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Module 4 · L4 of 15 ~35 min +95 XP available

Exponential Modelling and Equations

Populations expand, investments compound, substances decay — all following the same mathematical shape. Setting up $y = Ae^{kt}$ and solving it with logarithms is the engine behind every prediction in science, finance, and medicine. Master these two steps and a huge slice of real-world mathematics opens up.

Today's hook — A population grows from $200$ to $800$ in 6 years. Without doing the algebra: estimate roughly how long until it reaches $3200$. Trust your intuition. Hold that answer — you'll check it in the Revisit card.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

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A population grows from $200$ to $800$ in 6 years. Without doing the algebra: estimate roughly how long until it reaches $3200$. Trust your gut — you'll revisit this at the end.

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The two moves

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Every exponential modelling question — no matter how dressed-up — comes down to two moves. Master these and nothing can hide from you.

Move 1 — Model: identify the initial value $A$ and growth/decay rate $k$. Write $y = Ae^{kt}$. A second data point lets you solve for $k$.

Move 2 — Solve: once the model is set, substitute the target and take logarithms of both sides to bring the exponent down. Then isolate the unknown.

$$y = Ae^{kt}$$

Find $k$ from a second point

Substitute $(t, y)$ known data → divide by $A$ → take $\ln$ → divide by $t$.

Take logarithms to solve for $t$

$e^{kt} = c \Rightarrow kt = \ln c \Rightarrow t = \frac{\ln c}{k}$. Keep intermediate values exact.

Context check always

Negative time is impossible. Populations can't be fractional. Always sanity-check against the problem.

What you'll master

Know

Key facts

The general exponential model $y = Ae^{kt}$
$k > 0$ means growth; $k < 0$ means decay
How to solve same-base equations by equating exponents

Understand

Concepts

Why logarithms "undo" exponentials
How to interpret $A$, $k$, and $t$ in context
Why doubling time is constant in exponential growth

Can do

Skills

Solve exponential equations algebraically and graphically
Set up a model from two data points
Predict future values and find time to reach a target

Key terms

Exponential modelAn equation of the form $y = A \cdot b^x$ or $y = Ae^{kx}$ used to describe growth or decay.

Initial value ($A$)The value of $y$ when $x = 0$; always read straight from the problem as "at the start".

Growth/decay rate ($k$)The constant in $y = Ae^{kx}$; positive for growth, negative for decay.

Half-lifeTime for a decaying quantity to halve; found by setting $y = \frac{A}{2}$ and solving for $t$.

Doubling timeTime for a growing quantity to double; constant in exponential growth.

Natural logarithm ($\ln$)Logarithm base $e$; the inverse of $e^x$. Use it to solve for exponents: $e^x = c \Rightarrow x = \ln c$.

Exponential models — the big picture

core concept

Exponential models have the form $y = Ae^{kt}$ where $A$ is the initial value and $k$ determines the rate and direction of change. To solve for an unknown exponent, take the natural logarithm of both sides — this brings the exponent down as a multiplier.

$$y = Ae^{kt} \qquad \text{or} \qquad y = A \cdot b^t$$

$A$ = initial value, $k$ = continuous rate, $b$ = growth factor per unit time

To solve exponential equations: when both sides can be written with the same base, equate exponents directly. When they can't (e.g. $e^x = 7$), take logarithms: $x = \ln 7$. For $2^x = 5$: $x = \frac{\ln 5}{\ln 2}$ using the change-of-base rule.

The doubling-time insight. In exponential growth, the time to double is always $T_2 = \frac{\ln 2}{k}$, regardless of the starting value. This is why in the hook question the population doubles every 3 years — so from $200 \to 3200$ requires 4 doublings = 12 years.

Exponential model: $y = Ae^{kt}$, where $A$ = initial value, $k$ = rate ($k > 0$ growth, $k < 0$ decay); To solve $e^{kt} = c$: take $\ln$ both sides → $kt = \ln c$ → $t = \frac{\ln c}{k}$

Pause — copy the exponential model $y = Ae^{kt}$ (where $A$ = initial value, $k > 0$ growth, $k < 0$ decay) and the solving procedure (take $\ln$ both sides, use power law, isolate the unknown) into your book.

Did you get this? True or false: in the model $y = Ae^{kt}$, a negative value of $k$ describes exponential decay.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SAME-BASE EQUATION

Solve $3^x = 27$.

$27 = 3^3$

Express 27 as a power of 3.

PROBLEM 2 · LOGARITHM METHOD

Solve $e^{2x} = 10$.

$\ln(e^{2x}) = \ln 10$

Take the natural logarithm of both sides.

PROBLEM 3 · FULL MODELLING QUESTION

A population grows from 500 to 800 in 3 years. Model with $P = Ae^{kt}$ and predict when it reaches 2000.

$A = 500$ (initial population at $t = 0$)

Read the initial value directly from the problem.

Quick check: To solve $e^{3x} = 15$, which is the correct first step?

Common errors · the 3 traps that cost marks

Trap 01

Taking $\ln$ of individual terms instead of the whole side

When solving $e^{2x} = 3 \times 5$, you must write $\ln(e^{2x}) = \ln(15)$. You cannot take logarithms of $3$ and $5$ separately then add. Always apply $\ln$ to the entire expression on each side.

Trap 02

Misusing log laws — $\ln(a + b) \neq \ln a + \ln b$

Only $\ln(ab) = \ln a + \ln b$ and $\ln\!\left(\frac{a}{b}\right) = \ln a - \ln b$ are valid. The logarithm of a sum has no simple form. Students lose marks every year for this.

Trap 03

Using the wrong logarithm base

For $e^x = 5$, use natural log: $x = \ln 5$. For $2^x = 5$: $x = \frac{\ln 5}{\ln 2}$ or $\frac{\log 5}{\log 2}$. Both give the same answer. But never mix — using $\log$ on an $e^x$ equation adds messy constants.

Complete: The half-life of a substance with decay model $M = M_0 e^{-0.05t}$ is found by solving $e^{-0.05t} = [\,]$.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Solve $2^x = 32$.

Solve $5^{x-1} = 25$.

Solve $e^x = 7$ to 2 decimal places.

A car worth $\$30{,}000$ depreciates to $\$20{,}000$ in 2 years using $V = 30000e^{kt}$. Find $k$.

The mass of a radioactive substance is $M = 100e^{-0.05t}$ grams. Find the half-life (time to reach 50 g).

Match it: Which equation represents exponential decay (not growth)?

Revisit your thinking

Earlier you estimated when a population would hit $3200$ if it grew $200 \to 800$ in 6 years. The model is $P = 200 \times 2^{t/3}$ — it doubles every 3 years. From $200$ to $3200$ requires 4 doublings, so $4 \times 3 = 12$ years. Doubling time is constant in exponential growth, which is why each successive doubling takes the same 3 years regardless of the starting amount.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. Solve $4^{x+1} = 64$. Show all working. (2 marks)

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ApplyBand 43 marks

Q2. The temperature of a cooling cup of coffee is modelled by $T = 85e^{-0.1t} + 20$ where $t$ is in minutes and $T$ is in °C. Find (a) the initial temperature and (b) the temperature after 10 minutes. (3 marks)

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AnalyseBand 54 marks

Q3. A culture of bacteria grows from 200 to 500 in 4 hours. Model the population as $P = Ae^{kt}$ and predict when the population reaches 2000. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $x = 5$ · 2: $x = 3$ · 3: $x = \ln 7 \approx 1.95$ · 4: $k = \frac{\ln(2/3)}{2} \approx -0.203$ · 5: $t = \frac{\ln 2}{0.05} \approx 13.9$ years

Q1 (2 marks): $64 = 4^3$ [0.5]. $4^{x+1} = 4^3 \Rightarrow x+1 = 3 \Rightarrow x = 2$ [1.5].

Q2 (3 marks): (a) $T(0) = 85e^0 + 20 = 85 + 20 = 105\,^\circ$C [1]. (b) $T(10) = 85e^{-1} + 20 \approx 85 \times 0.368 + 20 \approx 51.3\,^\circ$C [2].

Q3 (4 marks): $A = 200$ [0.5]. $500 = 200e^{4k} \Rightarrow e^{4k} = 2.5$ [0.5]. $k = \frac{\ln 2.5}{4} \approx 0.229$ [1]. $2000 = 200e^{0.229t} \Rightarrow e^{0.229t} = 10 \Rightarrow t = \frac{\ln 10}{0.229} \approx 10.1$ hours [2].

Boss battle · The Modeller

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering exponential modelling questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 3 · The Number e and Natural Exponentials Lesson 5 · Introduction to Logarithms →

Module overview · Maths Advanced · Checkpoint 1