VCE Mathematical Methods (Year 12) · Functions, Relations and Graphs · 22m read
Exponential and Logarithmic Functions
Ever wondered how your phone battery drains or how scientists date ancient fossils? That's exponential and logarithmic functions at work. These two are like mirror images of each other—exponentials describe rapid growth or decay, while logarithms help us solve the puzzles they create. You'll learn to sketch them, transform them, and solve real equations, which unlocks everything from finance to medicine and makes you fluent in the language of change itself.
Learning objective: Sketch, transform and analyse exponential and logarithmic functions; apply index and log laws; solve exponential and logarithmic equations in exact form.
By the end of this lesson you can:
- I sketch $y = a \cdot b^{x} + c$ identifying asymptote, intercepts and key features.
- I apply all index laws and logarithm laws fluently.
- I solve equations of the form $a^{f(x)} = b$ using logarithms.
- I convert between exponential and logarithmic form: $a^x = y \iff x = \log_a y$.
- I find the equation of an exponential model given two points.
Exponentials and logs are mirror images. Reflect y = eˣ in the line y = x and you get y = ln x — each undoes the other.
Exponential functions — shape, features and transformations
### The base form $y = a^x$
An exponential function has the variable in the exponent, not the base. For a base $a > 0$, $a \neq 1$:
$$y = a^x, \quad x \in \mathbb{R}, \quad y > 0$$
Key features:
- Domain: $\mathbb{R}$ (all real numbers)
- Range: $(0, \infty)$ — always positive
- Horizontal asymptote: $y = 0$ (the $x$-axis)
- Passes through $(0, 1)$ — any number to the power 0 is 1
- Passes through $(1, a)$
| Base $a$ | Behaviour |
|----------|-----------|
| $a > 1$ | Growth — increases left to right |
| $0 < a < 1$ | Decay — decreases left to right |
| $a = 1$ | Constant $y = 1$ (degenerate) |
### The natural exponential $e^x$
The most important base is Euler's number:
$$e \approx 2.71828\ldots$$
$y = e^x$ has the special property that its gradient equals its value everywhere:
$$\frac{d}{dx}(e^x) = e^x$$
This makes it the natural choice for calculus-based models.
### Transformations of $y = a^x$
The general form is:
$$y = A \cdot a^{b(x-h)} + k$$
| Parameter | Effect |
|-----------|--------|
| $A$ | Vertical dilation by $|A|$; reflection in $x$-axis if $A < 0$ |
| $b$ | Horizontal dilation by $1/|b|$; reflection in $y$-axis if $b < 0$ |
| $h$ | Translation right by $h$ |
| $k$ | Translation up by $k$; shifts asymptote to $y = k$ |
Critical: The asymptote shifts to $y = k$.
Logarithmic functions — the inverse of exponentials
### Definition and notation
The logarithm answers: "What power of $a$ gives $y$?"
$$\log_a y = x \iff a^x = y$$
$$\ln x \equiv \log_e x \ \quad \text{(natural logarithm)}$$
$$\log x \equiv \log_{10} x \ \quad \text{(common logarithm, assumed in some contexts)}$$
### Key features of $y = \log_a x$
- Domain: $(0, \infty)$ — only defined for positive $x$
- Range: $\mathbb{R}$
- Vertical asymptote: $x = 0$ (the $y$-axis)
- Passes through $(1, 0)$ — since $a^0 = 1$
- Passes through $(a, 1)$ — since $a^1 = a$
### Inverse relationship
$y = e^x$ and $y = \ln x$ are inverses of each other:
$$e^{\ln x} = x \quad \text{and} \quad \ln(e^x) = x$$
Their graphs are reflections of each other across the line $y = x$.
### Transformations of $y = \log_a x$
General form: $y = A \log_a(b(x - h)) + k$
The asymptote shifts to $x = h$ (vertical asymptote). Domain becomes $x > h$.
Index Laws and Logarithm Laws
- Index: product:
a^m \cdot a^n = a^{m+n}(Same base: add exponents) - Index: quotient:
\dfrac{a^m}{a^n} = a^{m-n}(Same base: subtract exponents) - Index: power:
(a^m)^n = a^{mn}(Raise to power: multiply) - Index: zero:
a^0 = 1 \ (a \neq 0)(Any non-zero base to 0 = 1) - Index: negative:
a^{-n} = \dfrac{1}{a^n}(Negative exponent = reciprocal) - Log: product:
\log_a(xy) = \log_a x + \log_a y(Product → sum) - Log: quotient:
\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y(Quotient → difference) - Log: power:
\log_a(x^n) = n \log_a x(Power → coefficient) - Change of base:
\log_a x = \dfrac{\ln x}{\ln a}(Convert any log to natural log) - Inverse cancel:
e^{\ln x} = x, \ \ln(e^x) = x(e and ln are inverses)
Solving exponential and log equations
- Isolate the exponential: Get $a^{f(x)}$ alone on one side before taking logs.
- Take ln both sides: Apply $\ln$ to both sides: $\ln(a^{f(x)}) = \ln b \Rightarrow f(x)\ln a = \ln b$.
- Isolate the log: Get $\log_a(f(x))$ alone, then convert: $f(x) = a^{\text{value}}$.
- Domain check: After solving, check: argument of any log must be $> 0$.
- Exact form: Leave answer as $\ln(\ldots)$ or $\log(\ldots)$ unless a decimal is requested.
- Two solutions?: Quadratic in disguise: let $u = a^x$, solve the quadratic, then $a^x = u$.
Worked example (Easy)
Simplify $\log_2 32 + \log_2 \frac{1}{4}$.
- Rewrite each term. Express as powers of 2:
\log_2 32 = \log_2 2^5 = 5 - Second term.
\log_2 \tfrac{1}{4} = \log_2 2^{-2} = -2 - Add.
5 + (-2) = 3
Answer: $\log_2 32 + \log_2 \frac{1}{4} = 3$
Worked example (Medium)
Solve $\log_2(x + 6) - \log_2(x) = 3$ for $x$.
- Combine logs. Quotient law.
\log_2\!\left(\dfrac{x+6}{x}\right) = 3 - Convert to index form. $\log_2 A = 3 \Rightarrow A = 2^3$.
\dfrac{x+6}{x} = 8 - Solve. Multiply by $x$.
x + 6 = 8x \Rightarrow x = \dfrac{6}{7} - Domain check. Need $x > 0$ and $x + 6 > 0$.
x = 6/7 satisfies both
Answer: $x = \dfrac{6}{7}$.
Worked example (Medium)
Solve $3e^{2x} = 48$, giving your answer in exact form.
- Isolate the exponential. Divide both sides by 3:
e^{2x} = 16 - Take the natural log. $\ln$ is the inverse of $e^x$:
\ln(e^{2x}) = \ln 16 \implies 2x = \ln 16 - Simplify $\ln 16$. $16 = 2^4$, so $\ln 16 = 4 \ln 2$:
2x = 4 \ln 2 \implies x = 2 \ln 2
Answer: $x = 2\ln 2$ (exact form). Decimal: $x \approx 1.386$.
Worked example (Hard)
A radioactive substance has initial mass $M_0 = 200 \ \text{g}$. After $5$ hours, its mass is $160 \ \text{g}$. The decay model is $M = M_0 e^{-kt}$. Find $k$ and determine when the mass will be $50 \ \text{g}$.
- Substitute the data point to find $k$. At $t = 5$, $M = 160$:
160 = 200 e^{-5k} - Isolate and take ln.
e^{-5k} = \frac{160}{200} = 0.8 \implies -5k = \ln(0.8) \implies k = -\frac{\ln 0.8}{5} - Evaluate $k$ (exact and decimal).
k = \frac{\ln(5/4)}{5} \approx 0.04463 \ \text{h}^{-1} - Find $t$ when $M = 50$. Substitute $M = 50$, $M_0 = 200$:
50 = 200 e^{-kt} \implies e^{-kt} = 0.25 \implies -kt = \ln(0.25) - Solve for $t$.
t = \frac{\ln(0.25)}{-k} = \frac{-\ln 4}{-\frac{\ln(5/4)}{5}} = \frac{5\ln 4}{\ln(5/4)} \approx 31.1 \ \text{hours}
Answer: $k = \frac{\ln(5/4)}{5} \approx 0.0446 \ \text{h}^{-1}$. The mass reaches $50 \ \text{g}$ after approximately $31.1$ hours.
Common mistakes
- $\ln(a + b) = \ln a + \ln b$ The logarithm product law applies to $\ln(ab)$, not $\ln(a+b)$. There is no simplification for the log of a sum. Fix: Log laws only apply to products, quotients and powers inside the argument: $\ln(ab) = \ln a + \ln b$ but $\ln(a+b)$ cannot be split.
- Forgetting to check the domain after solving a logarithmic equation If you solve $\ln(x-3) = 2$ and get $x = e^2 + 3$, that's fine. But if you solve $\ln(x) + \ln(x-2) = 0$ and get $x = -1$ or $x = 3$, then $x = -1$ is invalid since $\ln(-1)$ is undefined. Fix: After solving, substitute each solution back and check that every argument of every log in the original equation is strictly positive.
- Not isolating the exponential before taking logs Applying $\ln$ directly to $3e^{2x} = 48$ gives $\ln(3e^{2x}) = \ln 48 \Rightarrow \ln 3 + 2x = \ln 48$, which is actually valid but unnecessarily complicated. It's easier to divide by 3 first. Fix: Always isolate the exponential term before applying $\ln$: $e^{2x} = 16 \Rightarrow 2x = \ln 16$.
Why is $e$ the "natural" base? The definition via calculus.
The number $e$ arises naturally from asking: "What base $a$ makes $\frac{d}{dx}(a^x) = a^x$ — i.e., what function is its own derivative?"
Using the limit definition of the derivative:
$$\frac{d}{dx}(a^x) = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}$$
For this to equal $a^x$, we need:
$$\lim_{h \to 0} \frac{a^h - 1}{h} = 1$$
The unique base $a$ for which this holds is defined to be $e$. Numerically:
$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828\ldots$$
This also equals the sum of the infinite series:
$$e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots$$
This is why $e^x$ is so important in calculus: it is the exponential function that is its own derivative (and integral), making it the natural exponential.
Practice questions
Which of the following equals $\log_5(125)$?
- $3$
- $25$
- $5$
- $\ln 125$
(1 marks)
Simplify $\log_2 8 - \log_2 2$.
- $4$
- $2$
- $3$
- $1$
(2 marks)
Solve $2^{x+1} = 16$.
- $x = 3$
- $x = 4$
- $x = 15$
- $x = 8$
(2 marks)
Solve $\ln(2x - 1) = 3$ for $x$. Give exact answer.
(3 marks)
Solve $e^{2x} - 5e^x + 6 = 0$.
(3 marks)
The population of bacteria follows $P = P_0 e^{kt}$. At $t = 0$, $P = 500$. At $t = 3$ hours, $P = 2000$. Find $k$ (exact) and the time when $P = 10000$.
(4 marks)
Summary
- $y = a^x$: domain $\mathbb{R}$, range $(0,\infty)$, passes through $(0,1)$ and $(1,a)$, asymptote $y=0$.
- $y = \log_a x$: domain $(0,\infty)$, range $\mathbb{R}$, passes through $(1,0)$, asymptote $x=0$.
- $a^x = y \iff x = \log_a y$ — always able to switch between forms.
- Log laws: product → sum, quotient → difference, power → coefficient.
- Solve $a^{f(x)} = b$: isolate, take $\ln$ both sides, apply log power law.
- Quadratic substitution: for $a^{2x} + pa^x + q = 0$, let $u = a^x$ and solve the quadratic.
- Domain: argument of any logarithm must be $> 0$. Check solutions after solving.