3.2 Logarithms and Logarithmic Functions

3.2 Logarithms and Logarithmic Functions

Logarithmic functions are the algebraic inverses of exponential functions. Introduced by John Napier in the 17th century to simplify complex astronomical calculations, logarithms convert multiplicative operations into additive ones. In modern Gaokao Mathematics, logarithms play a critical role in solving exponential equations, analyzing function growth rates, investigating derivative behavior in calculus, and establishing algebraic inequalities.

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1. Definition of Logarithms and Fundamental Operational Laws

Mathematical Definition of Logarithm

If $a^b = N$ ($a > 0$ and $a \neq 1$), then the exponent $b$ is called the logarithm of $N$ to the base $a$, written as:

$$\log_a N = b$$

• $a$ is called the base ($a > 0, a \neq 1$).
• $N$ is called the argument (or number whose logarithm is taken), requiring $N > 0$.
• $b$ is the logarithm value ($b \in \mathbb{R}$).

Special Logarithmic Bases in Gaokao

1. Common Logarithm: Base $a = 10$, denoted as $\lg N = \log_{10} N$.
2. Natural Logarithm: Base $a = e \approx 2.718281828459...$, denoted as $\ln N = \log_e N$.

Core Identity: $\log_a 1 = 0$, $\log_a a = 1$, and $a^{\log_a N} = N$ ($N > 0$).

Operational Rules of Logarithms

For any valid base $a > 0, a \neq 1$ and positive numbers $M > 0, N > 0$:

$$\begin{aligned} 1.\quad \& \log_a(M \cdot N) = \log_a M + \log_a N \\ 2.\quad \& \log_a\left(\frac{M}{N}\right) = \log_a M - \log_a N \\ 3.\quad \& \log_a\left(M^n\right) = n \log_a M \quad (n \in \mathbb{R}) \\ 4.\quad \& \log_{a^m}\left(N^n\right) = \frac{n}{m} \log_a N \quad (m, n \in \mathbb{R}, m \neq 0) \end{aligned}$$

Change of Base Formula

For bases $a > 0, a \neq 1$ and $c > 0, c \neq 1$, and argument $b > 0$:

$$\log_a b = \frac{\log_c b}{\log_c a}$$

Derived Corollaries:

• Reciprocal Rule: $\log_a b = \frac{1}{\log_b a}$
• Chain Multiplication Rule: $\log_a b \cdot \log_b c \cdot \log_c d = \log_a d$

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2. Logarithmic Functions $y = \log_a x$ and Analytic Properties

Definition

A function of the form $f(x) = \log_a x$ ($a > 0, a \neq 1$) is defined as a logarithmic function with base $a$.

Analytic Properties Summary Table

Inverse Function Symmetry

The exponential function $y = a^x$ and logarithmic function $y = \log_a x$ are inverse functions of each other:

• Their domains and ranges swap: Domain of $a^x$ is Range of $\log_a x$, and vice versa.
• Graphs of $y = a^x$ and $y = \log_a x$ are mirror images across the main diagonal line $y = x$.

3. Solving Logarithmic Equations and Inequalities

The Non-Negotiable Golden Rule: Domain First!

Before manipulating logarithmic equations or inequalities using logarithmic rules (e.g. combining $\log_a u + \log_a v = \log_a(uv)$), you MUST establish domain bounds. Combining terms can inadvertently expand the domain (e.g., $u > 0, v > 0$ becomes $uv > 0$, which allows $u < 0$ and $v < 0$, introducing extraneous false solutions!).

Equivalence Systems for Logarithmic Inequalities

For $\log_a f(x) > \log_a g(x)$:

• If $a > 1$: $$\begin{cases} f(x) > g(x) \\ g(x) > 0 \end{cases}$$
• If $0 < a < 1$: $$\begin{cases} 0 < f(x) < g(x) \\ f(x) > 0 \end{cases}$$

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4. Worked Gaokao Exam Examples

Example 1: Solving a Quadratic Logarithmic Inequality

Problem: Solve the logarithmic inequality $\log_2(x^2 - 3x) \le 2$.

Detailed Solution:

1. Establish Domain Condition: Arguments of logarithms must be strictly positive: $$x^2 - 3x > 0 \implies x(x-3) > 0 \implies x \in (-\infty, 0) \cup (3, +\infty)$$
2. Express Right-Hand Side with Base 2: $$2 = \log_2\left(2^2\right) = \log_2 4$$ Thus, the inequality becomes: $$\log_2(x^2 - 3x) \le \log_2 4$$
3. Apply Base Monotonicity: Since base $a = 2 > 1$, the function $y = \log_2 u$ is strictly increasing, maintaining the inequality direction: $$x^2 - 3x \le 4 \implies x^2 - 3x - 4 \le 0 \implies (x-4)(x+1) \le 0 \implies x \in [-1, 4]$$
4. Intersect Solution with Domain Bounds: $$\text{Solution Set} = [-1, 4] \cap \left((-\infty, 0) \cup (3, +\infty)\right) = [-1, 0) \cup (3, 4]$$

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Example 2: Domain and Composite Monotonicity Analysis

Problem: Given function $f(x) = \log_{\frac{1}{2}}(x^2 - 2x - 3)$, determine its domain and find its strictly increasing interval.

Detailed Solution:

1. Determine Domain: The inner expression must be strictly positive: $$u(x) = x^2 - 2x - 3 > 0 \implies (x-3)(x+1) > 0$$ Thus, domain $D = (-\infty, -1) \cup (3, +\infty)$.
2. Analyze Inner Quadratic Function $u(x)$: $$u(x) = (x-1)^2 - 4$$
   • Axis of symmetry is $x = 1$.
   • On $(-\infty, -1)$, $u(x)$ is strictly decreasing and $u(x) > 0$.
   • On $(3, +\infty)$, $u(x)$ is strictly increasing and $u(x) > 0$.
3. Analyze Outer Function $g(u)$:
   • $g(u) = \log_{\frac{1}{2}} u$ has base $a = \frac{1}{2} \in (0, 1)$, so $g(u)$ is strictly decreasing on $(0, +\infty)$.
4. Apply Composite Monotonicity Rule:
   • On $(-\infty, -1)$, $u(x)$ is decreasing and $g(u)$ is decreasing, so $f(x) = g(u(x))$ is strictly increasing.
   • On $(3, +\infty)$, $u(x)$ is increasing and $g(u)$ is decreasing, so $f(x) = g(u(x))$ is strictly decreasing.
5. Conclusion: The domain of $f(x)$ is $(-\infty, -1) \cup (3, +\infty)$, and its strictly increasing interval is $(-\infty, -1)$.