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Module 4 · L15 of 15 ~45 min ⚡ +95 XP available

Module Synthesis

$e^x$ and $\ln x$ are inverse partners that describe growth, decay, and information. This final lesson ties together definitions, graphs, laws, derivatives, and applications — from the central identity $\frac{d}{dx}(e^x) = e^x$ outward to every technique in the module.

Today's challenge — $e^x$ is the only function that is its own derivative. That single fact generates the entire module: the natural log, all the log laws, growth and decay models, and every optimisation technique you've learned. Can you trace the chain?

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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

+5 XP warm-up

In one sentence, what is the relationship between $e^x$, $\ln x$, and everything else in Module 4? Write your version before reading on.

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The two moves

+5 XP to read

The whole module connects through one inverse pair. Everything else is a corollary.

Move 1 — Use the inverse relationship: $\ln(e^x) = x$ and $e^{\ln x} = x$. When stuck, rewrite in terms of $e$ and $\ln$.
Move 2 — Connect the derivatives: $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$. These two are the engine of the entire module.

$$\ln(e^x) = x \qquad e^{\ln x} = x$$

Simplify logs first

Expand $\ln\!\left(\frac{x^2}{x+1}\right)$ to $2\ln x - \ln(x+1)$ before differentiating.

Use $u$-substitution

Equations like $e^{2x} - 5e^x + 6 = 0$ are quadratics in disguise: let $u = e^x$.

Be explicit about bases

Write $\ln$ for base $e$ and $\log_{10}$ when you mean base 10. Never leave the base ambiguous in an HSC response.

What you'll master

Know

Key facts

$e^x$ and $\ln x$ are inverse functions
All log laws and change of base formula
Key derivatives: $\frac{d}{dx}e^x = e^x$, $\frac{d}{dx}\ln x = \frac{1}{x}$

Understand

Concepts

Why $e$ is the natural base for calculus
How log laws arise from exponent laws
The connections between all Module 4 topics

Can do

Skills

Solve exponential equations using substitution
Differentiate complex exponential and log expressions
Solve growth, decay, and optimisation problems

Key terms

Exponential function$y = a^x$ where $a > 0$, $a \neq 1$. Domain: all reals; range: positive reals.

Logarithmic function$y = \log_a x$, the inverse of $y = a^x$. Domain: positive reals; range: all reals.

Natural exponential$y = e^x$ where $e \approx 2.71828$; unique property: $\frac{d}{dx}(e^x) = e^x$.

Natural logarithm$y = \ln x = \log_e x$; derivative is $\frac{1}{x}$.

Change of base$\log_a x = \dfrac{\ln x}{\ln a}$ — converts any base to natural logarithms.

Growth/decay model$P = P_0 e^{kt}$; $k > 0$ for growth, $k < 0$ for decay.

Module 4 in one picture

synthesis

Everything in Module 4 radiates from one central fact: $\frac{d}{dx}(e^x) = e^x$. This is why $e$ is the natural base. The natural logarithm $\ln x$ is its inverse, giving $\frac{d}{dx}(\ln x) = \frac{1}{x}$. Log laws (product, quotient, power) are exponent laws in disguise. Growth and decay models $P = P_0 e^{kt}$ arise directly from the differential equation $\frac{dP}{dt} = kP$. And optimisation uses the fact that $e^x$ never vanishes to factor and solve cleanly.

$$\begin{aligned} \dfrac{d}{dx}(e^x) &= e^x & \dfrac{d}{dx}(\ln x) &= \dfrac{1}{x} \\[6pt] \dfrac{d}{dx}(a^x) &= a^x \ln a & \dfrac{d}{dx}(\log_a x) &= \dfrac{1}{x \ln a} \\[6pt] \log_a(xy) &= \log_a x + \log_a y & \log_a\!\left(\dfrac{x}{y}\right) &= \log_a x - \log_a y \\[6pt] P &= P_0 e^{kt} & \log_a x &= \dfrac{\ln x}{\ln a} \end{aligned}$$

Exponentials and logarithms are inverse functions · $e \approx 2.71828$

Central identity: $\frac{d}{dx}(e^x) = e^x$ — this is WHY $e$ is special; Inverse pair: $\ln(e^x) = x$ and $e^{\ln x} = x$ for all valid $x$

Pause — copy the central identity $\frac{d}{dx}(e^x) = e^x$ and the inverse pair $\ln(e^x) = x$, $e^{\ln x} = x$ — these are the two facts that connect the whole module into your book.

Did you get this? True or false: $\dfrac{d}{dx}(\ln x) = e^x$.

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

PROBLEM 1 · SOLVE EXPONENTIAL EQUATION

Solve $e^{2x} - 5e^x + 6 = 0$.

Let $u = e^x$. The equation becomes $u^2 - 5u + 6 = 0$.

Substitute to reveal the hidden quadratic structure.

PROBLEM 2 · DIFFERENTIATE USING LOG LAWS

Differentiate $y = \ln\!\left(\dfrac{x^2}{x + 1}\right)$.

$y = \ln(x^2) - \ln(x + 1) = 2\ln x - \ln(x + 1)$

Apply log laws before differentiating — this is always easier.

PROBLEM 3 · DOUBLING TIME FROM DATA

A population grows from 1000 to 3000 in 5 years. Find the doubling time.

$3000 = 1000e^{5k}$, so $e^{5k} = 3$, giving $k = \dfrac{\ln 3}{5} \approx 0.220$

Set up the model and solve for $k$.

Quick check: Which is the correct derivative of $y = \ln(x^2 + 1)$?

Common errors · the 3 traps that cost marks

Trap 01

Mixing up derivatives of $e^x$ and $\ln x$

$\frac{d}{dx}(e^x) = e^x$ but $\frac{d}{dx}(\ln x) = \frac{1}{x}$. These look different but are related: $\ln x$ is the inverse of $e^x$. Don't confuse them — they appear in virtually every question in this module.

Trap 02

Differentiating complex logs without expanding first

$\ln\!\left(\frac{x^2}{x+1}\right)$ is much easier to differentiate after expanding to $2\ln x - \ln(x+1)$. Trying to apply the chain rule directly on the compound expression leads to algebra errors. Always simplify using log laws first.

Trap 03

Using $\log$ when you mean $\ln$

In calculus, $\ln x$ (base $e$) is the standard. $\log x$ without a base usually means $\log_{10} x$ in HSC contexts. Be explicit: write $\ln$ when you mean the natural logarithm, or $\log_{10}$ when you mean base 10.

Fill in the gap: To differentiate $y = \ln(3x^2)$ efficiently, first expand using log laws to get $y = \ln 3 +$ , then differentiate to get $\frac{dy}{dx} =$ .

Work mode · how are you completing this lesson?

Quick-fire practice · 5 mixed problems

Solve $\ln(x + 2) = 3$.

Simplify $\log_2 12 + \log_2 3 - \log_2 9$.

Differentiate $y = e^{x^2} \ln x$.

Find the half-life if a substance decays from 200 g to 50 g in 20 days.

Find the maximum of $f(x) = xe^{-x/2}$ for $x \ge 0$.

Odd one out: Three of these correctly apply log laws. Which one is wrong?

Revisit your thinking

Earlier you summarised Module 4 in one sentence. The central thread is that $e^x$ and $\ln x$ are inverse functions, and $e$ is the natural base because it makes calculus simplest — $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$. Every other rule in this module (general $a^x$, $\log_a x$, growth and decay, optimisation) is a corollary of those two fundamental derivatives. If you understand why $e$ is special, you can reconstruct everything else.

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Multiple choice

+5 XP per correct · +25 XP all-correct

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 43 marks

Q1. Differentiate $y = \dfrac{e^{2x}}{x + 1}$. (3 marks)

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ApplyBand 43 marks

Q2. Solve $\log_3(x - 1) + \log_3(x + 1) = 2$. (3 marks)

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AnalyseBand 54 marks

Q3. A bacterial culture grows according to $P = 500e^{0.2t}$. Find the rate of growth when $t = 5$ and interpret your answer. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $\ln(x+2)=3 \Rightarrow x+2=e^3 \Rightarrow x = e^3 - 2 \approx 18.09$

Drill 2: $\log_2(12 \times 3 \div 9) = \log_2 4 = 2$

Drill 3: Product rule: $\frac{dy}{dx} = 2xe^{x^2}\ln x + e^{x^2}\cdot\frac{1}{x} = e^{x^2}\!\left(2x\ln x + \frac{1}{x}\right)$

Drill 4: $200\to50$ is 2 halvings in 20 days, so half-life = 10 days

Drill 5: $f'(x) = e^{-x/2}\!\left(1 - \frac{x}{2}\right) = 0 \Rightarrow x=2$; $f(2) = 2e^{-1} = 2/e$

Q1 (3 marks): Quotient rule: $u=e^{2x}$, $v=x+1$; $u'=2e^{2x}$, $v'=1$; $\frac{dy}{dx} = \frac{2e^{2x}(x+1)-e^{2x}}{(x+1)^2} = \frac{e^{2x}(2x+1)}{(x+1)^2}$ [3]

Q2 (3 marks): $\log_3[(x-1)(x+1)]=2$ [0.5]; $(x^2-1)=9 \Rightarrow x^2=10$ [1]; $x=\sqrt{10}$ (reject $x=-\sqrt{10}$, need $x>1$) [1.5]

Q3 (4 marks): $\frac{dP}{dt} = 100e^{0.2t}$ [1]; at $t=5$: $100e^1 \approx 272$ bacteria/hour [1.5]; interpretation: at 5 hours the population is growing at approximately 272 bacteria per hour [1.5]

Boss battle · Module 4 Final

earn bronze · silver · gold

Five timed questions spanning all of Module 4. 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 Module 4 mixed questions — the final synthesis challenge.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 14 · Optimisation with Exponentials Maths Advanced Year 11 Overview →

Module overview · Maths Advanced