M
hscscience Maths Adv · Y11
0/100daily goal
0
0
0 due
0
L1 · 0 XP
KJ
Your weak spots
Insights load after your first practice round.
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?
0/5QUESTS
1
You’re here

Orient and recall

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

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
01
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.

auto-saved
2
You’re here

The two moves

Work through the core explanation before applying it.

02
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.
3
You’re here

What you'll master

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

03
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
Beyond the syllabus. This page is labelled a module synthesis, but most of what it synthesises sits outside MAV-11-07 and MAV-11-08. Deriving and using $\frac{d}{dx}(\ln x)$, differentiating products and quotients of logarithms, growth and decay models, half-life arithmetic and optimisation are all Year 12. The Year 11 focus is the graphs of $y = ka^x$ and $y = ka^{-x}$, the gradient investigation that identifies $e$, the result $\frac{d}{dx}(e^x) = e^x$, the definition of a logarithm, its inverse relationship with the exponential, the log laws, change of base, and exponential and logarithmic equations. Those are what you will be examined on. Work through this page as a preview of Year 12 if you want to, but the drills on half-life and on the logarithm derivative are not a Year 11 mastery gate.
04
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.
4
You’re here

Module 4 in one picture

Work through the core explanation before applying it.

05
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: $\dfrac{d}{dx}(\ln x) = e^x$.

Concept map summarising exponential and logarithmic functions, their derivative rules, growth and decay, and optimisation

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

5
You’re here

Work examples end to end

Follow the reasoning through complete worked solutions.

PROBLEM 1 · SOLVE EXPONENTIAL EQUATION

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

1
Let $u = e^x$. The equation becomes $u^2 - 5u + 6 = 0$.
Substitute to reveal the hidden quadratic structure.
1 of 3
PROBLEM 2 · DIFFERENTIATE USING LOG LAWS

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

1
$y = \ln(x^2) - \ln(x + 1) = 2\ln x - \ln(x + 1)$
Apply log laws before differentiating, this is always easier.
1 of 3
PROBLEM 3 · DOUBLING TIME FROM DATA

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

1
$3000 = 1000e^{5k}$, so $e^{5k} = 3$, giving $k = \dfrac{\ln 3}{5} \approx 0.220$
Set up the model and solve for $k$.
1 of 3

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

Common errors · the 3 traps that cost marks

6
You’re here

Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

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 $\frac{dy}{dx} =$ .

Quick-fire practice · 5 mixed problems

7
You’re here

Drill it, then lock it in

Run the quick drill and copy the summary into your book.

1

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

2

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

3

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

4

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

5

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?

11
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.

auto-saved
1
You’re here

Multiple choice

Answer the drill bank and rate your confidence.

01
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.

2
You’re here

Short answer

Write full responses, then check them against the model answers.

02
Short answer
ApplyBand 43 marks

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

auto-saved
ApplyBand 43 marks

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

auto-saved
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)

auto-saved
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]

Module overview · Maths Advanced