TASMath MethodsSyllabus dot point

How do exponential and logarithmic functions behave, and how do we differentiate and apply them?

Exponential and logarithmic functions: graphs, laws, the number e, derivatives and growth and decay models

Index and log laws, the natural base e, derivatives of e^x and ln x, and exponential growth and decay modelling for TCE Mathematics Methods Unit 3.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

Exponential and logarithmic functions describe anything that grows or shrinks at a rate proportional to its current size: populations, money at compound interest, radioactive decay, cooling. This dot point develops the algebra, the calculus, and the modelling.

Functions and their inverses

The exponential function $y = a^x$ (with $a > 0$ , $a \ne 1$ ) and the logarithm $y = \log_a x$ are inverses: $a^x = y \iff x = \log_a y$ . Their graphs are reflections in the line $y = x$ . The exponential has a horizontal asymptote $y = 0$ ; the logarithm has a vertical asymptote $x = 0$ and is only defined for $x > 0$ .

Index and logarithm laws

The index laws ( $a^m a^n = a^{m+n}$ , $\dfrac{a^m}{a^n} = a^{m-n}$ , $(a^m)^n = a^{mn}$ ) translate directly into log laws:

\log_a(xy) = \log_a x + \log_a y, \qquad \log_a\!\left(\frac{x}{y}\right) = \log_a x - \log_a y, \qquad \log_a(x^n) = n\log_a x.

The change-of-base rule $\log_a x = \dfrac{\ln x}{\ln a}$ lets you evaluate any logarithm with the natural log on a calculator.

Derivatives

The standard derivatives are:

\frac{d}{dx}(e^x) = e^x, \qquad \frac{d}{dx}(\ln x) = \frac{1}{x}.

With the chain rule these generalise:

\frac{d}{dx}\big(e^{f(x)}\big) = f'(x)\,e^{f(x)}, \qquad \frac{d}{dx}\big(\ln f(x)\big) = \frac{f'(x)}{f(x)}.

Solving an exponential equation and finding a rate

Problem

A culture of bacteria grows according to $N(t) = 500\,e^{0.4t}$ , where $t$ is in hours. Find when the population reaches $2000$ , and the rate of growth at that time.

Solution

Set $N(t) = 2000$ :

500\,e^{0.4t} = 2000 \quad\Rightarrow\quad e^{0.4t} = 4.

Take the natural log of both sides:

0.4t = \ln 4 \quad\Rightarrow\quad t = \frac{\ln 4}{0.4} \approx 3.47 \text{ hours}.

The growth rate is the derivative:

N'(t) = 500 \times 0.4\,e^{0.4t} = 200\,e^{0.4t}.

At the moment $N = 2000$ we have $e^{0.4t} = 4$ , so

N'(t) = 200 \times 4 = 800 \text{ bacteria per hour}.

Growth and decay models

Exponential models take the form $A(t) = A_0 e^{kt}$ . A positive $k$ is growth; a negative $k$ is decay. The constant $A_0$ is the initial amount (at $t = 0$ ). For radioactive decay the half-life $T$ satisfies $e^{kT} = \tfrac12$ , giving $k = \dfrac{\ln(1/2)}{T}$ . Always read whether the question gives you a rate, a doubling time, or a half-life, and convert it to $k$ first.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC5 marksThe number of Carbon-14 atoms in a sample is given by A = A0 e^(-kt), where A0 is the initial number of atoms, k is a constant, and A is the number of atoms remaining t years later. a) If A is equal to half the value of A0 after 5730 years, find the value of k to 6 decimal places. b) Hence find the age of a sample if 70% of the initial number remains.

Show worked answer →

a) Half remaining means A = A0/2 at t = 5730. Substitute and cancel A0:
A0/2 = A0 e^(-5730 k), so 1/2 = e^(-5730 k).
Take natural logs: ln(1/2) = -5730 k, so k = -ln(1/2)/5730 = ln(2)/5730.
k = 0.693147.../5730 = 0.000121 (to 6 decimal places, k = 0.000121).

b) 70% remaining means A/A0 = 0.7, so 0.7 = e^(-kt).
ln(0.7) = -kt, so t = -ln(0.7)/k = -(-0.356675)/0.000121 = 2948 years (approximately).
Three marks in a) for setting up the equation, taking logs and the 6 dp value; two in b) for substituting 0.7 and solving for t.

2024 TASC3 marksSolve 2 log3(x) - log3(x - 1) = log3(x + 3). Show all algebraic working.

Show worked answer →

Use the log laws to combine the left side into a single logarithm. Recall 2 log3(x) = log3(x^2) and log3(a) - log3(b) = log3(a/b):
log3(x^2) - log3(x - 1) = log3(x^2/(x - 1)).

Now equate the arguments since the bases match:
x^2/(x - 1) = x + 3.
Multiply out: x^2 = (x + 3)(x - 1) = x^2 + 2x - 3.
Cancel x^2: 0 = 2x - 3, so x = 3/2.

Check the domain: logs require x > 0 and x - 1 > 0, i.e. x > 1. Since 3/2 > 1, x = 3/2 is valid. Always state the domain check to secure full marks.

2023 TASC4 marksFor the equation log2(x + 3) + log2(x + 4) = log2(6x + 18): a) Verify algebraically that x = 2 is a solution. b) Solve the equation algebraically and explain why x = 2 is the only real solution.

Show worked answer →

a) Substitute x = 2. Left: log2(5) + log2(6) = log2(30). Right: log2(6 times 2 + 18) = log2(30). The two sides are equal, so x = 2 is a solution.

b) Combine the left side: log2((x + 3)(x + 4)) = log2(6x + 18). Equate arguments:
(x + 3)(x + 4) = 6x + 18, so x^2 + 7x + 12 = 6x + 18.
x^2 + x - 6 = 0, which factors as (x + 3)(x - 2) = 0, giving x = -3 or x = 2.
Reject x = -3 because log2(x + 3) = log2(0) is undefined (the argument must be positive). Hence x = 2 is the only real solution.