Axiom Learning·AP PrecalculusChapter 7 · Logarithmic Functions
对数函数 — Expressions, Inverses, Laws, Equations & Modeling
本章学习目标 Learning Objectives
- 理解对数的定义并求对数表达式的值 · Evaluate logarithmic expressions
- 掌握对数函数作为指数函数反函数的关系 · Understand logarithm as inverse of exponential
- 分析对数函数的性质：定义域、值域、端点行为、凹凸性 · Analyze properties of log functions
- 运用对数运算法则化简表达式 · Apply logarithm laws
- 求解指数方程与对数方程及不等式 · Solve exponential/logarithmic equations
- 用对数模型建模实际数据（两点、比例与零点、LnReg），用半对数图线性化数据 · Build log models; linearize data with semi-log plots
- 理解对数刻度、对数法则的图像含义，求变换后函数的反函数 · Log scales, graphical meaning of the laws, inverses of transformed functions
| Term 术语 | Definition / Form 定义/形式 | Chinese 中文 |
|---|---|---|
| Logarithm | $\log_b x = y \Leftrightarrow b^y = x$ | 对数 |
| Base | $b$ in $\log_b x$; requires $b>0,b\neq1$ | 底数 |
| Argument | $x$ in $\log_b x$; requires $x>0$ | 真数 |
| Common Logarithm | $\log x = \log_{10} x$ | 常用对数 |
| Natural Logarithm | $\ln x = \log_e x$, $e\approx2.718$ | 自然对数 |
| Vertical Asymptote | $x=0$ for parent $f(x)=\log_b x$ | 垂直渐近线 |
| Extraneous Solution | Algebraic solution making argument $\leq0$; must reject | 增根（外来根） |
| Interpolation | Predicting within the data range | 插值 |
| Extrapolation | Predicting outside the data range (less reliable) | 外推 |
| Semilog Plot | One log axis, one linear axis | 半对数图 |
7.1 Logarithmic Expressions 对数表达式
A logarithm answers: "What exponent should the base be raised to in order to get the given number?"
$$\log_b x = y \quad\Longleftrightarrow\quad b^y = x$$
where $b>0,\;b\neq1$ and $x>0$.
Logarithmic Form 对数形式
$\log_b x = y$
⇔
Exponential Form 指数形式
$b^y = x$
Basic Values 基本值:
- $\log_b 1 = 0$ (because $b^0=1$) 任何底数的对数，真数为1时值为0
- $\log_b b = 1$ (because $b^1=b$) 底数等于真数时，对数值为1
- $\log_b b^x = x$ (inverse identity) 对数与指数互为逆运算
Special Notations 特殊记号:
$\log x = \log_{10} x$ (common log 常用对数, base 10)
$\ln x = \log_e x$ (natural log 自然对数, base $e\approx2.718$)
Change of Base 换底公式: $\log_b x = \dfrac{\log x}{\log b} = \dfrac{\ln x}{\ln b}$
$\log x = \log_{10} x$ (common log 常用对数, base 10)
$\ln x = \log_e x$ (natural log 自然对数, base $e\approx2.718$)
Change of Base 换底公式: $\log_b x = \dfrac{\log x}{\log b} = \dfrac{\ln x}{\ln b}$
一起做 T-1 · Evaluate Logarithmic Expressions
Evaluate each expression exactly:
- $\log_2 32$
- $\log_{10} 1$
- $\log_{10} 1000$
- $\log_3 \dfrac{1}{27}$
(1) $\log_2 32$
Ask: $2^? = 32 = 2^5$ → answer is $\boxed{5}$
Ask: $2^? = 32 = 2^5$ → answer is $\boxed{5}$
(2) $\log_{10} 1$
$10^? = 1 = 10^0$ → answer is $\boxed{0}$ (basic value: $\log_b 1=0$)
$10^? = 1 = 10^0$ → answer is $\boxed{0}$ (basic value: $\log_b 1=0$)
(3) $\log_{10} 1000$
$10^? = 1000 = 10^3$ → answer is $\boxed{3}$
$10^? = 1000 = 10^3$ → answer is $\boxed{3}$
(4) $\log_3 \dfrac{1}{27}$
$3^? = \dfrac{1}{27} = 3^{-3}$ → answer is $\boxed{-3}$
$3^? = \dfrac{1}{27} = 3^{-3}$ → answer is $\boxed{-3}$
一起做 T-2 · Change of Base — Evaluate with Calculator
Evaluate $\log_5 72$ to the nearest hundredth.
Apply change of base formula
$\log_5 72 = \dfrac{\log 72}{\log 5} = \dfrac{\ln 72}{\ln 5}$
$\log_5 72 = \dfrac{\log 72}{\log 5} = \dfrac{\ln 72}{\ln 5}$
Calculate
$\dfrac{\log 72}{\log 5} = \dfrac{1.8573}{0.6990} \approx \boxed{2.66}$
Verify: $5^{2.66}\approx72$ ✓
$\dfrac{\log 72}{\log 5} = \dfrac{1.8573}{0.6990} \approx \boxed{2.66}$
Verify: $5^{2.66}\approx72$ ✓
练习 E-1 · Evaluate Logarithms
Evaluate: (a) $\log_4 64$ (b) $\log_{1/2} 8$ (c) $\ln e^5$ (d) $\log_6\dfrac{1}{36}$ (e) $\log_7 1$
(a) $4^?=64=4^3$ → $\boxed{3}$
(b) $\left(\frac{1}{2}\right)^?=8=\left(\frac{1}{2}\right)^{-3}$ → $\boxed{-3}$
(c) $\ln e^5 = 5$ (inverse identity $\log_b b^x=x$) → $\boxed{5}$
(d) $6^?=\frac{1}{36}=6^{-2}$ → $\boxed{-2}$
(e) $\log_b 1=0$ always → $\boxed{0}$
(b) $\left(\frac{1}{2}\right)^?=8=\left(\frac{1}{2}\right)^{-3}$ → $\boxed{-3}$
(c) $\ln e^5 = 5$ (inverse identity $\log_b b^x=x$) → $\boxed{5}$
(d) $6^?=\frac{1}{36}=6^{-2}$ → $\boxed{-2}$
(e) $\log_b 1=0$ always → $\boxed{0}$
Logarithmic Scale 对数刻度
On a logarithmic scale (base 10), equally spaced marks stand for $10^0, 10^1, 10^2, \dots = 1, 10, 100, \dots$: each unit represents multiplying by the base, not adding 1 (CED 2.9.A.3). A value $x$ is placed at position $\log x$.
对数刻度上"走一格"等于乘以底数。数 $x$ 画在位置 $\log x$ 处，所以 500 在 100 和 1000 之间约 70% 的位置（$\log500\approx2.699$）。
练习 E-2 · Reading a Log Scale 读对数刻度
(a) On a base-10 log scale, what value lies exactly halfway between 1 and 10,000? Between 10 and 1000?
(b) The pH of a solution is $\text{pH}=-\log[\text{H}^+]$. Solution A has $[\text{H}^+]=10^{-3}$ and solution B has $[\text{H}^+]=10^{-6}$. Find each pH. How many times more acidic (larger $[\text{H}^+]$) is A than B?
(a) Positions $\log1=0$ and $\log10000=4$; halfway is position 2 → $10^2=100$. Between 10 and 1000: position 2 → 100 (not 505!).
(b) pH of A $=3$, pH of B $=6$. A difference of 3 units on the log scale means a factor of $10^3=1000$: A is 1000 times more acidic.
(b) pH of A $=3$, pH of B $=6$. A difference of 3 units on the log scale means a factor of $10^3=1000$: A is 1000 times more acidic.
7.2 Logarithm as Inverse of Exponential 对数函数：指数函数的反函数
Key Idea: The logarithmic function $f^{-1}(x)=\log_b x$ is the inverse of the exponential function $f(x)=b^x$.
$$b^x = y \quad\Longleftrightarrow\quad \log_b y = x$$
Inverse Identities 反函数恒等式:
$$\log_b(b^x) = x \qquad b^{\log_b x} = x\;(x>0)$$
These show that $f$ and $f^{-1}$ undo each other. 互为逆运算，复合后还原。
图像关系 Graph Relationship
The graph of $y=\log_b x$ is the reflection of $y=b^x$ across the line $y=x$.对数函数的图像是指数函数图像关于直线 $y=x$ 的对称图像。
Swapped features: domain/range swap, x-int/y-int swap.
Exponential ↔ Logarithmic Inverse Explorer · 指数—对数反函数动画
Watch a point $(x,b^x)$ reflect across $y=x$ and become $(b^x,x)$. 观察坐标交换如何把指数函数上的点变成对数函数上的点。
Begin with point P on the exponential curve. 从指数函数上的点 P 开始。
Step 1 / 3
1 · Exponential Point 指数点
2 · Reflect over y = x 关于直线反射
3 · Logarithmic Point 对数点
Inverse Functions on the Same Coordinate Plane · 同一坐标系中的反函数
y = bx
y = logbx
y = x
Exponential Form 指数形式
21 = 2
Logarithmic Form 对数形式
log2 2 = 1
Coordinate Swap 坐标交换
(1, 2)
⇄
(2, 1)
What Swaps? 哪些性质交换？
Domain ↔ Range
x-intercept ↔ y-intercept
HA y = 0 ↔ VA x = 0
x-intercept ↔ y-intercept
HA y = 0 ↔ VA x = 0
Parent Logarithmic Function 基本对数函数 $f(x)=\log_b x$
$b > 1$ (e.g., $\log_2 x$, $\ln x$, $\log x$)
- Domain: $(0,+\infty)$
- Range: $(-\infty,+\infty)$
- x-intercept: $(1,0)$
- VA: $x=0$
- Increasing ↑ · Concave down ∩
- $\lim_{x\to0^+}\log_b x=-\infty$; $\lim_{x\to+\infty}\log_b x=+\infty$
$0 < b < 1$ (e.g., $\log_{1/2} x$)
- Domain: $(0,+\infty)$
- Range: $(-\infty,+\infty)$
- x-intercept: $(1,0)$
- VA: $x=0$
- Decreasing ↓ · Concave up ∪
- $\lim_{x\to0^+}\log_b x=+\infty$; $\lim_{x\to+\infty}\log_b x=-\infty$
⚠️ 重要：没有 y 轴截距 No y-intercept
对数函数 $f(x)=\log_b x$ 的定义域是 $(0,+\infty)$，不包含 $x=0$，因此没有 y-intercept。x-intercept 总是 $(1,0)$（因为 $\log_b 1=0$）。
垂直渐近线总是 $x=0$（y 轴）。
一起做 T-3 · From Exp Graph to Log Graph
Use the graph of $f(x)=2^x$ to describe the graph of $g(x)=\log_2 x$. State domain and range.
Relationship
$g(x)=\log_2 x$ is the inverse of $f(x)=2^x$.
The graph of $g$ is the reflection of $f$ over the line $y=x$.
$g(x)=\log_2 x$ is the inverse of $f(x)=2^x$.
The graph of $g$ is the reflection of $f$ over the line $y=x$.
Key features swap
$f(x)=2^x$: domain $(-\infty,+\infty)$, range $(0,+\infty)$, y-int $(0,1)$, HA $y=0$
$g(x)=\log_2 x$: domain $(0,+\infty)$, range $(-\infty,+\infty)$, x-int $(1,0)$, VA $x=0$
$f(x)=2^x$: domain $(-\infty,+\infty)$, range $(0,+\infty)$, y-int $(0,1)$, HA $y=0$
$g(x)=\log_2 x$: domain $(0,+\infty)$, range $(-\infty,+\infty)$, x-int $(1,0)$, VA $x=0$
Behavior
$g$ is increasing, concave down ($b=2>1$).
Key points: $(1,0),(2,1),(4,2),(8,3)$ — inputs double, output increases by 1.
$g$ is increasing, concave down ($b=2>1$).
Key points: $(1,0),(2,1),(4,2),(8,3)$ — inputs double, output increases by 1.
7.3 Properties & End Behavior 对数函数性质与端点行为
For $f(x)=\log_b x$:
| $b>1$ | $0<b<1$ | |
|---|---|---|
| $\displaystyle\lim_{x\to 0^+}\log_b x$ | $-\infty$ | $+\infty$ |
| $\displaystyle\lim_{x\to+\infty}\log_b x$ | $+\infty$ | $-\infty$ |
| Monotonicity | Increasing ↑ | Decreasing ↓ |
| Concavity | Concave down ∩ | Concave up ∪ |
理解端点行为
对数函数（$b>1$）：· 当 $x\to0^+$ 时（从右趋近0），$\log_b x\to-\infty$ — 趋近垂直渐近线 $x=0$，向下无界
· 当 $x\to+\infty$ 时，$\log_b x\to+\infty$ — 但增长极慢（比任何幂函数都慢）
一起做 T-4 · Evaluate Limits of Log Functions
Let $f(x)=\log_4 x$ and $g(x)=\log_{1/4} x$. Evaluate:
- $\displaystyle\lim_{x\to 0^+} f(x)$
- $\displaystyle\lim_{x\to+\infty} f(x)$
- $\displaystyle\lim_{x\to 0^+} g(x)$
- $\displaystyle\lim_{x\to+\infty} g(x)$
$f(x)=\log_4 x$: base $b=4>1$, increasing
(1) $\displaystyle\lim_{x\to0^+}\log_4 x = -\infty$ (VA at $x=0$, function drops to $-\infty$)
(2) $\displaystyle\lim_{x\to+\infty}\log_4 x = +\infty$
(1) $\displaystyle\lim_{x\to0^+}\log_4 x = -\infty$ (VA at $x=0$, function drops to $-\infty$)
(2) $\displaystyle\lim_{x\to+\infty}\log_4 x = +\infty$
$g(x)=\log_{1/4} x$: base $b=\frac{1}{4}<1$, decreasing
(3) $\displaystyle\lim_{x\to0^+}\log_{1/4} x = +\infty$ (function rises to $+\infty$ near VA)
(4) $\displaystyle\lim_{x\to+\infty}\log_{1/4} x = -\infty$
(3) $\displaystyle\lim_{x\to0^+}\log_{1/4} x = +\infty$ (function rises to $+\infty$ near VA)
(4) $\displaystyle\lim_{x\to+\infty}\log_{1/4} x = -\infty$
口诀
$b>1$：左边趋$-\infty$，右边趋$+\infty$（增函数）$0<b<1$：左边趋$+\infty$，右边趋$-\infty$（减函数）
General Form $a\log_b x$ & Identifying Logarithmic Data 一般式与从表格识别对数数据
General form: $f(x)=a\log_b x$ ($a\neq0$, $b>0$, $b\neq1$). Like exponential functions, it is always increasing or always decreasing, always concave up or always concave down, with no extrema (except on a closed interval) and no points of inflection (CED 2.11.A.2).
对数型数据：输入每次乘以同一个数，输出每次加同一个数（CED 2.10.A.2）。若输入要先加减一个常数才成比例，模型是 $a\log_b(x-h)$（CED 2.11.A.3）。
| Exponential $ab^x$ | Logarithmic $a\log_b x$ | |
|---|---|---|
| inputs | change additively (+1, +1, …) | change multiplicatively (×b, ×b, …) |
| outputs | change multiplicatively (×b) | change additively (+a) |
一起做 T-5 · Is the Table Logarithmic? 表格是不是对数型
| $x$ | 1 | 3 | 9 | 27 | 81 |
|---|---|---|---|---|---|
| $y$ | 2 | 5 | 8 | 11 | 14 |
Show that the data are logarithmic and write a model.
Pattern
Inputs are multiplied by 3 each time; outputs increase by 3 each time → logarithmic (inputs ×3, outputs +3).
Inputs are multiplied by 3 each time; outputs increase by 3 each time → logarithmic (inputs ×3, outputs +3).
Model
At $x=1$, $y=2$; each factor of 3 in $x$ adds 3 to $y$: $\;y=2+3\log_3 x$. Check: $x=27$: $2+3(3)=11$ ✓.
At $x=1$, $y=2$; each factor of 3 in $x$ adds 3 to $y$: $\;y=2+3\log_3 x$. Check: $x=27$: $2+3(3)=11$ ✓.
$y=2+3\log_3 x$
练习 E-3 · A Shifted Log Table 需要平移的对数表
| $x$ | 3 | 4 | 6 | 10 | 18 |
|---|---|---|---|---|---|
| $y$ | 0 | 1 | 2 | 3 | 4 |
The outputs increase by 1 each time, but the inputs are not proportional. Find a constant $h$ so that $x-h$ is proportional, and write a model. State its vertical asymptote.
$x-2$: $1, 2, 4, 8, 16$ → multiplied by 2 each time while $y$ increases by 1. So $y=\log_2(x-2)$. Vertical asymptote $x=2$ (the graph of $\log_2x$ shifted right 2).
7.4 Logarithm Laws 对数运算法则
| Law 法则 | Formula 公式 | Example 例子 | Chinese 中文 |
|---|---|---|---|
| Product Law 乘积法则 | $\log_b(MN)=\log_b M+\log_b N$ | $\log_2(8\cdot4)=\log_2 8+\log_2 4=3+2=5$ | 乘积的对数 = 对数之和 |
| Quotient Law 商法则 | $\log_b\dfrac{M}{N}=\log_b M-\log_b N$ | $\log_3\dfrac{81}{3}=\log_3 81-\log_3 3=4-1=3$ | 商的对数 = 对数之差 |
| Power Law 幂法则 | $\log_b(M^p)=p\cdot\log_b M$ | $\log_2(8^4)=4\log_2 8=4\cdot3=12$ | 幂的对数 = 指数×对数 |
| Change of Base 换底公式 | $\log_b x=\dfrac{\log x}{\log b}=\dfrac{\ln x}{\ln b}$ | $\log_5 72=\dfrac{\ln72}{\ln5}\approx2.66$ | 换底：除以底数的对数 |
Natural Logarithm Identities 自然对数恒等式:
$$\ln e = 1 \qquad \ln(e^x) = x \qquad e^{\ln x} = x\;(x>0) \qquad \ln 1 = 0$$
⚠️ Common Mistakes 常见错误:
$\log_b(M+N) \neq \log_b M + \log_b N$ ← 没有"和的对数"法则！
$\log_b(MN) \neq \log_b M \cdot \log_b N$ ← 乘积法则是相加，不是相乘
$\dfrac{\log_b M}{\log_b N} \neq \log_b\dfrac{M}{N}$ ← 商法则是对数内部相除
$\log_b(M+N) \neq \log_b M + \log_b N$ ← 没有"和的对数"法则！
$\log_b(MN) \neq \log_b M \cdot \log_b N$ ← 乘积法则是相加，不是相乘
$\dfrac{\log_b M}{\log_b N} \neq \log_b\dfrac{M}{N}$ ← 商法则是对数内部相除
一起做 T-6 · Expand Using Log Laws
Express $\log_5\dfrac{\sqrt[3]{x}}{25}$ in terms of $\log_5 x$.
Apply Quotient Law
$\log_5\dfrac{\sqrt[3]{x}}{25} = \log_5(\sqrt[3]{x}) - \log_5 25$
$\log_5\dfrac{\sqrt[3]{x}}{25} = \log_5(\sqrt[3]{x}) - \log_5 25$
Apply Power Law to each part
$\log_5(x^{1/3}) = \dfrac{1}{3}\log_5 x$
$\log_5 25 = \log_5 5^2 = 2$
$\log_5(x^{1/3}) = \dfrac{1}{3}\log_5 x$
$\log_5 25 = \log_5 5^2 = 2$
Combine
$$\log_5\frac{\sqrt[3]{x}}{25} = \frac{1}{3}\log_5 x - 2$$
$$\log_5\frac{\sqrt[3]{x}}{25} = \frac{1}{3}\log_5 x - 2$$
一起做 T-7 · Express in Terms of Known Values
If $\log 2 = a$ and $\log 3 = b$, express $\log 24$ in terms of $a$ and $b$.
Factor 24
$24 = 8 \times 3 = 2^3 \times 3$
$24 = 8 \times 3 = 2^3 \times 3$
Apply Product and Power Laws
$\log 24 = \log(2^3 \cdot 3)= \log 2^3 + \log 3 = 3\log 2 + \log 3$
$\log 24 = \log(2^3 \cdot 3)= \log 2^3 + \log 3 = 3\log 2 + \log 3$
Substitute
$= 3a + b$
$= 3a + b$
一起做 T-8 · Transformations of Log Functions
Describe the transformation from parent $y=\log_2 x$ to $g(x)=\log_2(2x)$. Are these equivalent to another transformation?
Direct interpretation
$\log_2(2x)=f(2x)$ where $f(x)=\log_2 x$.
→ Horizontal compression by factor $\dfrac{1}{2}$ 水平压缩为原来的 $\frac{1}{2}$
$\log_2(2x)=f(2x)$ where $f(x)=\log_2 x$.
→ Horizontal compression by factor $\dfrac{1}{2}$ 水平压缩为原来的 $\frac{1}{2}$
Equivalent form using Product Law
$\log_2(2x) = \log_2 2 + \log_2 x = 1 + \log_2 x$
→ Vertical shift UP by 1 垂直向上平移1单位
$\log_2(2x) = \log_2 2 + \log_2 x = 1 + \log_2 x$
→ Vertical shift UP by 1 垂直向上平移1单位
Conclusion
Horizontal compression by $\frac{1}{2}$ ≡ Vertical shift up 1.
这与指数函数中"水平平移 ≡ 垂直伸缩"的等价关系类似！
Horizontal compression by $\frac{1}{2}$ ≡ Vertical shift up 1.
这与指数函数中"水平平移 ≡ 垂直伸缩"的等价关系类似！
练习 E-4 · Log Laws Practice
Expand or condense each expression:
(a) Expand: $\ln\dfrac{x^2\sqrt{y}}{z^3}$
(b) Condense: $3\log_4 x - \frac{1}{2}\log_4 y + \log_4 5$
(c) If $\ln 2 = 0.693$, find $\ln 8$.
(a) $2\ln x + \frac{1}{2}\ln y - 3\ln z$
(b) $\log_4\dfrac{5x^3}{\sqrt{y}}$
(c) $\ln 8 = \ln 2^3 = 3\ln 2 = 3(0.693) = 2.079$
(b) $\log_4\dfrac{5x^3}{\sqrt{y}}$
(c) $\ln 8 = \ln 2^3 = 3\ln 2 = 3(0.693) = 2.079$
What the Laws Mean for Graphs 对数法则的图像含义
- Product law → a horizontal dilation equals a vertical translation: $\log_b(kx)=\log_b k+\log_b x$ (see T-8).
- Power law → raising the input to a power is a vertical dilation: $\log_b(x^k)=k\log_b x$ (for $x>0$).
- Change of base → $\log_b x=\dfrac{1}{\log_a b}\cdot\log_a x$, so every logarithmic function is a vertical dilation of every other one.
All pass through $(1,0)$ and have VA $x=0$; they differ only by a vertical dilation.
一起做 T-9 · Logs as Dilations of Each Other 对数函数互为垂直伸缩
(a) Describe $g(x)=\log_3(x^4)$ as a transformation of $f(x)=\log_3x$ (for $x>0$).
(b) Write $\log_2x$ as a vertical dilation of $\ln x$, and $\ln x$ as a vertical dilation of $\log x$.
(c) Describe $h(x)=\log_5(25x^3)$ as transformations of $\log_5x$.
(a) $\log_3(x^4)=4\log_3x$: vertical dilation by a factor of 4.
(b) $\log_2x=\dfrac{\ln x}{\ln2}\approx1.443\ln x$; $\ln x=\dfrac{\log x}{\log e}\approx2.303\log x$.
(c) $\log_5(25x^3)=\log_525+\log_5x^3=2+3\log_5x$: vertical dilation by 3, then shift up 2.
(a) vertical dilation ×4 · (b) $1.443\ln x$; $2.303\log x$ · (c) $3\log_5x+2$
7.5 Exponential & Logarithmic Equations 指数与对数方程
Strategy 1 — Solving Log Equations (same base both sides):
$$\log_b A = \log_b B \implies A = B$$
Always check: $A>0$ and $B>0$ after solving. Reject extraneous solutions that make any argument $\leq0$.
Strategy 2 — Solving Exponential Equations (bases can't match):
- Isolate the exponential expression
- Take $\ln$ (or $\log$) of both sides
- Use Power Law: bring exponent down
- Solve for the variable
Extraneous Solutions 增根: When solving log equations, always substitute your answer back. If the argument of any log becomes $\leq 0$, reject that solution.
解对数方程后必须验根！若代入后出现 $\log(\text{非正数})$，该根为增根，需舍去。
解对数方程后必须验根！若代入后出现 $\log(\text{非正数})$，该根为增根，需舍去。
一起做 T-10 · Solve Logarithmic Equation
Solve for $x$: $\log_5 x = 3\log_5 2 + \log_5 3$
Condense the right side using log laws
$3\log_5 2 + \log_5 3 = \log_5 2^3 + \log_5 3 = \log_5(8\cdot3) = \log_5 24$
$3\log_5 2 + \log_5 3 = \log_5 2^3 + \log_5 3 = \log_5(8\cdot3) = \log_5 24$
Set arguments equal (same base)
$\log_5 x = \log_5 24 \implies x = 24$
$\log_5 x = \log_5 24 \implies x = 24$
Check for extraneous solutions
$x=24>0$ ✓ Domain condition satisfied. Answer: $\boxed{x=24}$
$x=24>0$ ✓ Domain condition satisfied. Answer: $\boxed{x=24}$
一起做 T-11 · Solve with Natural Log
Solve for $x$: $\ln x - \ln 8 = \dfrac{1}{3}\ln 10$
Condense the left side
$\ln x - \ln 8 = \ln\dfrac{x}{8}$
$\ln x - \ln 8 = \ln\dfrac{x}{8}$
Simplify the right side
$\dfrac{1}{3}\ln 10 = \ln 10^{1/3} = \ln\sqrt[3]{10}$
$\dfrac{1}{3}\ln 10 = \ln 10^{1/3} = \ln\sqrt[3]{10}$
Equation becomes
$\ln\dfrac{x}{8} = \ln\sqrt[3]{10}$
Same base → $\dfrac{x}{8} = \sqrt[3]{10}$
$x = 8\sqrt[3]{10} \approx 8\times2.154 \approx 17.23$
$\ln\dfrac{x}{8} = \ln\sqrt[3]{10}$
Same base → $\dfrac{x}{8} = \sqrt[3]{10}$
$x = 8\sqrt[3]{10} \approx 8\times2.154 \approx 17.23$
Check
$x=8\sqrt[3]{10}>0$ ✓ → $\boxed{x=8\sqrt[3]{10}}$
$x=8\sqrt[3]{10}>0$ ✓ → $\boxed{x=8\sqrt[3]{10}}$
一起做 T-12 · Solve Exponential Equation
Solve for $x$: $2^{3x} = 6$
Bases can't match — take ln of both sides
$\ln(2^{3x}) = \ln 6$
$\ln(2^{3x}) = \ln 6$
Apply Power Law
$3x\cdot\ln 2 = \ln 6$
$3x\cdot\ln 2 = \ln 6$
Solve for $x$
$x = \dfrac{\ln 6}{3\ln 2} = \dfrac{\ln 6}{\ln 8}= \log_8 6$
Numerically: $x = \dfrac{\ln 6}{\ln 8} = 0.86165\ldots \approx \boxed{0.862}$ (round to 3 decimal places; do not truncate)
$x = \dfrac{\ln 6}{3\ln 2} = \dfrac{\ln 6}{\ln 8}= \log_8 6$
Numerically: $x = \dfrac{\ln 6}{\ln 8} = 0.86165\ldots \approx \boxed{0.862}$ (round to 3 decimal places; do not truncate)
一起做 T-13 · Solve Logarithmic Inequality
Solve the inequality: $\log_2(2x+3) > \log_2(3x)$
Domain restrictions first (both arguments must be positive)
$2x+3>0 \Rightarrow x>-\dfrac{3}{2}$ AND $3x>0 \Rightarrow x>0$
Combined domain: $x>0$
$2x+3>0 \Rightarrow x>-\dfrac{3}{2}$ AND $3x>0 \Rightarrow x>0$
Combined domain: $x>0$
Since $b=2>1$, $\log_2$ is increasing → inequality direction preserved
$\log_2(2x+3)>\log_2(3x) \implies 2x+3 > 3x$
$3 > x \implies x < 3$
$\log_2(2x+3)>\log_2(3x) \implies 2x+3 > 3x$
$3 > x \implies x < 3$
Combine with domain
$x>0$ AND $x<3$
Solution: $\boxed{0 < x < 3}$ or $(0,3)$
$x>0$ AND $x<3$
Solution: $\boxed{0 < x < 3}$ or $(0,3)$
练习 E-5 · Solve Equations
(a) $\log_3(x+5) + \log_3(x-1) = 3$
(b) $5e^{2x} = 80$
(c) $\log_4(x^2-3) = \log_4(2x)$
(a) $\log_3[(x+5)(x-1)]=3 \Rightarrow (x+5)(x-1)=27$
$x^2+4x-5=27 \Rightarrow x^2+4x-32=0 \Rightarrow (x+8)(x-4)=0$
$x=-8$: check $x-1=-9<0$ → extraneous, reject 增根
$x=4$: check $4+5=9>0$, $4-1=3>0$ ✓ → $\boxed{x=4}$
(b) $e^{2x}=16 \Rightarrow 2x=\ln16 \Rightarrow x=\dfrac{\ln16}{2}=\ln4\approx1.386$
(c) $x^2-3=2x \Rightarrow x^2-2x-3=0 \Rightarrow (x-3)(x+1)=0$
$x=-1$: $\log_4(1-3)=\log_4(-2)$ undefined → reject 增根
$x=3$: $\log_4(9-3)=\log_4 6$ ✓ and $\log_4(6)$ ✓ → $\boxed{x=3}$
$x^2+4x-5=27 \Rightarrow x^2+4x-32=0 \Rightarrow (x+8)(x-4)=0$
$x=-8$: check $x-1=-9<0$ → extraneous, reject 增根
$x=4$: check $4+5=9>0$, $4-1=3>0$ ✓ → $\boxed{x=4}$
(b) $e^{2x}=16 \Rightarrow 2x=\ln16 \Rightarrow x=\dfrac{\ln16}{2}=\ln4\approx1.386$
(c) $x^2-3=2x \Rightarrow x^2-2x-3=0 \Rightarrow (x-3)(x+1)=0$
$x=-1$: $\log_4(1-3)=\log_4(-2)$ undefined → reject 增根
$x=3$: $\log_4(9-3)=\log_4 6$ ✓ and $\log_4(6)$ ✓ → $\boxed{x=3}$
Rewriting an Exponential in Another Base 用对数改写指数的底
Because $b=c^{\log_c b}$, every exponential can be rewritten with any base: $\;b^x=c^{(\log_c b)x}$ (CED 2.13.A.3). Most useful: $\;b^x=e^{(\ln b)x}$ and $\;e^{kx}=\left(e^k\right)^x$.
常用来在"每期增长率"和"连续增长率"之间换算。
常用来在"每期增长率"和"连续增长率"之间换算。
一起做 T-14 · Growth Factor ↔ Continuous Rate 增长因子与连续增长率
(a) Rewrite $P(t)=5(1.04)^t$ in the form $5e^{kt}$. What is the continuous growth rate?
(b) Rewrite $A(t)=200e^{-0.3t}$ in the form $200b^t$. By what percent does $A$ decrease per unit of $t$?
(c) Rewrite $2^x$ as a power of 10.
(a) $1.04=e^{\ln1.04}$, so $P(t)=5e^{(\ln1.04)t}\approx5e^{0.039221t}$. Continuous rate $\approx3.922\%$ (slightly less than 4%).
(b) $e^{-0.3t}=\left(e^{-0.3}\right)^t\approx0.740818^t$, so $A(t)\approx200(0.740818)^t$: decreases about $25.918\%$ per unit.
(c) $2^x=10^{(\log2)x}\approx10^{0.301x}$.
(a) $5e^{0.039221t}$, 3.922% · (b) $200(0.740818)^t$, −25.918% · (c) $10^{0.301x}$
Inverses of Transformed Exponential & Log Functions 变换后函数的反函数
To invert $f(x)=a\,b^{\,x-h}+k$ or $f(x)=a\log_b(x-h)+k$, undo the operations in reverse order (subtract $k$, divide by $a$, apply $\log_b$ or $b^{(\ )}$, add $h$). The domain of $f^{-1}$ is the range of $f$.
按相反顺序"撤销"每一步运算；反函数的定义域 = 原函数的值域。
按相反顺序"撤销"每一步运算；反函数的定义域 = 原函数的值域。
一起做 T-15 · Invert Transformed Functions 求变换后函数的反函数
(a) $f(x)=3\cdot2^{\,x-1}+4$. (b) $g(x)=2\ln(x+5)-1$. Find each inverse and its domain and range.
(a) $x=3\cdot2^{\,y-1}+4 \Rightarrow \dfrac{x-4}{3}=2^{\,y-1} \Rightarrow y-1=\log_2\dfrac{x-4}{3}$.
$f^{-1}(x)=1+\log_2\dfrac{x-4}{3}$. Range of $f$ is $(4,\infty)$ (HA $y=4$), so $f^{-1}$ has domain $(4,\infty)$ and range all reals.
$f^{-1}(x)=1+\log_2\dfrac{x-4}{3}$. Range of $f$ is $(4,\infty)$ (HA $y=4$), so $f^{-1}$ has domain $(4,\infty)$ and range all reals.
(b) $x=2\ln(y+5)-1 \Rightarrow \dfrac{x+1}{2}=\ln(y+5) \Rightarrow y+5=e^{(x+1)/2}$.
$g^{-1}(x)=e^{(x+1)/2}-5$. Domain all reals; range $(-5,\infty)$ (the domain of $g$).
$g^{-1}(x)=e^{(x+1)/2}-5$. Domain all reals; range $(-5,\infty)$ (the domain of $g$).
$f^{-1}(x)=1+\log_2\frac{x-4}{3}$, $x>4$ | $g^{-1}(x)=e^{(x+1)/2}-5$
练习 E-6 · Graphical Solutions, Inequalities & Context 图像解法、不等式与情境
(a) Solve $2^x=x+3$ with a graphing calculator. (It cannot be solved with log laws alone.)
(b) Solve $3(1.05)^t>10$.
(c) A culture is started with 500 bacteria at $t=0$ hours and grows as $P(t)=500e^{0.02t}$. Solve $P(t)=400$ and interpret.
(a) Graph $Y_1=2^X$ and $Y_2=X+3$; intersect: $x\approx-2.863$ and $x\approx2.445$.
(b) $(1.05)^t>\tfrac{10}{3} \Rightarrow t\ln1.05>\ln\tfrac{10}{3}$; since $\ln1.05>0$, the direction stays: $t>\dfrac{\ln(10/3)}{\ln1.05}\approx24.677$.
(c) $e^{0.02t}=0.8\Rightarrow t=\dfrac{\ln0.8}{0.02}\approx-11.157$. The culture did not exist before $t=0$, so this solution is rejected by the context (CED 2.13.A.2): the population is never 400 in this situation.
(b) $(1.05)^t>\tfrac{10}{3} \Rightarrow t\ln1.05>\ln\tfrac{10}{3}$; since $\ln1.05>0$, the direction stays: $t>\dfrac{\ln(10/3)}{\ln1.05}\approx24.677$.
(c) $e^{0.02t}=0.8\Rightarrow t=\dfrac{\ln0.8}{0.02}\approx-11.157$. The culture did not exist before $t=0$, so this solution is rejected by the context (CED 2.13.A.2): the population is never 400 in this situation.
7.6 Logarithmic Function Modeling 对数函数建模
A logarithmic model fits data that:
- Increases quickly at first 增长初期很快
- Then continues increasing but at a slower rate 之后增速减缓
- Has no fixed upper bound 没有固定上限
插值与外推
Interpolation 插值：在已知数据范围内预测 — 通常更可靠Extrapolation 外推：在已知数据范围外预测 — 不太可靠，对数函数尤其要小心（它会持续增长但极慢）
一起做 T-16 · Evaluate a Logarithmic Model
A dataset is modeled by $y = 3.075 + 3.904\ln x$, with original $x$-values from 0.5 to 3.8.
(a) Evaluate $y(3)$ and decide if the prediction is reasonable.
(b) Use the same model to estimate $y(5)$. Is this prediction reliable?
(a) $y(3)$
$y(3)=3.075+3.904\ln3 = 3.075+3.904\times1.0986 \approx 3.075+4.289 \approx 7.364$
$x=3$ is within the data range $[0.5, 3.8]$ → Interpolation 插值 → prediction is reasonable ✓
$y(3)=3.075+3.904\ln3 = 3.075+3.904\times1.0986 \approx 3.075+4.289 \approx 7.364$
$x=3$ is within the data range $[0.5, 3.8]$ → Interpolation 插值 → prediction is reasonable ✓
(b) $y(5)$
$y(5)=3.075+3.904\ln5\approx3.075+6.283\approx9.358$ (keep full precision until the end)
$x=5$ is outside the data range $[0.5, 3.8]$ → Extrapolation 外推 → prediction is less reliable ⚠️
$y(5)=3.075+3.904\ln5\approx3.075+6.283\approx9.358$ (keep full precision until the end)
$x=5$ is outside the data range $[0.5, 3.8]$ → Extrapolation 外推 → prediction is less reliable ⚠️
练习 E-7 · Log Model Application
The loudness $L$ (in decibels) of a sound is modeled by $L=10\log\dfrac{I}{I_0}$, where $I$ is the sound intensity and $I_0=10^{-12}$ W/m² is the threshold of hearing.
(a) Find the loudness of a sound with intensity $I=10^{-5}$ W/m².
(b) If loudness increases by 20 dB, how does the intensity change?
(a) $L=10\log\dfrac{10^{-5}}{10^{-12}}=10\log10^7=10\times7=\boxed{70\text{ dB}}$
(b) Let $L_2=L_1+20$:
$L_1=10\log\dfrac{I_1}{I_0}$ and $L_2=10\log\dfrac{I_2}{I_0}$
$L_2-L_1=10\log\dfrac{I_2}{I_1}=20 \Rightarrow \log\dfrac{I_2}{I_1}=2 \Rightarrow \dfrac{I_2}{I_1}=100$
Intensity increases by a factor of 100×. 声强增大100倍。
(b) Let $L_2=L_1+20$:
$L_1=10\log\dfrac{I_1}{I_0}$ and $L_2=10\log\dfrac{I_2}{I_0}$
$L_2-L_1=10\log\dfrac{I_2}{I_1}=20 \Rightarrow \log\dfrac{I_2}{I_1}=2 \Rightarrow \dfrac{I_2}{I_1}=100$
Intensity increases by a factor of 100×. 声强增大100倍。
Building Logarithmic Models 构建对数模型
- Meaning: a log model fits situations where the inputs change proportionally while the outputs change by equal amounts. If $N=N_0\cdot2^t$, then $t=\log_2\frac{N}{N_0}$ counts how many times $N_0$ has been doubled (CED 2.14.A.1).
- From a proportion and a zero: "output rises by $a$ each time the input is multiplied by $b$, and the output is 0 at $x=x_0$" → $y=a\log_b\dfrac{x}{x_0}$.
- From two points: use $y=a+c\ln x$ (two unknowns, two equations), or recognize the proportion directly.
- Technology: LnReg fits $y=a+b\ln x$ (CED 2.14.A.4).
TI-84 · LnReg
- STAT → 1: Edit: inputs in L1 (must be positive), outputs in L2.
- STAT → CALC → 9: LnReg; Store RegEQ: Y1. The model is $y=a+b\ln x$.
一起做 T-17 · Build a Log Model Three Ways 三种方法建对数模型
(a) A tree's height increases by 4 m every time its age triples, and it is 2 m tall at age 1 year. Write a model $h(t)$ and find the age when it is 14 m tall.
(b) A log model $y=a+c\ln x$ passes through $(2,5)$ and $(8,11)$. Find it.
(c) The number of plant species $y$ found in a sample area of $x$ m²:
| $x$ (m²) | 1 | 2 | 4 | 8 | 16 | 32 |
|---|---|---|---|---|---|---|
| $y$ | 5 | 8 | 10 | 13 | 15 | 18 |
Use LnReg to find a model. Predict $y$ for 20 m² and for 100 m², and comment.
(a) Proportion + a value
Each factor of 3 in age adds 4 m: $h(t)=2+4\log_3t$. $2+4\log_3t=14\Rightarrow\log_3t=3\Rightarrow t=27$ years.
Each factor of 3 in age adds 4 m: $h(t)=2+4\log_3t$. $2+4\log_3t=14\Rightarrow\log_3t=3\Rightarrow t=27$ years.
(b) Two points
$c(\ln8-\ln2)=11-5 \Rightarrow c\ln4=6 \Rightarrow c=\dfrac{6}{\ln4}\approx4.328$; $a=5-c\ln2=5-3=2$.
So $y=2+\dfrac{6}{\ln4}\ln x=2+3\log_2x$ (inputs ×4 → outputs +6).
$c(\ln8-\ln2)=11-5 \Rightarrow c\ln4=6 \Rightarrow c=\dfrac{6}{\ln4}\approx4.328$; $a=5-c\ln2=5-3=2$.
So $y=2+\dfrac{6}{\ln4}\ln x=2+3\log_2x$ (inputs ×4 → outputs +6).
(c) LnReg
$y\approx5.143+3.669\ln x$. The area doubles while the species count rises by about 2.5 each time, a logarithmic pattern.
$y(20)\approx16.133$ species (interpolation, reasonable). $y(100)\approx22.037$ species (extrapolation beyond 32 m², less reliable).
$y\approx5.143+3.669\ln x$. The area doubles while the species count rises by about 2.5 each time, a logarithmic pattern.
$y(20)\approx16.133$ species (interpolation, reasonable). $y(100)\approx22.037$ species (extrapolation beyond 32 m², less reliable).
(a) $h(t)=2+4\log_3t$, 27 years · (b) $y=2+3\log_2x$ · (c) $y\approx5.143+3.669\ln x$
7.7 Semilog Plots & Linearizing Exponential Data 半对数图与指数数据线性化
A semilog plot has one logarithmic axis and one linear axis.
半对数图：一个轴是对数刻度，另一个是线性刻度。
Why useful: Exponential data that curves sharply on a regular graph appears approximately linear on a semilog plot.
半对数图：一个轴是对数刻度，另一个是线性刻度。
Why useful: Exponential data that curves sharply on a regular graph appears approximately linear on a semilog plot.
指数数据线性化 Linearizing Exponential Data
从指数模型 $y=ab^x$ 开始，取自然对数两边：
$$\ln y = \ln(ab^x) = \ln a + x\ln b$$
令 $Y=\ln y$，则 $Y = x\cdot(\ln b) + \ln a$ — 这是关于 $x$ 的线性方程！斜率 = $\ln b$；y截距 = $\ln a$
Regular Cartesian Plot 普通坐标图
Semilog Plot 半对数图 (log y-axis)
一起做 T-18 · Linearize an Exponential Model
Linearize the model $y = 5\cdot2^x$ using common logarithms ($\log_{10}$).
Take $\log$ of both sides
$\log y = \log(5\cdot2^x) = \log 5 + \log 2^x$
$\log y = \log(5\cdot2^x) = \log 5 + \log 2^x$
Apply Power Law
$\log y = x\log 2 + \log 5$
$\log y = x\log 2 + \log 5$
Identify linear form
Let $Y = \log y$:
$$Y = (\log 2)\cdot x + \log 5$$ This is linear in $x$:
Slope = $\log 2 \approx 0.301$; y-intercept = $\log 5 \approx 0.699$
在半对数图（y轴取 $\log y$）上，数据点应该近似落在一条直线上。
Let $Y = \log y$:
$$Y = (\log 2)\cdot x + \log 5$$ This is linear in $x$:
Slope = $\log 2 \approx 0.301$; y-intercept = $\log 5 \approx 0.699$
在半对数图（y轴取 $\log y$）上，数据点应该近似落在一条直线上。
一起做 T-19 · Linearize a Half-Life Model
A radioactive sample follows $A = A_0\cdot\left(\dfrac{1}{2}\right)^{t/H}$ where $H$ is the half-life. Linearize using $\log_2$.
Take $\log_2$ of both sides
$\log_2 A = \log_2\!\left[A_0\cdot\left(\tfrac{1}{2}\right)^{t/H}\right] = \log_2 A_0 + \log_2\left(\tfrac{1}{2}\right)^{t/H}$
$\log_2 A = \log_2\!\left[A_0\cdot\left(\tfrac{1}{2}\right)^{t/H}\right] = \log_2 A_0 + \log_2\left(\tfrac{1}{2}\right)^{t/H}$
Apply Power Law
$\log_2 A = \log_2 A_0 + \dfrac{t}{H}\cdot\log_2\tfrac{1}{2}$
Since $\log_2\dfrac{1}{2}=\log_2 2^{-1}=-1$:
$\log_2 A = -\dfrac{t}{H} + \log_2 A_0$
$\log_2 A = \log_2 A_0 + \dfrac{t}{H}\cdot\log_2\tfrac{1}{2}$
Since $\log_2\dfrac{1}{2}=\log_2 2^{-1}=-1$:
$\log_2 A = -\dfrac{t}{H} + \log_2 A_0$
Linear form
Let $Y=\log_2 A$:
$$Y = -\frac{1}{H}\cdot t + \log_2 A_0$$ Slope = $-\dfrac{1}{H}$ (negative, since decay); intercept = $\log_2 A_0$
从斜率可直接求半衰期：$H = -\dfrac{1}{\text{slope}}$
Let $Y=\log_2 A$:
$$Y = -\frac{1}{H}\cdot t + \log_2 A_0$$ Slope = $-\dfrac{1}{H}$ (negative, since decay); intercept = $\log_2 A_0$
从斜率可直接求半衰期：$H = -\dfrac{1}{\text{slope}}$
练习 E-8 · Semilog Interpretation
A semilog plot of $(\ln y)$ vs. $x$ shows a straight line with slope 0.4 and y-intercept 2.
(a) Write the linear equation for $\ln y$.
(b) Recover the original exponential model for $y$.
(a) $\ln y = 0.4x + 2$
(b) Exponentiate both sides: $y = e^{0.4x+2} = e^2\cdot e^{0.4x} \approx 7.389\cdot e^{0.4x}$
Or: $y = e^2\cdot(e^{0.4})^x \approx 7.389\cdot(1.492)^x$
This is exponential growth with $a=e^2\approx7.389$ and $b=e^{0.4}\approx1.492$.
(b) Exponentiate both sides: $y = e^{0.4x+2} = e^2\cdot e^{0.4x} \approx 7.389\cdot e^{0.4x}$
Or: $y = e^2\cdot(e^{0.4})^x \approx 7.389\cdot(1.492)^x$
This is exponential growth with $a=e^2\approx7.389$ and $b=e^{0.4}\approx1.492$.
Semi-log Plot from Data 由数据画半对数图并还原模型
Steps: (1) compute $\ln y$ (or $\log y$) for each data point; (2) check whether $(x,\ln y)$ is roughly linear (equal differences for equally spaced $x$); (3) fit a line $\ln y=mx+c$ with linear-model techniques (CED 2.15.B.1); (4) undo the log: $a=e^{c}$, $b=e^{m}$, so $y=ab^x$.
Advantage (CED 2.15.A.2): with a semi-log plot, a constant never needs to be added to the dependent variable to reveal that an exponential model is appropriate. 半对数图的优点：不需要先给因变量加减常数，就能看出指数模型是否合适。
Advantage (CED 2.15.A.2): with a semi-log plot, a constant never needs to be added to the dependent variable to reveal that an exponential model is appropriate. 半对数图的优点：不需要先给因变量加减常数，就能看出指数模型是否合适。
一起做 T-20 · Linearize Real Data 线性化真实数据（calculator 可用）
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $y$ | 3.2 | 5.1 | 7.9 | 12.6 | 19.8 | 31.5 |
(a) Compute $\ln y$ and decide whether an exponential model is appropriate. (b) Fit a line to $(x,\ln y)$ and convert it to an exponential model $y=ab^x$.
(a) $\ln y\approx1.163,\ 1.629,\ 2.067,\ 2.534,\ 2.986,\ 3.450$. Differences $\approx0.466,\ 0.438,\ 0.467,\ 0.452,\ 0.464$ are roughly constant, so $\ln y$ is roughly linear in $x$ → the data look linear on a semi-log plot → exponential model appropriate.
Standard axes: curved 普通坐标：曲线
Log-scaled $y$-axis: a straight line 对数纵轴：直线
Standard axes: curved 普通坐标：曲线
Log-scaled $y$-axis: a straight line 对数纵轴：直线
(b) LinReg on $(x,\ln y)$: $\ln y\approx0.456x+1.164$. Then $a=e^{1.164}\approx3.203$ and $b=e^{0.456}\approx1.578$: $\;y\approx3.203(1.578)^x$ (the same model ExpReg gives).
$\ln y\approx0.456x+1.164$ → $y\approx3.203(1.578)^x$
Chapter 7 Summary 本章总结
Key Facts & Formulas 核心知识
| Topic | Key Rule/Formula | Note 备注 |
|---|---|---|
| Definition | $\log_b x=y \Leftrightarrow b^y=x$ | $b>0,b\neq1,x>0$ |
| Basic values | $\log_b 1=0$; $\log_b b=1$; $\log_b b^x=x$ | Inverse identities |
| Parent domain/range | Domain $(0,+\infty)$; Range $(-\infty,+\infty)$ | No y-intercept |
| Vertical asymptote | $x=0$ always | Left boundary |
| Product Law | $\log_b(MN)=\log_b M+\log_b N$ | 乘积→加法 |
| Quotient Law | $\log_b(M/N)=\log_b M-\log_b N$ | 商→减法 |
| Power Law | $\log_b M^p=p\log_b M$ | 指数→系数 |
| Change of base | $\log_b x=\dfrac{\ln x}{\ln b}$ | Calculator friendly |
| Solve exp. eq. | Take $\ln$ both sides, use Power Law | Check domain after |
| Solve log eq. | Same base → arguments equal | Check for extraneous roots! |
| Linearize $y=ab^x$ | $\ln y = x\ln b + \ln a$ | Slope=$\ln b$, int=$\ln a$ |
| Log scale | Each unit = multiply by the base | $x$ is placed at $\log x$ |
| Log data | Inputs ×b, outputs +a → $y=a\log_b x$ | Shifted inputs → $a\log_b(x-h)$ |
| Laws as graphs | $\log_b(x^k)=k\log_b x$; change of base | All logs are vertical dilations of each other |
| Rewrite base | $b^x=e^{(\ln b)x}$ | Growth factor ↔ continuous rate |
| Inverse of $ab^{x-h}+k$ | Undo in reverse order | Domain of $f^{-1}$ = range of $f$ |
| Log models | Proportion + zero; two points; LnReg $y=a+b\ln x$ | Interpolate, don't extrapolate far |
口诀总结 Memory Tips
- 🔢 对数 = 指数问题：$\log_b x=y$ 就是问"$b$ 的几次方等于 $x$？"
- 🔄 对数与指数互为反函数：$\log_b(b^x)=x$，$b^{\log_b x}=x$
- 📏 对数函数定义域：真数必须 $>0$！解方程后必须验根
- ➕ 乘法→加法，除法→减法，幂→系数（三大对数法则）
- 📈 增长快后慢 → 对数模型；直线型 → 线性；等比增长 → 指数
- 📊 半对数图：指数数据变直线；斜率=$\ln b$，截距=$\ln a$
- 📐 对数刻度：走一格 = 乘以底数；对数型数据：输入乘、输出加
- 🔁 所有对数函数互为垂直伸缩；$b^x=e^{(\ln b)x}$ 可以换算连续增长率
- ⚠️ 解对数方程必须检验增根（extraneous solutions）！
Solving Strategy Quick Guide 解题策略速查:
指数方程 $b^x=c$ (bases can't match) → 取 $\ln$ 两边 → $x=\dfrac{\ln c}{\ln b}$
对数方程 $\log_b A = \log_b B$ → 令 $A=B$，再验域
对数不等式 $\log_b f(x) > \log_b g(x)$ → 若 $b>1$: $f>g$；若 $b<1$: $f<g$（方向翻转！）
指数方程 $b^x=c$ (bases can't match) → 取 $\ln$ 两边 → $x=\dfrac{\ln c}{\ln b}$
对数方程 $\log_b A = \log_b B$ → 令 $A=B$，再验域
对数不等式 $\log_b f(x) > \log_b g(x)$ → 若 $b>1$: $f>g$；若 $b<1$: $f<g$（方向翻转！）
Chapter 7 Review 章末习题
Part A选择题 · 不用计算器 Multiple Choice · No Calculator
R-1Halfway on a Logarithmic Scale — 对数刻度的中点改编自 Unit 2 Progress Check · MCQ Part B #1
On a base-$10$ logarithmic scale, two tick marks are labeled $4$ and $400$. Point $P$ is located exactly halfway between these two tick marks. What value does point $P$ represent?
(A) $20$
(B) $40$
(C) $100$
(D) $202$
答案：(B) 对数刻度上的位置是 $\log_{10}$ 值。$P$ 的位置 $=\dfrac{\log4+\log400}{2}=\log\sqrt{4\cdot400}=\log40$，所以 $P$ 表示 40。（等距的刻度对应相同的倍数：从 4 到 40 乘 10，从 40 到 400 也乘 10。）
(D) 202 是算术平均——那是普通刻度上的中点。
R-2Expanding a Logarithm — 对数展开改编自 Practice Exam 1 · MCQ Part A #5
Let $x$ and $y$ be positive constants. Which of the following is equivalent to $\log_3\left(\dfrac{9x^4}{y}\right)$?
(A) $9+4\log_3x-\log_3y$
(B) $2+\log_3(4x)-\log_3y$
(C) $\dfrac{2+4\log_3x}{\log_3y}$
(D) $2+4\log_3x-\log_3y$
答案：(D) 商 → 差，积 → 和，幂 → 系数：$\log_3(9x^4)-\log_3y=\log_39+\log_3x^4-\log_3y=2+4\log_3x-\log_3y$。
(A) $\log_39=2$，不是 9；(B) $\log_3x^4=4\log_3x$，不是 $\log_3(4x)$；(C) $\log_3\dfrac MN\ne\dfrac{\log_3M}{\log_3N}$。
R-3Points on the Inverse Graph — 反函数图像上的点改编自 Practice Exam 3 · MCQ Part A #5
The graph of the exponential function $k(x)=b^x$, where $b>0$ and $b\ne1$, passes through the point $(2,25)$. The logarithmic function $h$ is given by $h(x)=\log_bx$. Which of the following pairs of points are on the graph of $h$?
(A) $(25,2)$ and $\left(\tfrac15,-1\right)$
(B) $(2,25)$ and $\left(-1,\tfrac15\right)$
(C) $(25,2)$ and $\left(-1,\tfrac15\right)$
(D) $(25,2)$ and $(5,0)$
答案：(A) $b^2=25$，$b=5$。$h(x)=\log_5x$ 是 $k$ 的反函数，把 $k$ 上的点 $(x,y)$ 换成 $(y,x)$：$k$ 过 $(2,25)$、$(-1,\tfrac15)$，所以 $h$ 过 $(25,2)$、$(\tfrac15,-1)$。验证：$\log_525=2$，$\log_5\tfrac15=-1$ ✓。
(B) 是 $k$ 上的点；(C) 第二个点没交换坐标；(D) $\log_55=1\ne0$。
R-4Domain and Asymptote — 对数函数的定义域与渐近线改编自 2024 AP 真题（亚太卷）· MCQ Part A #27
The function $k$ is a transformation of a logarithmic function with base $3$. The domain of $k$ is all real numbers less than $3$. The graph of $k$ has a vertical asymptote at $x=3$ and a $y$-intercept at $(0,2)$. Which of the following could be an expression for $k(x)$?
(A) $\log_3(x-3)+1$
(B) $\log_3(3-x)$
(C) $\log_3(3-x)+1$
(D) $\log_3(x+3)+1$
答案：(C) 定义域 $x<3$ 要求真数 $3-x>0$，渐近线就是 $3-x=0$ 的 $x=3$。$y$ 截距：$\log_3(3-0)+1=1+1=2$ ✓。
(A) 定义域 $x>3$；(B) $y$ 截距是 $\log_33=1$；(D) $y$ 截距也是 2，但定义域是 $x>-3$，渐近线 $x=-3$。
R-5Exponential Data on a Semi-log Plot — 半对数图改编自 Practice Exam 3 · MCQ Part A #9
A data set that appears exponential is modeled by $k(t)=4(0.6)^t$. The data are graphed on a semi-log plot, where the vertical axis is logarithmically scaled with base $10$. Which of the following describes the appearance of the data in the semi-log plot?
(A) The data appear linear with a slope of $0.6$.
(B) The data appear linear with a slope of $\log_{10}0.6$.
(C) The data appear linear with a slope of $\log_{10}4$.
(D) The data appear as a decreasing curve that is concave up.
答案：(B) 两边取 $\log_{10}$：$\log k=\log4+t\log0.6$。以 $t$ 为横轴、$\log k$ 为纵轴，这是一条直线：斜率 $\log0.6\approx-0.222$（负的，所以递减），纵截距 $\log4$。
(A) 斜率是增长因子的对数，不是增长因子本身；(C) $\log4$ 是截距；(D) 半对数图的作用正是把指数数据"拉直"。
Part B选择题 · 可用计算器 Multiple Choice · Calculator Allowed
R-6Logarithmic vs. Linear Model — 对数模型与线性模型改编自 Practice Exam 1 · MCQ Part B #8
Two models are constructed for the height of a plant, in centimeters, $t$ weeks after it is planted: the logarithmic model $f(t)=3+2\ln(t+1)$ and the linear model $g(t)=3+0.5t$, for $t\ge0$. What is the first time $t$ at which the height predicted by $f$ is $1$ centimeter more than the height predicted by $g$?
(A) $1.258$
(B) $3.000$
(C) $5.467$
(D) $9.347$
答案：(A) 解 $f(t)-g(t)=1$，即 $2\ln(t+1)-0.5t=1$。用计算器画 $Y_1=f(x)-g(x)$ 和 $Y_2=1$，交点 $t\approx1.258$ 和 $t\approx5.467$，第一次是 $t\approx1.258$。
(B) $t=3$ 时两个模型相差最大；(C) 是第二次相差 1；(D) 是两个模型再次相等的时刻。
R-7Logarithmic Regression — 对数回归改编自 Practice Exam 2 · MCQ Part B #9
| Age $x$ (years) | $1$ | $2$ | $3$ | $5$ | $7$ | $10$ |
|---|---|---|---|---|---|---|
| Weight $y$ (kg) | $12$ | $30$ | $41$ | $55$ | $64$ | $73$ |
The table gives the weight $y$, in kilograms, of an animal at selected ages $x$, in years. A logarithmic regression of the form $y=a+b\ln x$ is used to model these data. What weight, in kilograms, does the model predict at age $4.5$ years?
(A) $44.773$
(B) $46.979$
(C) $51.786$
(D) $51.945$
答案：(D) STAT → CALC → 9: LnReg：$y\approx11.827+26.672\ln x$，代入 $x=4.5$ 得 $\approx51.945$ kg。
其他选项来自别的回归：(A) LinReg，(B) PwrReg，(C) QuadReg。
R-8Rewriting an Exponential in Another Base — 换底改写
The number of bacteria in a culture is modeled by $f(t)=200e^{0.35t}$, where $t$ is measured in hours. The model can be rewritten as $f(t)=200\cdot2^{t/k}$, where $k$ is the doubling time of the bacteria in hours. What is the value of $k$?
(A) $0.243$
(B) $0.505$
(C) $1.980$
(D) $2.857$
答案：(C) $e^{0.35t}=2^{t/k}$ 对所有 $t$ 成立。两边取 $\ln$：$0.35t=\dfrac tk\ln2$，所以 $k=\dfrac{\ln2}{0.35}\approx1.980$ 小时。（也可以直接解 $e^{0.35t}=2$。）
(B) 把分子分母弄反了；(D) $\dfrac1{0.35}$ 漏了 $\ln2$。
Part C简答题 Free Response · Show Your Work
R-9A Logarithmic Model from Two Data Points — 由两个数据建对数模型改编自 2024 AP 真题（亚太卷）· FRQ #2
On the day a new app is released $(t=0)$, it is downloaded $25$ thousand times. On day $63$ $(t=63)$, it is downloaded $61$ thousand times. The number of downloads on day $t$, in thousands, is modeled by $D(t)=a+b\ln(t+1)$. A calculator is allowed.
(a) (i) Use the given data to write two equations that can be used to find the values of $a$ and $b$.
(ii) Find the values of $a$ and $b$.
(b) (i) Find the average rate of change of the number of downloads, in thousands per day, from $t=0$ to $t=63$.
(ii) Use the average rate of change from (i) to estimate the number of downloads on day $30$.
(iii) The estimate in (ii) is less than $D(30)$. Explain why, using characteristics of the model $D$.
(c) According to the model, on what day $t$ do the downloads first reach $50$ thousand?
(a) (i) $D(0)=a+b\ln1=25$，$D(63)=a+b\ln64=61$。
(ii) $\ln1=0$，所以 $a=25$；$b=\dfrac{36}{\ln64}\approx8.656$。
(b) (i) $\dfrac{61-25}{63-0}=\dfrac{36}{63}\approx0.571$ 千次/天。
(ii) $25+0.571\times30\approx42.143$ 千次。
(iii) $b>0$，$D$ 递增且图像下凹（增长越来越慢），所以在 $t=0$ 到 $t=63$ 之间，连接两端点的割线在图像下方，用割线估计会偏小。（模型值 $D(30)\approx54.725$。）
(c) 解 $25+8.656\ln(t+1)=50$：$\ln(t+1)=\dfrac{25}{b}$，$t=e^{25/b}-1\approx16.96$，即大约第 17 天。
R-10Equations and an Inequality — 对数方程与不等式改编自 2024 AP 真题（亚太卷）· FRQ #4
Directions: Solve each part without a calculator. Give exact answers and show the work that leads to your answers.
(a) Solve $\log_2(x+6)+\log_2x=4$ for $x$.
(b) Solve $4\cdot3^{(x-1)}=20$ for $x$.
(c) Rewrite $2\log_5x-\log_5(x+1)+1$ as a single logarithm, for $x>0$.
(d) Solve the inequality $\log_3(2x-1)<2$. Write your answer in interval notation.
(a) 合并：$\log_2\big(x(x+6)\big)=4\Rightarrow x^2+6x=16\Rightarrow(x+8)(x-2)=0$。$x=-8$ 代回去 $\log_2(-8)$ 无定义，舍去。所以 $x=2$（检验：$\log_28+\log_22=3+1=4$ ✓）。
(b) $3^{x-1}=5\Rightarrow x-1=\log_35\Rightarrow x=1+\log_35$。
(c) $1=\log_55$：$\log_5x^2-\log_5(x+1)+\log_55=\log_5\dfrac{5x^2}{x+1}$。
(d) 定义域：$2x-1>0$，$x>\tfrac12$。底数 $3>1$，$\log_3$ 递增，不等号方向不变：$2x-1<3^2=9$，$x<5$。解集 $\left(\tfrac12,\,5\right)$。（不要漏掉定义域这一步。）
R-11Inverse of a Transformed Exponential — 变换后指数函数的反函数
The function $f$ is given by $f(x)=2\cdot3^{(x-1)}+4$.
(a) Find an expression for $f^{-1}(x)$.
(b) State the domain and range of $f^{-1}$, and give the equation of the vertical asymptote of the graph of $f^{-1}$. Explain how the asymptote is related to the graph of $f$.
(c) Find $f^{-1}(22)$ without a calculator, and verify your answer using $f$.
(d) Sketch the graphs of $f$ and $f^{-1}$ on the same axes.
(a) 设 $y=2\cdot3^{x-1}+4$，交换 $x,y$：$x=2\cdot3^{y-1}+4$。$3^{y-1}=\dfrac{x-4}{2}$，$y-1=\log_3\dfrac{x-4}{2}$，所以 $f^{-1}(x)=1+\log_3\left(\dfrac{x-4}{2}\right)$。
(b) $f$ 的值域是 $(4,\infty)$（$2\cdot3^{x-1}>0$），所以 $f^{-1}$ 的定义域是 $(4,\infty)$，值域是全体实数。$f^{-1}$ 的垂直渐近线 $x=4$，对应 $f$ 的水平渐近线 $y=4$——关于 $y=x$ 对称。
(c) $f^{-1}(22)=1+\log_3\dfrac{18}{2}=1+\log_39=3$。验证：$f(3)=2\cdot3^2+4=22$ ✓。
(d) 两条曲线关于 $y=x$ 对称；例如 $f$ 过 $(1,6)$，$f^{-1}$ 过 $(6,1)$：
R-12Linearizing Data with a Semi-log Plot — 用半对数图线性化数据
| $t$ (hours) | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $y$ | $2.0$ | $2.8$ | $3.9$ | $5.5$ | $7.7$ |
The table gives the area $y$, in square centimeters, of a bacteria colony $t$ hours after an experiment begins. A calculator is allowed.
(a) Find $\log_{10}y$ for each value in the table, to three decimal places.
(b) The points $(t,\log_{10}y)$ appear to lie on a line. Explain why this suggests that an exponential model is appropriate for the data.
(c) A linear regression of $\log_{10}y$ on $t$ gives $\log_{10}y=0.146t+0.300$. Use this equation to write an exponential model of the form $y=ab^t$.
(d) Interpret the value of $b$ in context, and use the model to predict the area at $t=6$.
(a) $0.301,\ 0.447,\ 0.591,\ 0.740,\ 0.886$。
(b) 如果 $y=ab^t$，两边取对数：$\log y=\log a+t\log b$，$\log y$ 是 $t$ 的线性函数。所以在半对数图上呈直线的数据，适合用指数模型。（这里 $\log y$ 每小时大约增加 $0.146$，增量近似不变。）
(c) $y=10^{0.146t+0.300}=10^{0.300}\cdot\left(10^{0.146}\right)^t\approx1.995(1.400)^t$。
(d) $b\approx1.400$：菌落面积每小时大约变为原来的 1.4 倍（增长 40%）。$y(6)=10^{0.146\times6+0.300}=10^{1.176}\approx14.997$ cm²（直接用回归直线算最准确）。
