Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Logarithms and Exponentials – Revision Notes

For full explanations and longer worked examples, use the study guide.

These notes cover section 2.2, Logarithmic and exponential functions, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.2 is part of Pure Mathematics 2 and is examined on Paper 2 (1 hour 15 minutes, 50 marks), the AS Level Pure Mathematics route. The same outcomes are section 3.2 of Pure Mathematics 3. A scientific calculator is allowed, but unsupported calculator answers earn no marks, so show each log step.

Course links: A Level Mathematics hub, printable 9709 checklist, practice questions for this unit, and the whole-paper Pure Mathematics 2 revision notes.

2.2 at a glance

Outcome	You must be able to
Logs and indices	Switch between logₐ b = x and aˣ = b; use the three laws (no change of base)
eˣ and ln x	Use e^(ln x) = x and ln(eˣ) = x; sketch y = eˣ, y = ln x and y = e^(kx) for k > 0 and k < 0
Unknown in an index	Take logs to solve equations and inequalities
Linear form	Change y = kxⁿ or y = k(aˣ) to a straight line; find constants from gradient and intercept

Definitions

• Logarithm: logₐ b = x ⇔ aˣ = b, with a > 0, a ≠ 1, b > 0.
• lg x = log₁₀ x. ln x = logₑ x, the natural logarithm. e ≈ 2.718.
• Inverse pair: e^(ln x) = x for x > 0, and ln(eˣ) = x for all x.
• Special values: logₐ 1 = 0, logₐ a = 1, ln 1 = 0, ln e = 1, e⁰ = 1.
• The log of zero or a negative number is undefined.

Formulas

Result	Form
Product law	log(xy) = log x + log y
Quotient law	log(x/y) = log x − log y
Power law	log(xⁿ) = n log x
Reciprocal (from the laws)	log(1/x) = −log x
Power model	y = kxⁿ ⇒ ln y = ln k + n ln x
Exponential model	y = k(aˣ) ⇒ ln y = ln k + x ln a

Change of base is not in the 9709 syllabus. To find a value such as log₉ 27, write both numbers as powers of 3: 9ˣ = 27 ⇒ 3^(2x) = 3³ ⇒ x = 3/2.

Graphs

	y = eˣ	y = ln x	y = e^(kx), k < 0
Passes through	(0, 1)	(1, 0)	(0, 1)
Asymptote	y = 0	x = 0	y = 0
Domain	all x	x > 0	all x
Range	y > 0	all y	y > 0
Direction	increasing	increasing	decreasing

y = eˣ and y = ln x are reflections of each other in y = x. For y = e^(kx), k > 0 gives growth (steeper as k increases) and k < 0 gives decay towards the x-axis.

Method in steps

A. Log equation (logs on one or both sides)

1. Use the power law first: 2 log x -> log(x^2)
2. Combine into one log on each side (product/quotient laws)
3. Remove logs:  log_a P = log_a Q  ->  P = Q
                 log_a P = c        ->  P = a^c
4. Solve (often a quadratic)
5. Reject any root that makes an original log argument <= 0

B. Unknown in an index, different bases

1. Take ln (or lg) of both sides
2. Bring the index down with the power law
3. Expand brackets, collect x terms on one side
4. Factorise out x and divide
5. Give an exact form if asked, then 3 s.f.

C. Hidden quadratic

1. Spot e^(2x) = (e^x)^2, or 9^x = (3^x)^2, or 3^(x+1) = 3 * 3^x
2. Let u = e^x (or 3^x) and solve the quadratic in u
3. Reject u <= 0, since e^x > 0 and a^x > 0
4. Take logs to find x

D. Inequality

1. Take logs of both sides
2. If you divide by ln of a number between 0 and 1, reverse the sign
3. For "smallest integer", round up the boundary; check both sides of it

E. Finding constants from a straight-line graph

1. Pick the model from the axes: ln y v ln x -> y = kx^n ;  ln y v x -> y = k(a^x)
2. Gradient from two points on the line
3. Substitute one point to get the intercept c
4. k = e^c ;  n = gradient  or  a = e^(gradient)
   (with lg:  k = 10^c,  a = 10^gradient)

Small worked reminders

• 3 ln 2 + ln 5 − ln 4 = ln 8 + ln 5 − ln 4 = ln(8 × 5/4) = ln 10.
• e^(3 ln 2) = e^(ln 8) = 8.
• log₂(x + 1) = 4 ⇒ x + 1 = 2⁴ ⇒ x = 15.
• 6ˣ = 11 ⇒ x ln 6 = ln 11 ⇒ x = ln 11/ln 6 = 1.34 (3 s.f.).
• ln y against x has gradient 0.2 and intercept 1.5 ⇒ a = e^0.2 = 1.22, k = e^1.5 = 4.48 (3 s.f.).

Must-know distinctions

• ln 5 − ln 3 vs (ln 5)/(ln 3): the first equals ln(5/3) = 0.511; the second is 1.46. They are not the same.
• log(x + y) vs log x + log y: only the second has a law; log(x + y) cannot be split.
• (ln x)² vs ln(x²): ln(x²) = 2 ln x, but (ln x)² is ln x times itself.
• y = kxⁿ vs y = k(aˣ): in the first the unknown power n is fixed and x is the base; in the second x is the index. They need different graphs.
• Intercept vs constant: the intercept on the ln y axis is ln k, so k = e^(intercept).
• Exact vs 3 s.f.: “exact” means leave ln 3, e², 2/(e − 1) in the answer; otherwise give 3 s.f.
• y = e^(kx), k > 0 vs k < 0: both pass through (0, 1); only the sign of k decides growth or decay.

Quick self-test

1. Find the exact value of log₄ 8.
2. Write 3 ln 2 + ln 7 − ln 14 as a single logarithm.
3. Solve e^(3x) = 20, giving x to 3 s.f.
4. Solve ln(2x − 1) = 3, giving x to 3 s.f.
5. Solve 2^(x + 3) = 7ˣ, giving x to 3 s.f.
6. Solve the inequality 0.6ˣ < 0.01.
7. Solve eˣ − 12e^(−x) = 1, giving an exact answer.
8. State the y-intercept and the asymptote of y = e^(−2x), and say whether it is increasing or decreasing.
9. y = kxⁿ. The graph of ln y against ln x has gradient 2.5 and intercept −0.7. Find k and n.
10. y = k(aˣ). The graph of ln y against x has gradient −0.3 and intercept 2. Find a and k.
11. Solve log₆ x + log₆(x + 5) = 2.
12. Simplify e^(2 ln 3).

Answers

1. 4ˣ = 8 ⇒ 2^(2x) = 2³ ⇒ 3/2.
2. ln(8 × 7/14) = ln 4.
3. 3x = ln 20 ⇒ x = (ln 20)/3 = 0.999.
4. 2x − 1 = e³ ⇒ x = (e³ + 1)/2 = 10.5.
5. (x + 3) ln 2 = x ln 7 ⇒ x(ln 7 − ln 2) = 3 ln 2 ⇒ x = 3 ln 2/ln 3.5 = 1.66.
6. x ln 0.6 < ln 0.01; ln 0.6 < 0 so the sign reverses: x > ln 0.01/ln 0.6, x > 9.02.
7. Multiply by eˣ: e^(2x) − eˣ − 12 = 0 ⇒ (eˣ − 4)(eˣ + 3) = 0. eˣ = −3 is impossible, so x = ln 4.
8. (0, 1), asymptote y = 0, decreasing (k = −2 < 0).
9. n = 2.5, ln k = −0.7 ⇒ k = 0.497.
10. ln a = −0.3 ⇒ a = 0.741; ln k = 2 ⇒ k = 7.39.
11. x(x + 5) = 6² ⇒ x² + 5x − 36 = 0 ⇒ (x + 9)(x − 4) = 0. x = −9 makes log₆ x undefined, so x = 4.
12. e^(ln 9) = 9.

Where marks are usually lost

• Removing logs term by term: ln(x + 4) − ln x = 2 does not give x + 4 − x = 2.
• Combining before using the power law, so 2 ln x − ln 3 becomes ln(2x/3) instead of ln(x²/3).
• Not reversing the inequality after dividing by ln 0.6, ln 0.85 or any other negative log.
• Keeping a root such as eˣ = −3 or x = −9 that makes the original expression undefined, or rejecting a root without giving the reason.
• Giving k as the intercept of the ln y graph rather than e^(intercept).
• Using a rounded gradient (0.4 instead of 0.405) and losing the accuracy mark on a = e^(gradient).
• Writing a decimal when the question says “exact”, or leaving ln 45/ln(25/3) unevaluated when it asks for 3 s.f.
• Sketches of y = e^(kx) that cross the x-axis, miss the label (0, 1) or show the wrong direction for the sign of k.
• A bare calculator answer to “solve 5^(2x − 1) = 3^(x + 2)” with no log line: the method mark is lost and the answer is then unsupported.

Next steps

Try the practice questions, then check your AS readiness with the free AS diagnostic or the 9709 self-check bank. For the rest of Paper 2, see the Pure Mathematics 2 guide. A Level candidates meet this content again in the Pure Mathematics 3 revision notes.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.2 Logarithmic and exponential functions.