Exponential and
Logarithmic Functions指数函数与对数函数

Identify.辨识参数。 Base $b = 2$, scale factor $a = 3$.底数 $b = 2$，缩放因子 $a = 3$。

(a) $y$-intercept.(a) $y$ 轴截距。 $f(0) = 3 \cdot 2^{0} = 3 \cdot 1 = 3$, so the intercept is $(0, 3)$.$f(0) = 3 \cdot 2^{0} = 3 \cdot 1 = 3$，故截距为 $(0, 3)$。

(b) Horizontal asymptote.(b) 水平渐近线。 As $x \to -\infty$, $2^{x} \to 0$, so $f(x) \to 0$. The asymptote is $y = 0$.当 $x \to -\infty$ 时，$2^{x} \to 0$，故 $f(x) \to 0$。渐近线为 $y = 0$。

(c) Evaluate.(c) 求值。

$$ f(4) \;=\; 3 \cdot 2^{4} \;=\; 3 \cdot 16 \;=\; 48. $$

(d) Growth or decay.(d) 增长或衰减。 Base $b = 2 > 1$, so $f$ is increasing: exponential growth.底数 $b = 2 > 1$，故 $f$ 递增：指数增长。

Evaluate.核验。 Sanity-check the equal-factors property: $f(0) = 3$, $f(1) = 6$, $f(2) = 12$, $f(3) = 24$ , each value doubles. That doubling is the equal-factor signature of an exponential with base $2$.用"等倍数"性质核验：$f(0) = 3$、$f(1) = 6$、$f(2) = 12$、$f(3) = 24$ , 每次翻倍。这正是底数为 $2$ 的指数函数的等倍数特征。

Rewrite to a common base.改写为同一底数。 Both $4$ and $32$ are powers of $2$: $4 = 2^{2}$ and $32 = 2^{5}$.$4$ 与 $32$ 都是 $2$ 的幂：$4 = 2^{2}$、$32 = 2^{5}$。

$$ (2^{2})^{x + 1} \;=\; 2^{5} \quad\Longrightarrow\quad 2^{2 (x + 1)} \;=\; 2^{5}. $$

Set exponents equal.令指数相等。

$$ 2 (x + 1) \;=\; 5 \quad\Longrightarrow\quad x \;=\; \tfrac{3}{2}. $$

Evaluate.核验。 Check: $4^{1 + 3/2} = 4^{5/2} = (4^{1/2})^{5} = 2^{5} = 32$. Confirmed.代回：$4^{1 + 3/2} = 4^{5/2} = (4^{1/2})^{5} = 2^{5} = 32$。验证通过。

Isolate the exponential.孤立指数项。

$$ 3^{x} \;=\; 16. $$

Take the natural log of both sides.两边取自然对数。

$$ \ln(3^{x}) \;=\; \ln 16 \quad\Longrightarrow\quad x \ln 3 \;=\; \ln 16. $$

Solve for $x$.解出 $x$。

$$ x \;=\; \frac{\ln 16}{\ln 3} \;\approx\; \frac{2.7726}{1.0986} \;\approx\; 2.524. $$

Evaluate.核验。 Sanity check: $3^{2} = 9$ and $3^{3} = 27$, so $x$ between $2$ and $3$ is plausible. We can also write $x = \log_{3} 16$ , the same answer in base-$3$ logarithm notation.合理性检查：$3^{2} = 9$、$3^{3} = 27$，故 $x$ 介于 $2$ 与 $3$ 之间合理。等价记法 $x = \log_{3} 16$，即同一答案的以 $3$ 为底对数表示。

(a) Exponential to log.(a) 指数转对数。 $2^{5} = 32$ says "$5$ is the exponent that takes $2$ to $32$." So $\log_{2} 32 = 5$.$2^{5} = 32$ 说"$5$ 是把 $2$ 取到 $32$ 的指数"，故 $\log_{2} 32 = 5$。

(b) Log to exponential.(b) 对数转指数。 $\log_{3} 81 = 4$ says "the exponent on $3$ that gives $81$ is $4$." So $3^{4} = 81$.$\log_{3} 81 = 4$ 说"$3$ 的某个指数得 $81$，该指数为 $4$"，故 $3^{4} = 81$。

(c) Exponential to log.(c) 指数转对数。 $e^{0} = 1$, so $\ln 1 = 0$. More generally, $\log_{b} 1 = 0$ for any base, since $b^{0} = 1$.$e^{0} = 1$，故 $\ln 1 = 0$。一般地，对任意底数 $b$ 都有 $\log_{b} 1 = 0$，因为 $b^{0} = 1$。

Evaluate.核验。 The translation has a fixed shape: "base on the same side, exponent and result swap." Memorise the pattern as a single move rather than three rules.互换有固定模式："底数留在原侧，指数与结果互换位置。"把它当作一个动作记住，而不是三条规则。

(a) Expand.(a) 展开。 Apply the quotient law first, then product on the numerator, then power on each piece:先用除法律，再对分子用乘法律，最后对每一项用幂律：

$$ \log_{2}\!\left( \frac{x^{3} \sqrt{y}}{z^{5}} \right) = \log_{2}(x^{3} \sqrt{y}) - \log_{2}(z^{5}) = \log_{2}(x^{3}) + \log_{2}(\sqrt{y}) - \log_{2}(z^{5}) $$ $$ = 3 \log_{2} x + \tfrac{1}{2} \log_{2} y - 5 \log_{2} z. $$

(b) Contract.(b) 合并。 Pull each coefficient inside as an exponent, then combine via product / quotient:把每个系数收回为指数，再用乘法律 / 除法律合并：

$$ 3 \ln a + \tfrac{1}{2} \ln b - 4 \ln c \;=\; \ln(a^{3}) + \ln(b^{1/2}) - \ln(c^{4}) \;=\; \ln\!\left( \frac{a^{3} \sqrt{b}}{c^{4}} \right). $$

Evaluate.核验。 Expansion and contraction are perfect inverses: a clean expansion should re-contract back to the original expression. Use that as a built-in sanity check.展开与合并互为反操作：一次干净的展开再合并，应回到原表达式。把这点当作内建的核验。

Combine the logs.合并对数。 Product law:用乘法律：

$$ \log_{2}\!\big( (x + 3)(x - 1) \big) \;=\; 5. $$

Convert to exponential form.转为指数形式。

$$ (x + 3)(x - 1) \;=\; 2^{5} \;=\; 32. $$

Expand and solve.展开并求解。

$$ x^{2} + 2 x - 3 \;=\; 32 \quad\Longrightarrow\quad x^{2} + 2 x - 35 \;=\; 0 \quad\Longrightarrow\quad (x + 7)(x - 5) \;=\; 0. $$

Candidates: $x = -7$ or $x = 5$.候选解：$x = -7$ 或 $x = 5$。

Check arguments.检验真数。 $x = -7$ makes $\log_{2}(x + 3) = \log_{2}(-4)$ undefined , extraneous. $x = 5$ gives $\log_{2} 8 + \log_{2} 4 = 3 + 2 = 5$ , valid.$x = -7$ 使 $\log_{2}(x + 3) = \log_{2}(-4)$ 无定义，为增根。$x = 5$ 代回 $\log_{2} 8 + \log_{2} 4 = 3 + 2 = 5$，合法。

Evaluate.核验。 Only $x = 5$ is a solution.仅 $x = 5$ 为解。

Isolate the exponential.孤立指数项。

$$ 3^{2 x - 1} \;=\; 27. $$

Recognise same base.识别同底。 $27 = 3^{3}$, so:$27 = 3^{3}$，故：

$$ 3^{2 x - 1} \;=\; 3^{3} \quad\Longrightarrow\quad 2 x - 1 \;=\; 3 \quad\Longrightarrow\quad x \;=\; 2. $$

Evaluate.核验。 Check: $2 \cdot 3^{3} = 2 \cdot 27 = 54$. Confirmed. (If the right-hand side had not been a clean power of $3$, we would have taken $\log$ of both sides instead.)代回：$2 \cdot 3^{3} = 2 \cdot 27 = 54$。验证通过。（若右侧不是 $3$ 的整数次幂，则应改用两边取 $\log$。）

(a) Annual compounding ($n = 1$).(a) 年复利（$n = 1$）。

$$ A \;=\; 1000 \left( 1 + 0.05 \right)^{10} \;=\; 1000 \cdot (1.05)^{10} \;\approx\; 1000 \cdot 1.6289 \;\approx\; \$1628.89. $$

(b) Continuous compounding.(b) 连续复利。

$$ A \;=\; 1000 \cdot e^{0.05 \cdot 10} \;=\; 1000 \cdot e^{0.5} \;\approx\; 1000 \cdot 1.6487 \;\approx\; \$1648.72. $$

Evaluate.核验。 The continuous-compounding value is higher by about $\$19.83$ (about $1.2\%$). At higher rates and longer times the gap widens; at $r = 0.05$ and $t = 10$ years the two formulas are very close because the rate is small. Continuous compounding is an upper bound on what any compounding frequency can deliver at a given nominal rate.连续复利约高出 $\$19.83$（约 $1.2\%$）。利率越高、时间越长，差距越大；本题因 $r = 0.05$、$t = 10$ 年且利率较低，两种公式结果非常接近。连续复利是任何复利频率在给定名义利率下所能达到的上界。

Set up.建立方程。 Doubling means $A = 2 P$:翻倍即 $A = 2 P$：

$$ 2 P \;=\; P e^{0.06 t} \quad\Longrightarrow\quad 2 \;=\; e^{0.06 t}. $$

Take $\ln$ of both sides.两边取 $\ln$。

$$ \ln 2 \;=\; 0.06 t \quad\Longrightarrow\quad t \;=\; \frac{\ln 2}{0.06} \;\approx\; \frac{0.6931}{0.06} \;\approx\; 11.55 \text{ years}. $$

Evaluate.核验。 The "rule of 72" approximation says doubling time $\approx 72 / (100 r) = 72 / 6 = 12$ years , close to the exact $11.55$. The rule of 72 is itself a Taylor-series approximation of the exact $\ln 2 / r$ formula, and works best for rates in the $4$ – $10$ percent range."七二法则"近似：倍增时间 $\approx 72 / (100 r) = 72 / 6 = 12$ 年 , 与精确值 $11.55$ 接近。七二法则本身就是精确公式 $\ln 2 / r$ 的泰勒展开近似，在利率 $4\%$ – $10\%$ 区间最准。

(a) Model.(a) 模型。 Doubling every $3$ hours means a base of $2$ on a time-scale $t/3$:每 $3$ 小时翻倍意味着以 $2$ 为底、时间尺度为 $t/3$：

$$ N(t) \;=\; 500 \cdot 2^{t / 3}. $$

(b) After $24$ hours.(b) $24$ 小时后。

$$ N(24) \;=\; 500 \cdot 2^{24/3} \;=\; 500 \cdot 2^{8} \;=\; 500 \cdot 256 \;=\; 128000 \text{ cells}. $$

(c) Time to exceed $100000$.(c) 超过 $100000$ 的时刻。 Solve $500 \cdot 2^{t/3} = 100000$:解 $500 \cdot 2^{t/3} = 100000$：

$$ 2^{t/3} \;=\; 200 \quad\Longrightarrow\quad (t/3) \ln 2 \;=\; \ln 200 \quad\Longrightarrow\quad t \;=\; \frac{3 \ln 200}{\ln 2} \;\approx\; \frac{3 \cdot 5.298}{0.693} \;\approx\; 22.93 \text{ hours}. $$

Evaluate.核验。 So the culture first exceeds $100000$ cells just before the $23$-hour mark, consistent with the $24$-hour value of $128000$ cells from part (b).菌群在第 $23$ 小时之前首次突破 $100000$，与 (b) 中 $24$ 小时时的 $128000$ 一致。

Model.模型。

$$ N(t) \;=\; N_{0} \cdot \left( \tfrac{1}{2} \right)^{t / 5730}. $$

Set up.建立方程。 $N(t) / N_{0} = 0.20$, so:$N(t) / N_{0} = 0.20$，故：

$$ \left( \tfrac{1}{2} \right)^{t / 5730} \;=\; 0.20. $$

Take $\ln$ of both sides.两边取 $\ln$。

$$ \frac{t}{5730} \cdot \ln \tfrac{1}{2} \;=\; \ln 0.20 \quad\Longrightarrow\quad t \;=\; 5730 \cdot \frac{\ln 0.20}{\ln (1/2)}. $$

Evaluate.核验。 $\ln 0.20 \approx -1.6094$ and $\ln(1/2) \approx -0.6931$, so $t \approx 5730 \cdot 2.3219 \approx 13305$ years. The fossil is about $13.3$ thousand years old, somewhat older than two half-lives (which would be $25\%$ remaining at $11460$ years) but less than three (which would be $12.5\%$ at $17190$ years).$\ln 0.20 \approx -1.6094$，$\ln(1/2) \approx -0.6931$，故 $t \approx 5730 \cdot 2.3219 \approx 13305$ 年。化石约 $1.33$ 万年龄，比两次半衰期（$25\%$ 残留、$11460$ 年）稍久，比三次半衰期（$12.5\%$、$17190$ 年）略短。