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.
• $\frac{d}{dx}(e^x) = e^x$ (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.
• $\frac{d}{dx}(\ln x) = \frac{1}{x}$ 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 $n$:

• $\ln(a b) = \ln(a) + \ln(b)$
• $\ln\!\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$
• $\ln(a^n) = n \ln(a)$
• $\ln(1) = 0$ and $\ln(e) = 1$
• Change of base: $\log_a(b) = \frac{\ln(b)}{\ln(a)}$

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:

• $\ln(e^x) = x$ for all $x \in \mathbb{R}$.
• $e^{\ln(x)} = x$ for all $x > 0$.

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

Examples in context

Example 1. Continuous population growth. A bacterial culture grows as $N = 500e^{0.2t}$, where $t$ is hours. To find when it reaches $2000$, solve $e^{0.2t} = 4$, so $0.2t = \ln 4$ and $t = \frac{\ln 4}{0.2} = 5\ln 4 \approx 6.93$ hours. Taking the natural log eliminates the exponential, using the identity $\ln(e^{0.2t}) = 0.2t$.

Example 2. pH on a logarithmic scale. Acidity is $\text{pH} = -\log_{10}[\text{H}^+]$, where $[\text{H}^+]$ is the hydrogen-ion concentration. A solution with $[\text{H}^+] = 10^{-3}$ has $\text{pH} = 3$. Diluting tenfold to $10^{-4}$ raises pH to $4$: each unit of pH represents a factor of $10$ in concentration, the defining property of a log scale.

Try this

Q1. Solve $e^{3x} = 5$ for $x$, giving the exact value. [2 marks]

• Cue. $3x = \ln 5$, so $x = \frac{\ln 5}{3}$.

Q2. Sketch features of $y = \ln(x + 3)$: state the asymptote and $x$-intercept. [3 marks]

• Cue. Vertical asymptote $x = -3$; $x$-intercept where $x + 3 = 1$, so $x = -2$.

Q3. Solve $e^{2x} - 5e^x + 6 = 0$ for $x$, exact form. [3 marks]

• Cue. Let $u = e^x$: $u^2 - 5u + 6 = (u - 2)(u - 3) = 0$; $x = \ln 2$ or $x = \ln 3$.