2000 \cdot e^{0.45}$. Compute $e^{0.45} \approx 1.568312$, so$P(10) = 2000 \cdot e^{0.045 \cdot 10} = 2000 \cdot e^{0.45}$。计算 $e^{0.45} \approx 1.568312$，所以 $$ P(10) \;\approx\; 2000 \cdot 1.568312 \;\approx\; 3136.62. $$ Balance: \$3{,}136.62.余额：\$3{,}136.62。

(b) Doubling time倍增时间 M1·A1·A1

Set $P(T) = 2 P_{0}$: $P_{0} e^{0.045 T} = 2 P_{0}$, divide by $P_{0}$ (note $P_{0}$ drops, the doubling time is principal-independent), giving $e^{0.045 T} = 2$. Take $\ln$:令 $P(T) = 2 P_{0}$：$P_{0} e^{0.045 T} = 2 P_{0}$，两边除以 $P_{0}$（注意 $P_{0}$ 被消去，倍增时间与本金无关），得 $e^{0.045 T} = 2$。取 $\ln$： $$ 0.045 T \;=\; \ln 2 \;\Longrightarrow\; T \;=\; \frac{\ln 2}{0.045}. $$ Numerically: $\ln 2 \approx 0.69315$, $T \approx 15.4033 \approx 15.40$ yr. (This is the continuous-time analogue of the "Rule of 72": $72/4.5 = 16$, a usable mental estimate.)数值：$\ln 2 \approx 0.69315$，$T \approx 15.4033 \approx 15.40$ 年。（这是"72 法则"的连续时间类比：$72/4.5 = 16$，可用作心算估计。）

(c) Parameter interpretation (HSF-LE.B.5)参数解释（HSF-LE.B.5） R1·R1

$P_{0}$ is the value of $P$ at $t = 0$, the initial principal deposited, in dollars. $0.045$ is the continuous nominal annual rate ($4.5\%$ per year), the constant rate of relative growth: $\tfrac{1}{P}\dfrac{d P}{d t} = 0.045$. As HSF-LE.B.5 directs, the constant in the exponent encodes the rate, and the multiplier out front encodes the initial value of the modeled quantity.是 $t = 0$ 时 $P$ 的值，即存入的初始本金（美元）。$0.045$ 是连续名义年利率（每年 $4.5\%$），即相对增长的恒定速率：$\tfrac{1}{P}\dfrac{d P}{d t} = 0.045$。正如 HSF-LE.B.5 所指，指数中的常数表征利率，前面的系数表征所建模量的初始值。

(d) Continuous vs. annually compounded连续复利与年复利的比较 M1·A1

Annual compounding formula: $P_{\text{ann}}(t) = P_{0} (1 + r)^{t}$ with $r = 0.045$.
$P_{\text{ann}}(10) = 2000 (1.045)^{10} \approx 2000 \cdot 1.552969 \approx 3105.94$. Balance: \$3{,}105.94.
Continuous exceeds annual by $3136.62 - 3105.94 = \$30.68$. (Continuous compounding always wins for $r > 0$: $e^{r t} > (1 + r)^{t}$ for $t > 0$, since $e^{r} > 1 + r$ by the Taylor series.)年复利公式：$P_{\text{ann}}(t) = P_{0} (1 + r)^{t}$，其中 $r = 0.045$。
$P_{\text{ann}}(10) = 2000 (1.045)^{10} \approx 2000 \cdot 1.552969 \approx 3105.94$。余额：\$3{,}105.94。
连续复利比年复利多 $3136.62 - 3105.94 = \$30.68$。（对 $r > 0$ 连续复利永远胜出：当 $t > 0$ 时 $e^{r t} > (1 + r)^{t}$，因为 Taylor 级数给出 $e^{r} > 1 + r$。）
Continuous form $P e^{r t}$ vs discrete form $P(1 + r/n)^{n t}$, and why the discrete tends to the continuous as $n \to \infty$.连续形式 $P e^{r t}$ 与离散形式 $P(1 + r/n)^{n t}$ 的对比，以及为何当 $n \to \infty$ 时离散趋近连续。 Annual compounding is the $n = 1$ case of the discrete formula: $(1 + r)^{t}$. Monthly is $n = 12$. As $n \to \infty$ (compounding "every instant"), the discrete formula converges to $e^{r t}$ by the limit $\lim_{n \to \infty} (1 + r/n)^{n t} = e^{r t}$, this is the definition of $e$ unpacked. The gap between continuous and any finite-$n$ compounding shrinks as $n$ grows: monthly is much closer to continuous than annual. AP Calculus picks this up later as the link between exponential growth modeled by an ODE ($P' = r P$) and exponential growth modeled by a recursion ($P_{n+1} = (1 + r) P_{n}$).年复利是离散公式的 $n = 1$ 情形：$(1 + r)^{t}$。月复利对应 $n = 12$。当 $n \to \infty$（"每瞬间"复利）时，离散公式由极限 $\lim_{n \to \infty} (1 + r/n)^{n t} = e^{r t}$ 收敛到 $e^{r t}$，这就是 $e$ 的定义。连续与任何有限 $n$ 复利之间的差距随 $n$ 增大而缩小：月复利远比年复利接近连续。AP Calculus 后续将其作为用微分方程 ($P' = r P$) 建模与用递推 ($P_{n+1} = (1 + r) P_{n}$) 建模指数增长之间的桥梁。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 答案Universal · 32 marks通用题型 · 共 32 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Population Growth人口增长 · HSF-LE.A.1c / HSF-LE.A.2 [10 marks][10 分]

Town population $2.4\%$/yr, $P(2020) = 48{,}000$. (a) $P(t) = a \cdot b^{t}$. (b) $P(10)$. (c) First year $P > 75{,}000$. (d) Cite HSF-LE.A.1c.城镇人口每年增长 $2.4\%$，$P(2020) = 48{,}000$。(a) $P(t) = a \cdot b^{t}$。(b) $P(10)$。(c) 首次 $P > 75{,}000$ 的年份。(d) 引用 HSF-LE.A.1c。

Answer:答案：  (a) $a = 48{,}000$, $b = 1.024$  ·  (b) $P(10) \approx 60{,}847$  ·  (c) $t = 19$, year $2039$$t = 19$，即 $2039$ 年  ·  (d) equal percent-per-year is the defining feature of exponential growth (vs. equal amount-per-year for linear).每年相同百分率是指数增长的定义性特征（线性增长则是每年相同的绝对量）。

(a) Build the explicit model建立显式模型 M1·A1·A1

$a = P(0) = 48{,}000$, the initial value in $2020$. The annual growth factor is $b = 1 + 0.024 = 1.024$ (each year the population is multiplied by $1.024$). Model:$a = P(0) = 48{,}000$，即 $2020$ 年的初始值。年增长倍数为 $b = 1 + 0.024 = 1.024$（每年人口乘以 $1.024$）。模型： $$ P(t) \;=\; 48{,}000 \cdot (1.024)^{t}. $$

(b) Population in 20302030 年的人口 M1·A1

$t = 10$ years after $2020$:距 $2020$ 年 $t = 10$ 年： $$ P(10) \;=\; 48{,}000 \cdot (1.024)^{10} \;\approx\; 48{,}000 \cdot 1.267651 \;\approx\; 60{,}847.24, $$ i.e. approximately $60{,}847$ people. (Round down, population counts whole people.)约为 $60{,}847$ 人。（向下取整，人口按整数计。）

(c) First year exceeding 75{,}000首次超过 $75{,}000$ 的年份 M1·A1·A1

Solve $48{,}000 \cdot (1.024)^{t} > 75{,}000$. Divide:求解 $48{,}000 \cdot (1.024)^{t} > 75{,}000$。两边除以 $48{,}000$： $$ (1.024)^{t} \;>\; \frac{75{,}000}{48{,}000} \;=\; 1.5625. $$ Take $\ln$ (both sides positive, $\ln$ strictly increasing):两边取 $\ln$（两侧均为正，$\ln$ 严格递增）： $$ t \cdot \ln(1.024) \;>\; \ln(1.5625) \;\Longrightarrow\; t \;>\; \frac{\ln 1.5625}{\ln 1.024}. $$ Numerically: $\ln 1.5625 \approx 0.44629$, $\ln 1.024 \approx 0.023717$, so $t > 18.819$. Round up to the next whole year: $t = 19$, i.e. year $2020 + 19 = 2039$.数值：$\ln 1.5625 \approx 0.44629$，$\ln 1.024 \approx 0.023717$，故 $t > 18.819$。向上取整到下一整年：$t = 19$，即$2020 + 19 = 2039$ 年。

(d) Cite HSF-LE.A.1c引用 HSF-LE.A.1c R1·R1

HSF-LE.A.1c: "Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another." Because the town grows by a constant percent ($2.4\%$) per year, not a constant amount, each year's increase is proportional to the current population, which is the defining feature of exponential (multiplicative) rather than linear (additive) growth. Linear growth would say "grows by $1{,}152$ people per year" forever; exponential says "grows by $2.4\%$ of whatever the current population is."："识别一个量在每个单位间隔内以恒定百分率相对于另一量增长或衰减的情境。"由于城镇每年以恒定百分率（$2.4\%$）增长，而非恒定数量，每年的增长量与当年人口成正比，这正是指数（乘性）增长而非线性（加性）增长的定义性特征。线性增长会说"每年增加 $1{,}152$ 人"；指数增长则说"每年按当前人口的 $2.4\%$ 增长"。
Growth factor vs growth rate: $b = 1 + r$, with $r$ as a decimal.增长倍数 vs 增长率：$b = 1 + r$，其中 $r$ 为小数。 A percent rate of $r\%$ per year gives a growth factor $b = 1 + r/100$ (here $1 + 0.024 = 1.024$). Decay would be $b = 1 - r/100 < 1$. The fast diagnostic: $b > 1$ means growth, $b < 1$ means decay, $b = 1$ means constant. Common slip: using $b = 0.024$ (the rate, not the factor) and getting $P(10) \approx 48000 \cdot (0.024)^{10} \approx 0$, a tell that the student confused rate with factor. Always check that your model produces the original value at $t = 0$.每年百分率 $r\%$ 给出增长倍数 $b = 1 + r/100$（此处 $1 + 0.024 = 1.024$）。衰减则 $b = 1 - r/100 < 1$。快速判别：$b > 1$ 为增长，$b < 1$ 为衰减，$b = 1$ 为不变。常见错误：用 $b = 0.024$（这是利率，不是倍数）会得到 $P(10) \approx 48000 \cdot (0.024)^{10} \approx 0$，一看便知混淆了利率与倍数。模型代入 $t = 0$ 必须复现初始值。
Q11MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §7 Radioactive Half-Life放射性半衰期 · MHF4U Exp & Log指数与对数 [11 marks][11 分]

I-131, half-life $8.02$ d, initial $40$ mg. (a) Half-life form. (b) Mass after 30 d. (c) Time to $5$ mg. (d) Practical domain + growth/decay. (e) One modelling assumption.I-131，半衰期 $8.02$ 天，初始 $40$ mg。(a) 半衰期形式。(b) 30 天后的质量。(c) 衰减到 $5$ mg 的时间。(d) 实际定义域 + 判别增长 / 衰减。(e) 一个建模假设。

Answer:答案：  (a) $M(t) = 40 \cdot \left(\tfrac{1}{2}\right)^{t/8.02}$, $M_{0} = 40$, $h = 8.02$  ·  (b) $\approx 3.00$ mg  ·  (c) $\approx 24.1$ d天  ·  (d) $t \in [0, \infty)$; exponential decay$t \in [0, \infty)$；指数衰减  ·  (e) sample isolated, no replenishment, constant temperature, no measurement noise (any one).样本隔离、无补给、温度恒定、无测量噪声（任选其一）。

(a) Half-life form半衰期形式 A1·A1

The half-life form $M(t) = M_{0} \left(\tfrac{1}{2}\right)^{t/h}$ encodes "after one half-life $h$, the mass is multiplied by $1/2$." With initial mass $M_{0} = 40$ mg and $h = 8.02$ d:半衰期形式 $M(t) = M_{0} \left(\tfrac{1}{2}\right)^{t/h}$ 表示"经过一个半衰期 $h$ 后质量乘以 $1/2$"。初始质量 $M_{0} = 40$ mg，$h = 8.02$ 天： $$ M(t) \;=\; 40 \cdot \left(\tfrac{1}{2}\right)^{t/8.02}. $$

(b) Mass after 30 days30 天后的质量 M1·A1

$$ M(30) \;=\; 40 \cdot \left(\tfrac{1}{2}\right)^{30/8.02} \;=\; 40 \cdot \left(\tfrac{1}{2}\right)^{3.7406}. $$ Numerically: $\left(\tfrac{1}{2}\right)^{3.7406} = 2^{-3.7406} \approx 0.074897$, so $M(30) \approx 40 \cdot 0.074897 \approx 2.996 \approx 3.00$ mg.数值：$\left(\tfrac{1}{2}\right)^{3.7406} = 2^{-3.7406} \approx 0.074897$，故 $M(30) \approx 40 \cdot 0.074897 \approx 2.996 \approx 3.00$ mg。

(c) Time to decay to 5 mg衰减到 $5$ mg 所需时间 M1·A1·A1·A1

Set $M(t) = 5$ and solve:令 $M(t) = 5$ 求解： $$ 40 \cdot \left(\tfrac{1}{2}\right)^{t/8.02} \;=\; 5 \;\Longrightarrow\; \left(\tfrac{1}{2}\right)^{t/8.02} \;=\; \tfrac{5}{40} \;=\; \tfrac{1}{8}. $$ Since $\tfrac{1}{8} = \left(\tfrac{1}{2}\right)^{3}$ (same-base attack works here): $\dfrac{t}{8.02} = 3$, so $t = 3 \cdot 8.02 = 24.06 \approx 24.1$ d.
Alternative via $\ln$ (general method):由于 $\tfrac{1}{8} = \left(\tfrac{1}{2}\right)^{3}$（此处可用同底法）：$\dfrac{t}{8.02} = 3$，所以 $t = 3 \cdot 8.02 = 24.06 \approx 24.1$ 天。
用 $\ln$ 的通法： $$ \tfrac{t}{8.02} \cdot \ln\!\left(\tfrac{1}{2}\right) \;=\; \ln\!\left(\tfrac{1}{8}\right) \;\Longrightarrow\; t \;=\; 8.02 \cdot \frac{\ln(1/8)}{\ln(1/2)} \;=\; 8.02 \cdot \frac{-\ln 8}{-\ln 2} \;=\; 8.02 \cdot \frac{3 \ln 2}{\ln 2} \;=\; 24.06. $$ So $t \approx 24.1$ d.故 $t \approx 24.1$ 天。

(d) Practical domain and growth/decay实际定义域与增长 / 衰减判别 A1·A1

Practical domain: $t \ge 0$, the model starts at the moment the sample is prepared; negative time is not in scope. Write as $[0, \infty)$ (the sample never fully vanishes mathematically, only asymptotically).
Since $b = 1/2 < 1$, the model is exponential decay.实际定义域：$t \ge 0$，模型从样本制备的时刻开始；负时间不在情境范围内。写作 $[0, \infty)$（样本在数学上永远不会完全消失，只是渐近趋近于零）。
由于 $b = 1/2 < 1$，本模型为指数衰减。

(e) One modelling assumption一个建模假设 R1

Any one of: the sample is isolated (no influx of new I-131 from another source); decay is purely first-order with a constant decay rate (no daughter-product feedback); environmental conditions (temperature, pressure) do not alter the decay constant; measurement is continuous and exact. In practice, none of these hold exactly, radiometric assays have measurement noise, samples are contaminated, and "half-life $= 8.02$ days" is itself an average measured to limited precision.任选其一：样本是隔离的（无外来 I-131 补充）；衰减是纯一阶且衰减常数不变（无子产物反馈）；环境条件（温度、压力）不影响衰减常数；测量连续且精确。实际中无一严格成立，放射性测量有噪声、样本会被污染，"半衰期 $= 8.02$ 天"本身也是有限精度下的平均值。
Half-life form derivation: $M(t) = M_{0} e^{-k t}$ with $k = (\ln 2) / h$.半衰期形式推导：$M(t) = M_{0} e^{-k t}$，其中 $k = (\ln 2) / h$。 Both forms describe the same decay; same-base ($1/2$) keeps the half-life $h$ visible, while base $e$ keeps the calculus visible. Conversion: $\left(\tfrac{1}{2}\right)^{t/h} = e^{(t/h) \ln(1/2)} = e^{-t (\ln 2)/h}$, hence the decay constant $k = (\ln 2)/h$. For I-131: $k = (\ln 2)/8.02 \approx 0.0864$ per day, so equivalently $M(t) = 40 e^{-0.0864 t}$. Habit move on AP Calc / BC PC12 papers: use the half-life form when the half-life is given (Q11, decay), and the $P e^{r t}$ form when a continuous rate is given (Q9, growth). Both are correct; the choice is about which constants are already known.两种形式描述同一衰减；同底 ($1/2$) 形式让半衰期 $h$ 一目了然，底数 $e$ 形式让微积分直观。互换：$\left(\tfrac{1}{2}\right)^{t/h} = e^{(t/h) \ln(1/2)} = e^{-t (\ln 2)/h}$，故衰减常数 $k = (\ln 2)/h$。对 I-131：$k = (\ln 2)/8.02 \approx 0.0864$ 每天，等价地 $M(t) = 40 e^{-0.0864 t}$。AP Calc / BC PC12 卷面习惯：已知半衰期就用半衰期形式（Q11，衰减），已知连续利率就用 $P e^{r t}$ 形式（Q9，增长）。两者均正确；选择取决于已知哪些常数。
Q12HARD难 🇨🇦 BC卑 🇺🇸 US美 BC Provincial-style卑诗省考风格 §7 Compound Interest复利 · BC PC12 / HSF-LE.A.3 [11 marks][11 分]

Option A: $A(t) = 5000(1 + 0.06/12)^{12 t}$. Option B: $B(t) = 5000 + 320 t$. (a) $A(5)$, $B(5)$, larger and gap. (b) Doubling time of A. (c) HSF-LE.A.3 + first year $A > B$. (d) Situational restriction + reason to prefer B.方案 A：$A(t) = 5000(1 + 0.06/12)^{12 t}$。方案 B：$B(t) = 5000 + 320 t$。(a) $A(5)$、$B(5)$，谁更大及差额。(b) 方案 A 的倍增时间。(c) HSF-LE.A.3 + $A > B$ 的首个年份。(d) 情境限制 + 选 B 的理由。

Answer:答案：  (a) $A(5) \approx \$6{,}744.25$, $B(5) = \$6{,}600.00$; A larger by \$144.25$A(5) \approx \$6{,}744.25$，$B(5) = \$6{,}600.00$；A 多 \$144.25  ·  (b) $T = \dfrac{\ln 2}{12 \ln(1.005)} \approx 11.58$ yr年  ·  (c) exponential eventually dominates linear; first $A > B$ at $t = 1$ (see table)指数终将超过线性；$A > B$ 首次出现在 $t = 1$（见表）  ·  (d) e.g. $0 \le t \le T_{\text{maturity}}$; B preferred when liquidity / capital protection is paramount.如 $0 \le t \le T_{\text{到期}}$；流动性 / 本金保护优先时偏好 B。

(a) Five-year balances五年余额 M1·A1·A1

Option A monthly rate: $0.06/12 = 0.005$ per month, total $60$ compounding periods:方案 A 月利率：$0.06/12 = 0.005$ 每月，共 $60$ 个计息期： $$ A(5) \;=\; 5000 (1.005)^{60} \;\approx\; 5000 \cdot 1.348850 \;\approx\; 6744.25. $$ Option B linear: $B(5) = 5000 + 320 \cdot 5 = 5000 + 1600 = 6600.00$.
Difference: $A(5) - B(5) \approx 6744.25 - 6600.00 = \$144.25$. Option A is larger by about \$144.25 after 5 years.方案 B 线性：$B(5) = 5000 + 320 \cdot 5 = 5000 + 1600 = 6600.00$。
差额：$A(5) - B(5) \approx 6744.25 - 6600.00 = \$144.25$。5 年后方案 A 多约 \$144.25。

(b) Doubling time of Option A方案 A 的倍增时间 M1·A1·A1

Set $A(T) = 10{,}000$:令 $A(T) = 10{,}000$： $$ 5000 (1.005)^{12 T} \;=\; 10{,}000 \;\Longrightarrow\; (1.005)^{12 T} \;=\; 2. $$ Take $\ln$: $12 T \cdot \ln(1.005) = \ln 2$, hence取 $\ln$：$12 T \cdot \ln(1.005) = \ln 2$，故 $$ T \;=\; \frac{\ln 2}{12 \ln(1.005)}. $$ Numerically: $\ln 2 \approx 0.69315$, $\ln 1.005 \approx 0.0049875$, so $12 \ln 1.005 \approx 0.05985$. Then $T \approx 0.69315 / 0.05985 \approx 11.582 \approx 11.58$ yr. (Rule-of-72 sanity check: $72/6 = 12$, close.)数值：$\ln 2 \approx 0.69315$，$\ln 1.005 \approx 0.0049875$，故 $12 \ln 1.005 \approx 0.05985$。$T \approx 0.69315 / 0.05985 \approx 11.582 \approx 11.58$ 年。（72 法则验算：$72/6 = 12$，接近。）

(c) HSF-LE.A.3 + first $A > B$HSF-LE.A.3 + $A > B$ 的首个时点 M1·A1·A1

HSF-LE.A.3: "Observe… that a quantity increasing exponentially eventually exceeds a quantity increasing… as a polynomial function." Because $A$ grows by a multiplicative factor each compounding period while $B$ grows by a fixed additive amount, $A(t)/B(t) \to \infty$ as $t \to \infty$, so $A$ must overtake $B$ at some finite $t$, no matter the initial gap. To find it, tabulate (both start at $5000$ when $t = 0$):："观察指数增长的量最终会超过多项式增长的量。"由于 $A$ 每个计息期按乘性因子增长而 $B$ 按固定加性量增长，当 $t \to \infty$ 时 $A(t)/B(t) \to \infty$，因此无论初始差距多大，$A$ 必在某个有限 $t$ 处赶超 $B$。列表查找（$t = 0$ 时两者都是 $5000$）：
  • $t = 0$: $A = 5000$, $B = 5000$, equal.$t = 0$：$A = 5000$，$B = 5000$，相等。
  • $t = 1$: $A = 5000(1.005)^{12} \approx 5{,}308.39$, $B = 5{,}320.00$, $B$ slightly ahead.$t = 1$：$A = 5000(1.005)^{12} \approx 5{,}308.39$，$B = 5{,}320.00$，$B$ 略领先。
  • $t = 2$: $A \approx 5{,}635.80$, $B = 5{,}640.00$, essentially tied; $B$ a hair ahead.$t = 2$：$A \approx 5{,}635.80$，$B = 5{,}640.00$，几乎并列；$B$ 略领先。
  • $t = 3$: $A \approx 5{,}983.40$, $B = 5{,}960.00$, $A$ first exceeds $B$ during year 3.$t = 3$：$A \approx 5{,}983.40$，$B = 5{,}960.00$，$A$ 首次在第 3 年超过 $B$。
Approximate crossover: between $t \approx 2$ and $t = 3$. (Exact solution requires solving $5000(1.005)^{12 t} = 5000 + 320 t$ numerically, no closed form.)交叉点约在 $t \approx 2$ 与 $t = 3$ 之间。（精确解需数值求 $5000(1.005)^{12 t} = 5000 + 320 t$，无闭式。）

(d) Restriction + reason to prefer B情境限制 + 偏好 B 的理由 R1·R1

Restriction:限制： $t \ge 0$ (no negative time) and typically $t \le T_{\text{maturity}}$ contractually, bonds and term deposits have a fixed redemption date. Also $t$ might be restricted to whole years if the bond pays only at year-end.
Reason to prefer B short-term: Option B is principal-protected (simple-interest bonds usually return at least the principal regardless of market conditions) and pays a predictable cash coupon, useful for a saver who needs guaranteed liquidity. In the first $2$–$3$ years B even leads A in this scenario, and the certainty of the cash flow may outweigh the eventual exponential dominance.$t \ge 0$（无负时间），合同上通常 $t \le T_{\text{到期}}$，债券和定存有固定赎回日。若仅年末付息，$t$ 还可能限制为整年。
短期偏好 B 的理由：方案 B 是本金保护型（单利债券通常无论市场如何至少返还本金），且支付可预测的现金利息，对需要保证流动性的储户有用。在本情境的前 $2$–$3$ 年 B 甚至领先 A，确定的现金流可能比指数终将占优更重要。
Exponential vs linear: the exponential always wins eventually, but "eventually" can be longer than your time horizon.指数 vs 线性：指数终将胜出，但"终将"可能长于你的时间视野。 HSF-LE.A.3 is a long-run statement. The cross-over time depends on the constants: a small exponential rate vs a large linear rate can leave the linear ahead for a decade or more, even though the exponential is guaranteed to overtake it. Here Option B's lead is small and short ($\approx 2$ yr), then A's compounding pulls ahead and the gap widens exponentially: by $t = 10$ years, $A(10) \approx \$9{,}083.48$ vs $B(10) = \$8{,}200.00$, a gap of about \$883. The lesson AP graders reward: state the long-run inevitability and respect the short-run details. Both matter in a real financial decision.HSF-LE.A.3 是长期陈述。交叉时间取决于常数：小指数率对大线性率可让线性领先十年以上，尽管指数终将超越。此处方案 B 的领先既小又短（约 $2$ 年），随后 A 的复利拉开差距，差额按指数扩大：到 $t = 10$ 年，$A(10) \approx \$9{,}083.48$ 而 $B(10) = \$8{,}200.00$，差约 \$883。AP 阅卷奖励的要点：既说明长期必然性，又尊重短期细节。两者在真实金融决策中都重要。