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

1. Core Logarithmic Definition and Properties

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

$\log_a x = y \Leftrightarrow a^y = x$

Fundamental Restrictions: Domain requires $x > 0$. The base must be $a > 0$ and $a \ne 1$.

$\log_a a = 1$
$\log_a 1 = 0$
$\log_a a^x = x$
$a^{\log_a x} = x$

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

Evaluate the following elementary logarithms base 2:

$\log_2 32 = 5$ (because $2^5 = 32$)
$\log_2 2^{100} = 100$ (using the property $\log_a a^x = x$)
$\log_2 1 = 0$ (because $2^0 = 1$)

EXAMPLE 2

Evaluate various base logarithms algebraically:

$\log 1000000 = 6$ (because $10^6 = 1000000$)
$\log 0.001 = -3$ (because $10^{-3} = 0.001$)
$\log_{25} 5 = \dfrac{1}{2}$ (because $25^{1/2} = \sqrt{25} = 5$)
$\log_{27} 9 = \dfrac{2}{3}$ (because $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$)

2. Four Universal Algebraic Laws

Multiplication: $\log(xy) = \log x + \log y$
Division: $\log\left(\dfrac{x}{y}\right) = \log x - \log y$
Exponentiation: $\log(x^n) = n\log x$
Inversion: $\log\left(\dfrac{1}{x}\right) = -\log x$

Notice: Variables can be strategically collected or expanded algebraically using these rules simultaneously:
$2\log A + 3\log B - 4\log C = \log\left(\dfrac{A^2 B^3}{C^4}\right)$

EXAMPLE 3

Given $\ln x = a$, $\ln y = b$, and $\ln z = c$, expand and substitute to evaluate the following:

(a) $\ln\left(\dfrac{x^3y}{z^2}\right) = 3\ln x + \ln y - 2\ln z = \mathbf{3a + b - 2c}$

(b) $\ln\sqrt{x} = \ln x^{1/2} = \dfrac{1}{2}\ln x = \mathbf{\dfrac{a}{2}}$

EXAMPLE 4

Given $\ln 2 = m$ and $\ln 5 = n$, write the following strictly in terms of $m$ and $n$:

(a) $\ln 50 = \ln(2 \cdot 25) = \ln(2 \cdot 5^2) = \ln 2 + 2\ln 5 = \mathbf{m + 2n}$

(b) $\ln 2.5 = \ln\left(\dfrac{5}{2}\right) = \ln 5 - \ln 2 = \mathbf{n - m}$

(d) $\ln 20 = \ln(4 \cdot 5) = \ln(2^2 \cdot 5) = 2\ln 2 + \ln 5 = \mathbf{2m + n}$

3. Logarithmic Equations

To solve logarithmic equations algebraically, isolate the logarithm and rewrite the system into an exponential form, or collapse multiple logarithms into a single log to utilize the one-to-one property.

EXAMPLE 5 (Simple Logarithmic Equations)

(a) $\log_2(x+2) = 3 \Rightarrow x+2 = 2^3 \Rightarrow x+2 = 8 \Rightarrow \mathbf{x = 6}$

(b) $\ln(x+2) = 3 \Rightarrow x+2 = e^3 \Rightarrow \mathbf{x = e^3 - 2}$

EXAMPLE 6 (Combined Logarithms)

(a) Solve $\log_2 x + \log_2 (x+2) = 3$

Collapse using the multiplication rule: $\log_2(x(x+2)) = 3$
Execute base exponentiation: $x(x+2) = 2^3 \Rightarrow x^2 + 2x - 8 = 0$
Factor the resulting quadratic: $(x+4)(x-2) = 0 \Rightarrow x=2 \text{ or } x=-4$.
Reject $x=-4$ because the original domain strictly requires $x > 0$. Solution: $\mathbf{x=2}$.

(b) Solve $\ln x + \ln(x-1) = \ln 6$

Collapse the left side: $\ln(x(x-1)) = \ln 6$
Equate the arguments (one-to-one property): $x^2 - x = 6 \Rightarrow x^2 - x - 6 = 0$
Factor the quadratic: $(x-3)(x+2) = 0 \Rightarrow x=3 \text{ or } x=-2$.
Reject $x=-2$. Solution: $\mathbf{x=3}$.

4. Change of Base Formula

Any operating logarithm can cleanly convert to matching or convenient bases using the following quotient formula:

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

Notice: A highly useful consequence of this formula is the reciprocal inversion property: $\log_a b = \dfrac{1}{\log_b a}$.

EXAMPLE 7

Evaluate $\log_4 8$ algebraically without a calculator.

Convert to the common base of 2: $\log_4 8 = \dfrac{\log_2 8}{\log_2 4} = \mathbf{\dfrac{3}{2}}$

EXAMPLE 8

(a) Write $\log_2 5 + \log_4 7$ as a single combined logarithm.

Convert base 4 to base 2: $\log_4 7 = \dfrac{\log_2 7}{\log_2 4} = \dfrac{\log_2 7}{2} = \dfrac{1}{2}\log_2 7 = \log_2(7^{1/2}) = \log_2\sqrt{7}$
Combine mathematically: $\log_2 5 + \log_2\sqrt{7} = \mathbf{\log_2(5\sqrt{7})}$

(b) Write $\log x - \log_2 y$ as an expression utilizing strictly base 10 logarithms.

Convert base 2 to base 10: $\log_2 y = \dfrac{\log y}{\log 2}$
Substitute into the original expression: $\mathbf{\log x - \dfrac{\log y}{\log 2}}$

EXAMPLE 9 (Variable Base Systems)

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

1) Evaluate base 4 into strictly base 2: $\dfrac{\log_2(x+12)}{\log_2 4} = 1 + \dfrac{1}{2}\log_2 x$

2) Since $\log_2 4 = 2$, this resolves to: $\dfrac{\log_2(x+12)}{2} = 1 + \dfrac{\log_2 x}{2}$

3) Multiply universally by 2 to clear fractions: $\log_2(x+12) = 2 + \log_2 x$

4) Subtract and collapse cleanly: $\log_2(x+12) - \log_2 x = 2 \Rightarrow \log_2\left(\dfrac{x+12}{x}\right) = 2$

5) Execute exponential conversion: $\dfrac{x+12}{x} = 2^2 = 4$

6) Final linear calculation: $x+12 = 4x \Rightarrow 3x = 12 \Rightarrow \mathbf{x = 4}$

5. The Graph of $f(x) = \log_a x$

Domain & Range: The Domain is strictly bounded to $x > 0$. The Range covers $y \in \mathbb{R}$.
Asymptotes: The y-axis acts as an unbreakable Vertical Asymptote ($x = 0$). There is no horizontal asymptote.
Intercepts: It inherently intercepts the x-axis at $(1, 0)$. There is no y-intercept.
Monotony: The function strictly increases if $a > 1$, and strictly decreases if $0 < a < 1$.
Inverse Relation: The logarithmic function $y = \log_a x$ is the exact geometric reflection of the exponential function $y = a^x$ across the diagonal line $y = x$.

EXPONENTIAL FUNCTIONS AND LOGARITHMS AS INVERSES

The exponential function $y = a^x$ and the logarithmic function $y = \log_a x$ are inverse functions, so their graphs are reflections of each other in the line $y=x$. Below are two representative cases.

(a) $y = 2^x$ and $y = \log_2 x$

Blue: $y=2^x$

Green: $y=\log_2 x$

Gray dashed: $y=x$

(b) $y = (0.5)^x$ and $y = \log_{0.5} x$

Blue: $y=(0.5)^x$

Green: $y=\log_{0.5} x$

Gray dashed: $y=x$

If $a>1$, then both $y=a^x$ and $y=\log_a x$ are increasing.
If $0<a<1$, then both $y=a^x$ and $y=\log_a x$ are decreasing.
Each exponential graph and logarithmic graph are reflections of each other in the line $y=x$.

EXAMPLE 10 (Graphing Logarithms)

Find the domain, asymptotes, and intercepts, then sketch the following functions:

(a) $y = \ln x$

Domain: $x > 0$
V.A: $x = 0$
x-int: $(1, 0)$

(b) $y = \ln(x-2)$

Domain: $x > 2$
V.A: $x = 2$
x-int: $(3, 0)$

Domain: $x > 0$
V.A: $x = 0$
x-int: $(e^{-2}, 0)$