Graphs of exponential functions $f(x) = a^x$ (in particular $f(x) = e^x$) and logarithmic functions $f(x) = \log_a(x)$ (in particular $f(x) = \ln(x)$), including their key features and the inverse relationship

What this dot point is asking

VCAA wants the graphs and properties of exponential functions $f(x) = a^x$ (with focus on $f(x) = e^x$) and logarithmic functions $f(x) = \log_a(x)$ (with focus on $f(x) = \ln(x)$). You need to recognise key features (asymptotes, intercepts, domain, range), apply transformations, use log laws to manipulate expressions, and understand that the two functions are inverses of each other.

The natural exponential

$f(x) = e^x$ where $e \approx 2.718$ is Euler's number.

Key features.

• Domain: $\mathbb{R}$. Range: $(0, \infty)$.
• Horizontal asymptote: $y = 0$ as $x \to -\infty$.
• Strictly increasing. Always positive.
• y-intercept: $(0, 1)$. No x-intercept.
• IMATH_11 (the function is its own derivative).

For general bases $a > 0$ with $a \neq 1$, $f(x) = a^x$ has the same shape (increasing if $a > 1$, decreasing if $0 < a < 1$). The change of base is $a^x = e^{x \ln a}$.

The natural logarithm

$f(x) = \ln(x) = \log_e(x)$ is the inverse of $e^x$.

Key features.

• Domain: $(0, \infty)$. Range: $\mathbb{R}$.
• Vertical asymptote: $x = 0$.
• Strictly increasing.
• x-intercept: $(1, 0)$. No y-intercept.
• IMATH_24 for $x > 0$.

The graph of $\ln(x)$ is the reflection of $e^x$ in the line $y = x$.

Log laws

These are Paper 1 staples. For positive $a, b$ and any real $n$:

• IMATH_31
• IMATH_32
• IMATH_33
• IMATH_34 and $\ln(e) = 1$
• Change of base: IMATH_36

These rules also hold for $\log_a$ with any valid base.

Transformations

A transformed exponential takes the form $f(x) = A e^{b(x - h)} + k$. The key features shift accordingly:

• Horizontal asymptote moves from $y = 0$ to $y = k$.
• y-intercept is $f(0) = A e^{-bh} + k$.
• The graph still has no x-intercept unless $A$ and $k$ have opposite signs.

A transformed log takes the form $f(x) = A \ln(b(x - h)) + k$, valid when $b(x - h) > 0$.

• Vertical asymptote moves from $x = 0$ to $x = h$ (when $b > 0$).
• Domain becomes $(h, \infty)$ for $b > 0$, or $(-\infty, h)$ for $b < 0$.

Worked example: combining laws and solving

Solve $\ln(2x) + \ln(x - 1) = \ln(6)$ for $x$.

Combine the left-hand side using $\ln(a) + \ln(b) = \ln(ab)$: $\ln(2x(x-1)) = \ln(6)$.

Exponentiate both sides: $2x(x - 1) = 6$, so $2x^2 - 2x - 6 = 0$ and $x^2 - x - 3 = 0$.

Quadratic formula: $x = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}$.

Reject negative solutions (the original equation needs $2x > 0$ and $x - 1 > 0$, i.e. $x > 1$). So $x = \frac{1 + \sqrt{13}}{2}$.

The inverse relationship

$\ln$ and $\exp$ are inverses, which gives two useful identities:

• IMATH_67 for all $x \in \mathbb{R}$.
• IMATH_69 for all $x > 0$.

These let you eliminate a log by exponentiating, or eliminate an exponential by taking logs.

Common Paper 1 traps

Treating $\ln(a + b)$ as $\ln(a) + \ln(b)$. This is wrong. The product law $\ln(ab) = \ln(a) + \ln(b)$ applies only to products, not sums.

Forgetting domain restrictions when solving log equations. After solving, check that all $\log$ arguments are positive. Otherwise you may include spurious solutions.

Confusing the vertical and horizontal asymptotes. Exponentials have a horizontal asymptote at $y = 0$; logs have a vertical asymptote at $x = 0$.

Treating $e$ like an unknown variable. $e$ is a constant ($e \approx 2.718$); $\ln(e) = 1$, not $\ln(e^x)$.

Decimal approximations on Paper 1. When the question asks for exact form, give answers like $\ln 3$ or $\frac{1 + \sqrt{13}}{2}$, not $1.10$ or $2.30$.

In one sentence

The natural exponential $e^x$ and natural logarithm $\ln x$ are mutually inverse functions whose graphs reflect across $y = x$, with $e^x$ having domain $\mathbb{R}$ and range $(0, \infty)$ and $\ln x$ having domain $(0, \infty)$ and range $\mathbb{R}$; the log laws and the inverse identities $\ln(e^x) = x$ and $e^{\ln x} = x$ are essential Paper 1 algebra.