3.4 Logarithmic Functions, Laws & Equations

3.4 Logarithmic Functions, Laws & Equations

Logarithmic functions are the mathematical inverses of exponential functions. On the Bocconi Admission Test, logarithmic problems appear as algebraic equation solving, base-change simplifications, domain checks, and inequality analysis. Because logarithms impose strict domain restrictions, precision in tracking signs and discarding extraneous roots is essential for maximizing exam scores.

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1. Definition and Fundamental Properties of the Logarithm

For any positive base $b > 0$ with $b \neq 1$, the logarithm to base $b$ of a positive real number $x > 0$ is defined as the unique exponent $y$ to which $b$ must be raised to yield $x$:

$$y = \log_b(x) \iff b^y = x$$

Analytical Characteristics of $y = \log_b(x)$

1. Natural Domain: $\text{Dom}(f) = (0, \infty)$. The argument must be strictly positive. Logarithms of zero or negative numbers are undefined in the real number system $\mathbb{R}$.
2. Range: $\text{Ran}(f) = \mathbb{R} = (-\infty, \infty)$. A logarithm can take any real value (positive, zero, or negative).
3. Universal $x$-Intercept: Because $b^0 = 1$, every logarithmic curve passes through $(1, 0)$: $$\log_b(1) = 0$$
4. Base Point: Because $b^1 = b$, every logarithmic curve passes through $(b, 1)$: $$\log_b(b) = 1$$
5. Vertical Asymptote: The line $x = 0$ (the $y$-axis) is a vertical asymptote for all logarithmic curves:
   • If $b > 1$: $\lim_{x \to 0^+} \log_b(x) = -\infty$ and $\lim_{x \to \infty} \log_b(x) = \infty$ (strictly increasing).
   • If $0 < b < 1$: $\lim_{x \to 0^+} \log_b(x) = \infty$ and $\lim_{x \to \infty} \log_b(x) = -\infty$ (strictly decreasing).

Canonical Notations

• Natural Logarithm: Base $e \approx 2.71828$, denoted $\ln(x) = \log_e(x)$.
• Common Logarithm: Base 10, denoted $\log(x) = \log_{10}(x)$.

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2. Fundamental Laws of Logarithms

Because logarithms are exponents, each logarithmic law corresponds directly to an exponential property. These laws hold if and only if all constituent arguments are strictly positive ($M > 0, N > 0$):

Critical Bocconi Trap: Logarithms do NOT distribute across addition or subtraction: $\log_b(M + N) \neq \log_b(M) + \log_b(N) \qquad \text{and} \qquad \log_b(M - N) \neq \frac{\log_b(M)}{\log_b(N)}$ Furthermore, powers of logarithms are not powers of arguments: $[\log_b(x)]^k \neq k \log_b(x)$.

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3. Base Conversion Identities & Fast Mental Shortcuts

On the Bocconi exam, expressions frequently mix logarithms with different bases (e.g., base 2, 4, 8, or base 3, 9, 27). Converting to a uniform base is required.

The Change of Base Formula

For any positive bases $b, c > 0$ with $b, c \neq 1$ and argument $x > 0$:

$$\log_b(x) = \frac{\log_c(x)}{\log_c(b)} = \frac{\ln(x)}{\ln(b)}$$

Derived Power and Reciprocal Identities

1. Reciprocal Base Identity: Setting $x = c$ in the change of base formula: $$\log_b(a) = \frac{1}{\log_a(b)}$$
2. Power in the Base Identity: When the base is a power $b^k$: $$\log_{b^k}(x) = \frac{\ln(x)}{\ln(b^k)} = \frac{\ln(x)}{k \ln(b)} = \frac{1}{k} \log_b(x)$$
3. General Power-Ratio Identity: $$\log_{b^k}(x^m) = \frac{m}{k} \log_b(x)$$ Speed Example: $\log_8(32) = \log_{2^3}(2^5) = \frac{5}{3} \log_2(2) = \frac{5}{3}$. This takes 3 seconds mentally.
4. Telescoping Logarithm Products: $$\log_a(b) \cdot \log_b(c) \cdot \log_c(d) = \frac{\ln b}{\ln a} \cdot \frac{\ln c}{\ln b} \cdot \frac{\ln d}{\ln c} = \frac{\ln d}{\ln a} = \log_a(d)$$

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4. Solving Logarithmic Equations: The Mandatory 3-Step Protocol

Algebraic operations on logarithms (such as applying the product law to combine terms) expand the domain of the expression. For example, the domain of $\log_2(x) + \log_2(x - 2)$ is $(2, \infty)$, but the combined expression $\log_2(x(x - 2))$ is defined on $(-\infty, 0) \cup (2, \infty)$. Consequently, solving combined equations often introduces extraneous roots. To prevent errors, follow this protocol:

Step 1: Establish the Natural Domain First

Before applying any logarithmic laws, write down the strict positivity condition for every argument:

$$\mathcal{D} = \{x \in \mathbb{R} \mid g_1(x) > 0, \, g_2(x) > 0, \dots\}$$

Step 2: Condense and Exponentiate

Use logarithm properties to condense the equation into either:

• Equal Bases: $\log_b(f(x)) = \log_b(g(x)) \implies f(x) = g(x)$
• Single Logarithm Equal to Constant: $\log_b(f(x)) = c \implies f(x) = b^c$

Step 3: Solve the Resulting Equation & Filter Candidate Roots

Solve the polynomial or algebraic equation for $x$, then test every candidate solution against the domain $\mathcal{D}$ established in Step 1. Discard any solution outside $\mathcal{D}$.

Fully Worked Example: Extraneous Root Detection

Solve for $x \in \mathbb{R}$:

$$\log_3(x - 2) + \log_3(x + 6) = 2$$

• Step 1 (Domain):

  • $x - 2 > 0 \implies x > 2$
  • $x + 6 > 0 \implies x > -6$
  • Combined natural domain: $\mathcal{D} = (2, \infty)$.

• Step 2 (Condense & Exponentiate):

  $$\log_3((x - 2)(x + 6)) = 2 \implies (x - 2)(x + 6) = 3^2 = 9$$

• Step 3 (Solve & Filter):

  $$x^2 + 4x - 12 = 9 \implies x^2 + 4x - 21 = 0 \implies (x + 7)(x - 3) = 0$$

  Candidate solutions: $x = -7$ and $x = 3$.

  • Check $x = 3$: $3 \in (2, \infty)$ is valid.
  • Check $x = -7$: $-7 \notin (2, \infty)$ is extraneous (it would evaluate to $\log_3(-9)$, which is undefined).

Therefore, the unique real solution is $x = 3$.

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5. Solving Logarithmic Inequalities

When solving inequalities of the form $\log_b(f(x)) < \log_b(g(x))$, the direction of the inequality depends strictly on base $b$, alongside domain restrictions:

$$\begin{cases} 0 < f(x) < g(x) & \text{if } b > 1 \quad (\text{preserves inequality direction}) \\ f(x) > g(x) > 0 & \text{if } 0 < b < 1 \quad (\text{reverses inequality direction}) \end{cases}$$

Worked Example: Inequality with Fractional Base

Find the solution set for the inequality:

$$\log_{1/3}(3x - 6) > -2$$

Step 1: Domain Restriction.

$$3x - 6 > 0 \implies 3x > 6 \implies x > 2$$

Step 2: Solve with Base Reversal. Because the base $b = 1/3$ satisfies $0 < b < 1$, exponentiating with base $1/3$ reverses the inequality sign:

$$3x - 6 < \left(\frac{1}{3}\right)^{-2} = (3^{-1})^{-2} = 3^2 = 9$$

Solve for $x$:

$$3x < 15 \implies x < 5$$

Step 3: Intersect with Domain. Intersecting $x < 5$ with the domain $x > 2$:

$$x \in (2, 5)$$

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6. Common Bocconi Traps & Exam Strategies

1. Omitting the Positivity Check: Solving $\log_2(x-1) = \log_2(2x-5)$ gives $x-1 = 2x-5 \implies x = 4$. Checking $x=4$ gives $4-1=3>0$ and $8-5=3>0$, so $x=4$ is valid. But if the algebra had yielded $x=0$, both arguments would be negative. Always check candidate roots in the original expressions.
2. Even Power Splitting Trap: Be cautious with $\log_b(x^2) = 6$. The solution is $x^2 = b^6 \implies x = \pm b^3$. If you carelessly rewrite $\log_b(x^2)$ as $2\log_b(x)$, you artificially restrict the domain to $x > 0$ and lose the valid negative solution $x = -b^3$.
3. Base Change Telescoping: When faced with products of logarithms such as $\log_2(3) \cdot \log_3(4) \cdots \log_{31}(32)$, immediately telescope the chain to $\log_2(32) = 5$.