Learn9709Paper 2 · Pure Mathematics 2Logarithmic and exponential functionsThe functions eˣ and ln x

Logarithmic and exponential functions · Subtopic 2 of 5

The Functions eˣ and ln x

Meet $e\approx2.718$ , the most natural base for growth, and its inverse $\ln x$ : their graphs, asymptotes and intercepts, transformations, inverses and exact answers.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Sketch $y=e^x$ , $y=e^{-x}$ and $y=\ln x$ , with intercepts and asymptotes.
Sketch transformations such as $y=2-e^{-x}$ and $y=\ln(x+3)$ .
Use $e^x$ and $\ln x$ as inverse functions to solve equations exactly.
Find inverses of functions involving $e^x$ and $\ln x$ .

Pause and think

A bank pays $100\%$ interest a year. Paid once, $1 becomes $2. Paid in two halves: $1.5^2=2.25$ . Monthly: $\left(1+\tfrac1{12}\right)^{12}$ . What happens if it is paid every second?

Section 2 of 14

Core idea

Start hereThe most natural number

Invest 1 dollar at 100% interest a year. Paid once: 2 dollars. Paid twice (50% each half): $1.5^2=2.25$ . Monthly: $(1+\frac1{12})^{12}=2.613$ . Every instant: $2.71828\ldots$ That limit is $e$ , and $e^x$ has a unique property: its gradient equals its own value everywhere.

The big idea

$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\approx2.718$ . $y=e^x$ satisfies $\frac{d}{dx}e^x=e^x$ . Its inverse is $\ln x=\log_ex$ . The graphs are reflections in $y=x$ : $e^x$ has range $y>0$ ; $\ln x$ needs $x>0$ .

1. Exponential graphs

Comparing bases: y = bˣ

base b = 1.50

The line

y=x+1

has gradient

1

. Only when

b=e\approx2.72

does the curve just touch it at

(0,1)

$y=b^x$ ( $b>1$ ): through $(0,1)$ ; grows rapidly; $y\to0$ as $x\to-\infty$ (asymptote $y=0$ ).
$0<b<1$ : decays.
For $2^x$ , gradient ≈ $0.69\times$ height; for $3^x$ , ≈ $1.10\times$ height; for $e^x$ , exactly $1\times$ .

2. ln x

eˣ and ln x

y=e^x

has asymptote

y=0

and is always positive;

y=\ln x

has asymptote

x=0

y=e^x\iff x=\ln y\qquad e^{\ln x}=x\ (x>0)\qquad\ln e^x=x

$\ln$: natural logarithm, $\log_e$ .

$\ln1=0$ , $\ln e=1$ , $\ln x<0$ for $0<x<1$ .
Domain $x>0$ ; range all reals; asymptote $x=0$ .

3. Transformations

Worked exampleSketch $y=3-2e^{-x}$

Find intercepts and asymptote.

Goal: Key features.
Strategy: Build from $e^x$ by transformations.

$e^x\to e^{-x}$
Reflect in the $y$ -axis: decay.
$\to-2e^{-x}$
Stretch ×2 and reflect in the $x$ -axis: approaches $0$ from below.
$\to3-2e^{-x}$
Up 3: asymptote $y=3$ .
$x=0:\ y=1;\quad y=0:\ e^{-x}=1.5\Rightarrow x=-\ln1.5\approx-0.405$
Intercepts.

AnswerAsymptote

y=3

;

y

-intercept

1

;

x

-intercept

-0.405

✓ CheckAs

x\to\infty

e^{-x}\to0

, so

y\to3

✓.

Edge caseDomain of $\ln(\ldots)$

y=\ln(2x-6)

needs

2x-6>0

x>3

Pause and think

Solve $e^{2x}=5$ .

Summary

$e^x$ : its own gradient.
$\ln$ is its inverse; they cancel.
Asymptotes and domains.

e: $2.71828\ldots$
Natural log: $\ln x=\log_ex$ .

HistoryDiscovered through interest. Jacob Bernoulli found the number

2.718\ldots

in 1683 while studying compound interest, as the limit of

\left(1+\tfrac1n\right)^n

. Leonhard Euler named it

e

around 1731 and showed how it connects to almost everything, including the famous identity

e^{i\pi}+1=0

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