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Logarithmic Function, Its Properties and Graph

Mathematics Grade 10 Difficulty 2/5

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📐 Logarithmic Function, Its Properties and Graph

Grade 10 Algebra — A Complete Interactive Lesson


1. Definition of the Logarithmic Function

Definition

A logarithmic function is a function of the form:

$$y = \log_a x$$

where $a$ is a fixed positive number called the base, with $a > 0$ and $a \neq 1$, and $x$ is the independent variable.

Examples of logarithmic functions:
$y = \log_3 x$
base $a = 3$
$y = \log_{1/2} x$
base $a = \tfrac{1}{2}$
$y = \lg x$
base $a = 10$
$y = \ln x$
base $a = e \approx 2.718$
⚠️ Why $a > 0$, $a \neq 1$? If $a = 1$, then $1^y = x$ gives $x = 1$ for all $y$, which is not a function. If $a \leq 0$, the exponential $a^y$ is not defined for all real $y$.

2. Properties of $y = \log_a x$

Property 1: Domain

The domain of $y = \log_a x$ is all positive real numbers: $\mathcal{D}(f) = (0, +\infty)$.

  • $\log_a x$ has meaning only for $x > 0$.
  • The $y$-axis (the line $x = 0$) is a vertical asymptote.
  • The domain does NOT include zero or negative numbers.

Property 2: Range

The range of $y = \log_a x$ is all real numbers: $\mathcal{R}(f) = \mathbb{R} = (-\infty, +\infty)$.

Proof: For any real number $b$, there exists a positive number $x = a^b$ such that $\log_a(a^b) = b$. Therefore every real number $b$ is attained as a value of the function. $\square$

Property 3: Not Bounded

The logarithmic function is unbounded — it takes arbitrarily large positive and arbitrarily large negative values.

This follows directly from Property 2: since the range is all of $\mathbb{R}$, no finite bound can contain all function values.

Property 4: Monotonicity

If $a > 1$:

$y = \log_a x$ is strictly increasing on $(0, +\infty)$.

Proof: Let $0 < x_1 < x_2$. Then $a^{\log_a x_1} < a^{\log_a x_2}$ (since $x_1 < x_2$). Because $a > 1$, the exponential function $a^t$ is increasing, so the exponents must satisfy $\log_a x_1 < \log_a x_2$. $\square$

If $0 < a < 1$:

$y = \log_a x$ is strictly decreasing on $(0, +\infty)$.

Proof: Let $0 < x_1 < x_2$. Then $a^{\log_a x_1} < a^{\log_a x_2}$. Because $0 < a < 1$, the exponential $a^t$ is decreasing, so the inequality of exponents reverses: $\log_a x_1 > \log_a x_2$. $\square$

Property 5: Sign of Function Values

When $a > 1$:
  • $\log_a x > 0$ for $x > 1$
  • $\log_a x < 0$ for $0 < x < 1$
  • $\log_a x = 0$ only at $x = 1$
When $0 < a < 1$:
  • $\log_a x > 0$ for $0 < x < 1$
  • $\log_a x < 0$ for $x > 1$
  • $\log_a x = 0$ only at $x = 1$
Key Observation: All logarithmic graphs pass through the fixed point $(1, 0)$, because $\log_a 1 = 0$ for every valid base $a$.

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