Study GuideYear 12Level 3 · ChallengeHSCVCE

Free Year 12 Exponential Growth and Decay

An HSC/VCE-style study guide to the model N = N₀eᵏᵗ: finding k from data, predicting future values, half-life and doubling time, the rate of change dN/dt = kN and a Newton's-law-of-cooling example. Every worked example keeps k unrounded until the final step and shows the logarithm step explicitly, with a six-question self-check and answers.

Download Free PDF

Instant download · No sign-up · Free for personal, classroom and homeschool use

Year: Year 12
Subject: Mathematics
Topic: Exponentials & Logarithms
Difficulty: Level 3 · Challenge
Estimated time: 30 minutes
Curriculum: Australian Curriculum
Answers: Not applicable
Format: PDF (A4) + print

Students will practise

finding the constant k in N = N₀eᵏᵗ from two data points
predicting a quantity at a later time and finding the time to reach a given value
calculating half-life, doubling time and the rate of change dN/dt = kN
converting between the forms N = N₀aᵗ and N = N₀eᵏᵗ

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Exponential Growth and Decay

More resources for this topic are coming soon.

Study guide

How to use this study guide

Read it together first, pausing at each worked example to try the step before reading the answer.
Attempt the "Check yourself" questions at the end without looking back.
Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 12 students (typically ages 17–18) working on exponentials & logarithms. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Explore more Year 12 mathematics resources.

Year 12 · Mathematics · Exponentials & Logarithms

Exponential Growth and Decay

Success Tutoring

Student name:

Date:

What you need to know

A quantity grows or decays exponentially when its rate of change is proportional to its current size: [frac dN/dt] = kN. The solution is N = N₀e[sup kt], where N₀ is the value at t = 0 and k is the growth constant. If k > 0 the quantity grows (populations, investments); if k < 0 it decays (radioactivity, drug concentration, cooling). The derivative of N₀e^kt is kN₀e^kt = kN, which is why this function fits.

Quantity	How to find it	Formula
k from two data points	divide to isolate e^kt, then take ln	k = ln ()
Doubling time (k > 0)	solve e^kt = 2	t =
Half-life (k < 0)	solve e^kt =	t = = −
Rate of change at time t	multiply k by the current value	= kN
Convert N₀a^t to N₀e^kt	a = e^{ln a}	k = ln a

A town's population is 5000 and grows to 8000 in 4 years. Find k, then predict the population after 10 years.

Model: N = 5000e^kt. At t = 4, N = 8000: 8000 = 5000e^4k
e^4k = 1.6, so 4k = ln 1.6 and k = = 0.1175… (keep this in the calculator)
At t = 10: N = 5000e^10k = 5000 × e^1.175… = 16 190.8…
Population after 10 years ≈ 16 200 (about 16 190 — round sensibly for people).

At what rate is that population growing after 10 years?

= kN, so use the value of N at t = 10, not N₀.
= 0.1175… × 16 190.8… = 1902.4…
The population is growing at about 1900 people per year at t = 10.

A radioactive sample of 100 g has a half-life of 30 years. Find k, the mass after 50 years, and the time until 10 g remains.

Half-life: 50 = 100e^30k, so e^30k = and k = = − = −0.02310…
After 50 years: M = 100e^{−0.02310… × 50} = 100e^−1.155… = 31.5 g (to 1 decimal place)
Time to 10 g: 10 = 100e^kt, so e^kt = 0.1 and t = = = 99.7 years
Sense check: 10 g is between 3 half-lives (12.5 g at 90 years) and 4 half-lives (6.25 g at 120 years). Yes.

Cooling: coffee at 80°C is left in a 22°C room. Its temperature follows T = 22 + 58e^−kt (t in minutes). After 5 minutes it is 60°C. When will it reach 40°C?

At t = 5: 60 = 22 + 58e^−5k, so 58e^−5k = 38 and e^−5k =
−5k = ln () = −0.4229…, so k = 0.08457…
For T = 40: 40 = 22 + 58e^−kt, so e^−kt =
−kt = ln () = −1.1701…, so t = = 13.8 minutes

Write N = 3 × 2^t in the form N = N₀e^kt

2 = e^{ln 2}, so 2^t = (e^{ln 2})^t = e^{(ln 2)t}
N = 3e^kt with k = ln 2 ≈ 0.693. The quantity doubles every unit of time, which matches t = = 1.

Common mistake: rounding k too early. Using k = 0.12 instead of 0.1175… in the population example gives 16 600 instead of 16 190 after 10 years. Store k in a calculator memory, or write it exactly as .

Common mistake: treating a percentage growth rate as k. "Grows by 5% per year" means N = N₀ × 1.05^t, so k = ln 1.05 = 0.0488, not 0.05. The two are close for small rates but not equal. Also check the sign: a decay question must produce a negative k.

Check yourself

For N = 200e^0.3t, state the initial value.
For N = 200e^0.3t, find N when t = 2 (1 decimal place).
A quantity doubles every 5 years. Find k (4 decimal places).
A substance has a half-life of 8 days. Find k (4 decimal places).
40 g of that substance is left for 24 days. How much remains?
How long does N = 500e^0.02t take to reach 1000? (1 decimal place)

Answers: 1. 200 2. 200e^0.6 ≈ 364.4 3. k = ≈ 0.1386 4. k = − ≈ −0.0866 5. 24 days is 3 half-lives: 40 → 20 → 10 → 5 g 6. e^0.02t = 2, so t = ≈ 34.7

Preview — the PDF contains the full resource.

Exponentials & Logarithms in other year levels

Index and Logarithm Laws
Study Guide · Year 11 · Exponentials & Logarithms · Level 2 · Standard

Was this resource helpful?

About this resource

Created by: Success Tutoring
Last reviewed: 1 October 2026
How it was made: Written by our team

Free to print and share for personal, classroom and homeschool use. Please don't resell. Spotted an error? Let us know.