IB DP Maths AA · Topic 1 Number and algebra · SL 1.5, 1.7 · SL 2.9 · ~7 min read

Exponents and logarithms

Laws of exponents, logarithms as inverses of exponentials, natural logs and e, the laws of logarithms and change of base, solving exponential and log equations, and exponential and logarithmic functions and models.

Laws of exponentsLogarithmsNatural log & eLaws of logsChange of baseExponential equationsExponential & log graphsGrowth & decay models

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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

A logarithm answers "what power?": a^x = b ⇔ x = log_a b. Logs undo exponentials, and the other way round.
The laws of logs turn products into sums, quotients into differences and powers into multipliers. Use them to bring an unknown down from an exponent.
e ≈ 2.718 is the natural base; ln x = log_e x. Exponential models have the form y = Ae^kt (growth if k > 0, decay if k < 0).
The graphs of y = a^x and y = log_a x are reflections in y = x, with asymptotes y = 0 and x = 0.
This note is the same at SL and HL, and unchanged for 2029.

By the end you can

Use the laws of exponents with integer and rational powers.
Convert between exponential and log form and use the laws of logarithms and change of base.
Solve exponential and logarithmic equations, analytically and with a GDC.
Sketch and interpret exponential and log graphs, and use exponential models.

1Laws of exponents

Law	Example
a^m × aⁿ = a^m+n	2³ × 2⁴ = 2⁷
a^m ÷ aⁿ = a^m−n	x⁵/x² = x³
(a^m)ⁿ = a^mn	(3²)⁴ = 3⁸
a⁰ = 1, a⁻ⁿ = 1/aⁿ	5⁻² = 1/25
a^m/n = (ⁿ√a)^m	8^2/3 = 2² = 4

Worked example

Simplify (27x⁶)^−2/3.

= 27^−2/3 × x⁻⁴ M1
= 19x⁴ A1

Same base trick: to solve 4^x+1 = 8^x, write both as powers of 2: 2^2x+2 = 2^3x, so 2x + 2 = 3x and x = 2.

2Logarithms

a^x = b ⇔ x = log_a b FORMULA BOOKLET

a > 0, b > 0, a ≠ 1

Special values: log_a 1 = 0, log_a a = 1, log_a a^x = x, a^log_a x = x. log x means log₁₀ x; ln x means log_e x.

Law of logarithms FORMULA BOOKLET	In words
log_a xy = log_a x + log_a y	product → sum
log_a (x/y) = log_a x − log_a y	quotient → difference
log_a x^m = m log_a x	power → multiplier

log_b a = log_c alog_c b FORMULA BOOKLET

change of base

Worked example

Write 2 ln 3 − ln 6 + ln 4 as a single logarithm.

= ln 9 − ln 6 + ln 4 M1
= ln 9 × 46 = ln 6 A1

Worked example

Evaluate log₈ 32 without a calculator.

log₈ 32 = log₂ 32log₂ 8 M1
= 53 A1

Common mistakeWriting log(x + y) = log x + log y. There is no law for the log of a sum. The product law is log(xy).

3Solving equations

Worked example

Solve 3^2x+1 = 20, giving x to 3 s.f.

take ln of both sides: (2x + 1) ln 3 = ln 20 M1
2x + 1 = ln 20/ln 3 = 2.727 A1
x = 0.863 A1

Worked example

Solve log₂ x + log₂(x − 2) = 3.

log₂ x(x − 2) = 3, so x(x − 2) = 2³ = 8 M1
x² − 2x − 8 = 0 ⇒ x = 4 or x = −2 A1
log₂(−2) is undefined, so x = 4 only R1

Worked example

Solve e^2x − 5e^x + 6 = 0.

let u = e^x: u² − 5u + 6 = 0 M1
u = 2 or 3 A1
x = ln 2 or x = ln 3 A1

Exam tipPaper 1 log equations almost always have a "reject" step: check every solution in the ORIGINAL equation, because logs of negative numbers and zero are undefined.

GDC (calculator) methodPaper 2: graph both sides (y = 3^2x+1 and y = 20) and use intersect, or use the numerical solver. Write the equation you solved and give the answer to 3 s.f.

4Exponential and logarithmic functions

e^x and ln x undo each other, so their graphs mirror in y = x.

	y = a^x (a > 1)	y = log_a x
domain	x ∈ ℝ	x > 0
range	y > 0	y ∈ ℝ
intercept	(0, 1)	(1, 0)
asymptote	y = 0	x = 0

e^x is special: its gradient at every point equals its value, which is why it is the natural choice for growth models (see Differentiation).

a^x = e^{x ln a} FORMULA BOOKLET

any exponential can be written with base e

Exponential models

Growth: N = N₀e^kt, k > 0. Decay: N = N₀e^−kt. Doubling time = ln 2/k; half-life = ln 2/k.

Worked example

A population is modelled by P = 500e^0.04t (t in years). Find the initial population and when it doubles.

t = 0: P = 500 A1
1000 = 500e^0.04t ⇒ e^0.04t = 2 M1
t = ln 2/0.04 = 17.3 years A1

Worked example

A cup of tea cools as T = 20 + 70e^−0.05t (°C, t in minutes). When is it 50 °C, and what is the long-term temperature?

30 = 70e^−0.05t ⇒ t = −ln(3/7)/0.05 = 16.9 min M1A1
as t → ∞, e^−0.05t → 0, so T → 20 °C (room temperature) A1

Linking questionLogs straighten exponential data: if y = Ae^kx then ln y = ln A + kx, a straight line. Physics uses the same trick for radioactive decay (E.3).

Idea for your IA / explorationFit an exponential (or logistic) model to real data: COVID cases, cooling coffee, a bouncing ball's heights, and compare models using a log-transformed graph.

Test yourself — cover the answers first

Quick check

Write log₃ 81 = 4 in exponential form.
Simplify log_a a⁵.
Solve 2^x = 32.
What is the domain of y = ln(x − 3)?
Expand ln(x²y/z).
Find the half-life for N = N₀e^−0.2t.

Exam-style

Solve 2 log₃ x − log₃(x + 6) = 1.
A car worth $30 000 loses 15 % of its value each year. After how many whole years is it first worth less than $10 000?
Show that e^{x ln 2} = 2^x and hence solve 2^x = e³.

Show answers

3⁴ = 81
5
x = 5
x > 3
2 ln x + ln y − ln z
ln 2/0.2 = 3.47
log₃(x²/(x + 6)) = 1
x² = 3x + 18 ⇒ x = 6 or x = −3
x = 6 (x = −3 rejected)
30 000 × 0.85ⁿ < 10 000
n > ln(1/3)/ln 0.85 = 6.76
7 years
e^{x ln 2} = (e^{ln 2})^x = 2^x
x ln 2 = 3
x = 3/ln 2 ≈ 4.33

Practise this chapter

IB Maths AA practice papers

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