IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~8 min read

Exponential & Logarithmic Functions

Exponentials and logarithms are inverse functions — that one fact does most of the work in this section. Exponentials grow (or decay) by repeatedly multiplying. Logs undo them. Knowing the standard graphs, the key features, and how the two are linked sets you up for everything from compound interest to half-life problems to differential equations later on.

📘 What you need to know

The exponential function — y = ax

Exponential function f(x) = ax,   a > 0
domain: x ∈ ℝ   ·   range: f(x) > 0

Key features

y-intercept
(0, 1)
always — because a0 = 1
Anchor point
(1, a)
tells you the base from the graph
Horizontal asymptote
y = 0
graph hugs but never touches the x-axis
y = aˣ — two cases depending on a
a > 1 — growth O (0, 1) y = 0 (asymptote)0 < a < 1 — decay O (0, 1) y = 0 (asymptote)
If a > 1, the function grows ever faster as x increases. If 0 < a < 1, the function decays toward zero as x increases. Either way, the curve never crosses the x-axis — output is always positive.

The natural exponential

The most-used exponential is y = ex, where e ≈ 2.718 is Euler’s number. It’s the one that pops up in calculus (its own derivative), in compound interest (continuously compounded), and almost any natural growth/decay process.

Convert any base to e ax = ex ln a ✓ in formula booklet

The logarithmic function — y = loga x

Logarithmic function f(x) = loga x,   x > 0
domain: x > 0   ·   range: f(x) ∈ ℝ

Key features

x-intercept
(1, 0)
always — because loga 1 = 0
Anchor point
(a, 1)
tells you the base from the graph
Vertical asymptote
x = 0
graph never touches the y-axis

The natural logarithm

Just as ex is the natural exponential, ln x = loge x is the natural logarithm. They’re inverses of each other:

Inverse property of e and ln ln(ex) = x   and   eln x = x
Change of base loga x = ln xln a ✓ in formula booklet

The inverse relationship — reflections in y = x

Since logs and exponentials are inverses of each other, their graphs are mirror images across the line y = x. Flipping all the (x, y) pairs swaps the key features:

Exponential y = ax
(0, 1) and (1, a)
y-intercept (0, 1); horizontal asymptote y = 0
Logarithmic y = loga x
(1, 0) and (a, 1)
x-intercept (1, 0); vertical asymptote x = 0
y = eˣ and y = ln x — reflections in y = x
x y y = x y = eˣ (0,1) y = ln x (1,0)

🤔 Why does this matter?

Whenever you have an exponential equation you can’t solve algebraically directly (e.g. ex = 7), apply the inverse — take ln of both sides — and the unknown comes loose: x = ln 7. That trick alone solves a huge slice of exam problems.

Worked examples

WE 1

Identify features of an exponential function

For f(x) = 5x, state: (a) the y-intercept, (b) the horizontal asymptote, (c) the value of f(2) and f(−1).

(a) y-intercept: substitute x = 0 f(0) = 5⁰ = 1 → (0, 1) (b) Horizontal asymptote: as x → −∞, 5ˣ → 0 y = 0 (c) Substitute x = 2 and x = −1 f(2) = 25,   f(−1) = 1/5 (a) (0, 1);   (b) y = 0;   (c) f(2) = 25, f(−1) = 1/5 negative exponents give reciprocals: 5⁻¹ = 1/5
WE 2

Rewrite an exponential using base e

Express y = 7x in the form y = ekx for some constant k.

Use the identity aˣ = eˣ ln a 7ˣ = eˣ ln 7 y = e^(x ln 7),   so k = ln 7 this trick is essential for calculus — derivatives of bases other than e require this conversion
WE 3

Identify features of a logarithmic function

For f(x) = log4 x, state: (a) the x-intercept, (b) the vertical asymptote, (c) the value of f(16) and f(1/4).

(a) x-intercept: solve log₄ x = 0 x = 4⁰ = 1 → (1, 0) (b) Vertical asymptote: log only defined for x > 0 x = 0 (c) Use definition: log₄ y = z ⟺ 4ᶻ = y f(16) = log₄ 16 = 2 (since 4² = 16) f(1/4) = log₄(1/4) = −1 (since 4⁻¹ = 1/4) (a) (1, 0);   (b) x = 0;   (c) f(16) = 2, f(1/4) = −1 log of a number less than 1 (but positive) is always negative
WE 4

Find the inverse of a logarithmic function

The function f(x) = log3 x has domain x > 0. Find f−1(x) and state its domain.

Step 1: Set y = log₃ x and swap y = log₃ x → swap: x = log₃ y Step 2: Rearrange to solve for y log₃ y = x ⟺ y = 3ˣ Step 3: Domain of f⁻¹ = range of f = ℝ f⁻¹(x) = 3ˣ,   domain x ∈ ℝ log and exponential of the same base are perfect inverses — always
WE 5

Evaluate a log using the change of base formula

Without a calculator, evaluate log8 32 using the change of base formula.

Convert to base 2 (since both 8 and 32 are powers of 2) log₈ 32 = log₂ 32 / log₂ 8 = 5 / 3 log₈ 32 = 5/3 picking a useful base (where the numbers simplify cleanly) makes this much faster than going through ln
WE 6

Identify a function from its key features

A function passes through (0, 1) and (2, 9), and has a horizontal asymptote at y = 0. Determine the equation of the function in the form y = ax.

Step 1: Identify the type — y-intercept (0, 1) and HA y = 0 → exponential Step 2: Use (2, 9) to find a 9 = a² a = 3 (must be positive) y = 3ˣ the (1, a) anchor point also confirms — at x = 1 the function gives y = 3 ✓

💡 Top tips

⚠ Common mistakes

Now you’ve got the building blocks. The next note covers solving equations analytically — using the inverse relationship to turn exponential and logarithmic equations into something solvable. After that, the focus shifts to graphical solving when algebra runs out of moves.

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