2.9 Logarithms: The Logarithmic Function $y=\log_a x$

1. Core Logarithmic Definition

Logarithmic operations reverse standard exponential mathematics. Evaluating a logarithm mathematically asks: "To what target exponent must the fixed base value be raised to perfectly yield the target argument?"

$\log_a x = y \Leftrightarrow a^y = x$
  • Common Logarithm: A base of exactly 10 is implied if visibly omitted. $\log(x)$ is mathematically interpreted as $\log_{10}(x)$.
  • Natural Logarithm: A base evaluated strictly at $e$ utilizes specialized notation. $\ln(x)$ identically operates as $\log_e(x)$.

EXAMPLE 1 & 2

$\log_2 32 = 5$ (because $2^5 = 32$). $\log_2 2^{100} = 100$. $\log_2 1 = 0$.
$\log 1000000 = 6$ (because $10^6 = 1000000$). $\log 0.001 = -3$.

2. Four Universal Algebraic Laws

  1. Multiplication: $\log(xy) = \log x + \log y$
  2. Division: $\log\left(\frac{x}{y}\right) = \log x - \log y$
  3. Exponentiation: $\log(x^n) = n\log x$
  4. Inversion: $\log\left(\frac{1}{x}\right) = -\log x$

Notice: Variables can be strategically collected algebraically: $2\log A + 3\log B - 4\log C = \log\left(\frac{A^2 B^3}{C^4}\right)$.

EXAMPLE 3 & 4

Given $\ln x = a$, $\ln y = b$, $\ln z = c$:

$\ln\left(\frac{x^3y}{z^2}\right) = 3\ln x + \ln y - 2\ln z = 3a + b - 2c$.
$\ln\sqrt{x} = \ln x^{1/2} = \frac{1}{2}\ln x = \frac{a}{2}$.

Given $\ln 2 = m$, $\ln 5 = n$:

$\ln 50 = \ln(2 \cdot 5^2) = \ln 2 + 2\ln 5 = m + 2n$.
$\ln 2.5 = \ln\left(\frac{5}{2}\right) = \ln 5 - \ln 2 = n - m$.

3. Logarithmic Equations and Change of Base

Change of Base Formula: Any operating logarithm can cleanly convert matching bases using:

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

EXAMPLE 5 & 6 (Solving Logarithmic Systems)

Simple Form: $\log_2(x+2) = 3 \Rightarrow x+2 = 2^3 = 8 \Rightarrow x = 6$.
$\ln(x+2) = 3 \Rightarrow x+2 = e^3 \Rightarrow x = e^3 - 2$.
Combined Logarithms: $\log_2 x + \log_2 (x+2) = 3$
Collapse utilizing the multiplication rule: $\log_2(x(x+2)) = 3$
Execute base exponentiation: $x(x+2) = 2^3 = 8$
Factor resulting quadratic: $x^2 + 2x - 8 = 0 \Rightarrow x=2$ or $x=-4$. Negative bound rejects $-4$ explicitly, maintaining strictly $\mathbf{x=2}$.

EXAMPLE 9 (Variable Base Systems)

Solve $\log_4(x+12) = 1 + \frac{1}{2}\log_2 x$.

Evaluate base 4 into purely base 2: $\frac{\log_2(x+12)}{\log_2 4} = 1 + \frac{1}{2}\log_2 x$.
Since $\log_2 4 = 2$, this resolves to $\frac{\log_2(x+12)}{2} = 1 + \frac{1}{2}\log_2 x$.
Multiply universally by 2: $\log_2(x+12) = 2 + \log_2 x$.
Subtract and collapse cleanly: $\log_2\left(\frac{x+12}{x}\right) = 2 \Rightarrow \frac{x+12}{x} = 2^2 = 4$.
Final calculation: $x+12 = 4x \Rightarrow 3x = 12 \Rightarrow \mathbf{x = 4}$.