A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.2 (also Paper 2) · ~8 min read

Logarithms and exponential functions

Logarithms as inverses of indices, the laws of logarithms, eˣ and ln x and their graphs, solving equations and inequalities with the unknown in an index, and turning y = kxⁿ and y = kaˣ into straight lines to find constants.

Logs & indicesLaws of logarithmseˣ and ln xGraphs of e^{kx}Index equationsInequalitiesHidden quadraticsLinearising with logs

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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

A logarithm is an index: log_a b = c means aᶜ = b. ln x is log_e x, so ln and e are inverse functions.
Three laws: log(xy) = log x + log y; log(x/y) = log x − log y; log(xⁿ) = n log x. There is no law for log(x + y).
Unknown in an index? Take ln of both sides and bring the power down.
Straight-line graphs from curves: y = kxⁿ gives ln y = ln k + n ln x (plot ln y against ln x); y = kaˣ gives ln y = ln k + x ln a (plot ln y against x).

By the end you can

Relate logarithms to indices and use the laws of logarithms.
Use the properties and graphs of eˣ and ln x, including y = e^kx for positive and negative k.
Solve equations and inequalities with the unknown in an index.
Reduce y = kxⁿ and y = kaˣ to linear form and find the constants from a graph.

1Logarithms

Definition and laws

Logarithm — log_a b is the power to which a must be raised to give b: log_a b = c ⇔ aᶜ = b (a > 0, a ≠ 1, b > 0).

log(xy) = log x + log y ; log(x/y) = log x − log y ; log(xⁿ) = n log x LEARN IT

any base, x, y > 0

Why the laws hold: they are the index laws written backwards. If x = aᵖ and y = a^q, then xy = a^p+q, so log xy = p + q.
Special values: log_a 1 = 0, log_a a = 1, ln 1 = 0, ln e = 1, e^{ln x} = x, ln(eˣ) = x.

Common mistakelog(x + y) = log x + log y is false. So is (log x)/(log y) = log(x − y). Combine logs only with the three laws.

Common mistakeLosing brackets: 3 + 2 ln 5x is not ln(3 × (5x)²). The June 2024 Paper 2 report found many errors like this. Write each step separately.

eˣ and ln x

y = eˣ and y = ln x are reflections of each other in y = x.

	y = eˣ	y = ln x
Domain	all real x	x > 0
Range	y > 0	all real y
Key point	(0, 1)	(1, 0)
Asymptote	y = 0 (as x → −∞)	x = 0 (as x → 0⁺)

y = e^kx: k > 0 gives growth; k < 0 gives decay towards y = 0 (e.g. e^−2x). Both pass through (0, 1).
Transformations apply as usual: y = 3 − e^−x has asymptote y = 3 and y-intercept 2.

2Solving equations and inequalities

Unknown in an index

Worked example

Solve 5^{2x − 1} = 8, giving x to 3 s.f.

(2x − 1) ln 5 = ln 8 M1
2x − 1 = ln 8/ln 5 = 1.292 A1
x = 1.15 A1

Worked example

Solve 3^{x + 1} = 4ˣ.

(x + 1) ln 3 = x ln 4 M1
x(ln 4 − ln 3) = ln 3 ⇒ x = ln 3/(ln 4 − ln 3) M1
x = 3.82 A1

Worked example

Solve 2^2x − 5(2ˣ) + 6 = 0.

let u = 2ˣ: u² − 5u + 6 = 0 ⇒ u = 2 or u = 3 M1 A1
2ˣ = 2 ⇒ x = 1; 2ˣ = 3 ⇒ x = ln 3/ln 2 = 1.585 A1
x = 1 or x = 1.58

Worked example

Solve e^2x − 4eˣ − 5 = 0.

(eˣ − 5)(eˣ + 1) = 0 M1
eˣ = −1 has no solution (eˣ > 0) B1
x = ln 5 A1

Log equations

Worked example

Solve ln(x + 1) + ln(x − 1) = ln 3.

ln[(x + 1)(x − 1)] = ln 3 ⇒ x² − 1 = 3 M1
x = ±2 A1
x − 1 must be positive, so x = 2 (reject −2) A1

Common mistakeNot rejecting a root that makes a logarithm's argument zero or negative. The 2024 Paper 3 reports flagged this.

Inequalities

Dividing by a negative log reverses the inequality. ln a < 0 when 0 < a < 1.

Worked example

Find the smallest integer n for which 0.8ⁿ < 0.1.

n ln 0.8 < ln 0.1 M1
ln 0.8 < 0, so dividing reverses the sign: n > ln 0.1/ln 0.8 = 10.3 M1
n = 11 A1

3Linearising with logarithms

Two models

Model	Take ln	Plot	Gradient	Intercept
y = kxⁿ	ln y = ln k + n ln x	ln y against ln x	n	ln k
y = kaˣ	ln y = ln k + x ln a	ln y against x	ln a	ln k

A power law becomes a straight line on a graph of log y against log x.

Finding constants: read two points on the line (or use the gradient and intercept), then undo the logs: k = e^intercept, a = e^gradient.
Name the variables in the straight-line form: "ln y = 0.8 + 2.6 ln x", then convert.

Worked example

The graph of ln y against ln x is a straight line through (0.5, 2.1) and (2.5, 7.3). Find y in terms of x.

gradient n = (7.3 − 2.1)/(2.5 − 0.5) = 2.6 M1
intercept: 2.1 = ln k + 2.6 × 0.5 ⇒ ln k = 0.8 M1
k = e^0.8 = 2.23: y = 2.23x^2.6 A1

Worked example

The graph of ln y against x is a straight line through (1, 1.9) and (4, 3.4). Find y = A × bˣ.

gradient ln b = 1.5/3 = 0.5 ⇒ b = e^0.5 = 1.65 M1 A1
ln A = 1.9 − 0.5 = 1.4 ⇒ A = e^1.4 = 4.06 A1
y = 4.06 × 1.65ˣ

Common mistakeSubstituting ln x values as x values when reading a log graph. In 2024 this stopped many candidates. If the axis is ln x, the point (0.5, 2.1) means ln x = 0.5, not x = 0.5.

Linking questioneˣ and ln x are differentiated and integrated in Further differentiation and Further integration; exponential growth and decay come from Differential equations; and inverse functions are explained in Functions and transformations.

Test yourself — cover the answers first

Quick check

Write log₂ 32 as an integer.
Simplify ln(e³).
Solve eˣ = 7.
Which graph is straight for y = 3 × 2ˣ?
Express 2 ln 3 − ln 6 as a single logarithm.

Exam-style

Solve 2 ln x − ln(x + 2) = ln 3.
The variables x and y satisfy y = Axⁿ. When ln y is plotted against ln x the line has gradient −1.5 and passes through (2, 0.4). Find A and n.
Solve the inequality 3ˣ > 2^x+2.

Show answers

5
3
x = ln 7
ln y against x
ln(9/6) = ln 1.5
ln(x²/(x + 2)) = ln 3 ⇒ x² − 3x − 6 = 0 ⇒ x = (3 + √33)/2 = 4.37 (reject the negative root)
n = −1.5; ln A = 0.4 + 3 = 3.4 ⇒ A = 30.0
x ln 3 > (x + 2) ln 2 ⇒ x(ln 3 − ln 2) > 2 ln 2 ⇒ x > 3.42

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