Exponentials & Logarithms — Drill Pack (Pure

Pure Mathematics — Paper 2 · AQA AS Mathematics (sample) · Drill 2 of 2

Why this topic matters. Exponentials & Logarithms appear on AQA Paper 1 or 2 in ~85% of sittings, typically 8–12 marks. The AQA 2023 examiner report specifically named writing log(x+8) as log(x)+log(8) as a very common error costing 2–3 marks. Eliminating this single mistake and mastering the convert-and-solve method recovers significant marks.

The scaffold to use (LCSE Framework)

Laws: product, quotient, power rules for logs — memorise and apply in order.
Convert: switch between logᵃ(b)=c and a^c=b forms as needed.
Substitute: let u=e^x (or u=a^x) to find hidden quadratics.
Exact: give answers as ln/log unless told to round.

Foundational — recall and single-step application

Q1. Write as a single logarithm: (a) log 5 + log 4; (b) log 20 − log 4; (c) 3 log 2.

Foundational

[3 marks]

M1 (a) log 5 + log 4 = log(20). (b) log 20 − log 4 = log(5). (c) 3 log 2 = log(8). AO1 — all three log laws Common mistake: Writing 3 log 2 = log 6 (multiplying instead of raising to the power).

A1 (a) log 20; (b) log 5; (c) log 8. AO1 — correct simplified forms

A1 Rule to fix in memory: log(a)+log(b) = log(ab) uses multiplication INSIDE the log, not addition. AO2 — distinguishing product rule from sum

Q2. Solve: log₂(x) = 5.

Foundational

[2 marks]

M1 log₂(x) = 5 ⇔ x = 2⁵ = 32. AO1 — converting log to exponential form

A1 x = 32. Check: log₂(32)=log₂(2⁵)=5 ✓. AO1 — correct answer

Q3. Solve: 3^x = 20. Give your answer to 3 significant figures.

Foundational

[3 marks]

M1 Take ln of both sides: x ln 3 = ln 20 ⇒ x = ln(20)/ln(3). AO1 — taking logs of both sides

A1 x = 2.996.../1.099... = 2.73 (3 s.f.). AO1 — correct numerical answer Common mistake: Writing x = log(20)/3 (dividing instead of dividing by log(3)).

A1 Check: 3^2.73 ≈ 20.0 ✓. AO2 — substitution check

Q4. Simplify: log 12 − log 4 + log 3.

Foundational

[3 marks]

M1 log 12 − log 4 = log(12/4) = log 3. AO1 — quotient rule

A1 log 3 + log 3 = log(3×3) = log 9. AO1 — product rule

A1 Answer: log 9 = 2 log 3. AO1 — final simplified form

Q5. Find: (a) ln(e³); (b) e^{ln 7}; (c) ln(1).

Foundational

[3 marks]

M1 (a) ln(e³) = 3. (b) e^{ln 7} = 7. (c) ln(1) = 0. AO1 — inverse identity ln and exp

A1 (a) 3; (b) 7; (c) 0. AO1 — all three correct

A1 Memory aid: e and ln are inverse functions — they cancel each other like squaring and square root. AO2 — inverse function concept

Q6. Solve: 2e^x = 14. Give your answer exactly.

Foundational

[3 marks]

M1 Divide by 2: e^x = 7. Take ln: x = ln 7. AO1 — isolate then take ln

A1 x = ln 7 (exact). Numerically: x ≈ 1.946. AO1 — exact form answer

A1 Note: write ln 7 not log 7 when the base is e. AO2 — notation precision

Mid-tier — application and analysis (exam-bulk questions)

Q7. Solve: log(3x + 4) = 2 (base 10).

Mid-tier

[4 marks]

M1 log(3x+4)=2 ⇔ 3x+4 = 10² = 100. AO1 — exponential form Common mistake: Writing 3x+4=2×10=20 instead of 10²=100.

A1 3x = 96 ⇒ x = 32. AO1 — solving for x

A1 Check: log(3(32)+4) = log(100) = 2 ✓. AO2 — verification

A1 x = 32. AO1 — final answer

Q8. Solve simultaneously: y = e^2x and y = 3e^x − 2.

Mid-tier

[5 marks]

M1 Equate: e^2x = 3e^x − 2. Let u=e^x: u² = 3u−2 ⇒ u²−3u+2=0. AO1 — hidden quadratic substitution Common mistake: Not recognising e^2x=(e^x)².

A1 (u−1)(u−2)=0 ⇒ u=1 or u=2. AO1 — solving quadratic in u

A1 e^x=1 ⇒ x=0; e^x=2 ⇒ x=ln2. AO1 — taking ln

A1 y-values: x=0 ⇒ y=1; x=ln2 ⇒ y=(e^ln2)²=4. AO1 — back-substituting for y

A1 Solutions: (0,1) and (ln2, 4). AO2 — both pairs clearly stated

Q9. Solve: log₂(x+3) + log₂(x−1) = 5.

Mid-tier

[5 marks]

M1 Product rule: log₂((x+3)(x−1)) = 5. AO1 — combining logs Common mistake: Writing log₂(x+3+x−1) = log₂(2x+2). This is WRONG. log(a)+log(b)=log(ab), not log(a+b).

M1 Convert: (x+3)(x−1) = 2⁵ = 32. Expand: x²+2x−3=32 ⇒ x²+2x−35=0. AO1 — exponential form and rearranging

A1 (x+7)(x−5)=0 ⇒ x=−7 or x=5. AO1 — solving the quadratic

A1 Domain check: x+3>0 and x−1>0 both require x>1. Reject x=−7. AO2 — domain check — critical step

A1 x = 5. AO1 — unique valid solution

Q10. Given log₄(a)=3 and log₄(b)=−1/2, find: (a) log₄(a²b); (b) log₄(a/√b).

Mid-tier

[4 marks]

M1 (a) log₄(a²b) = 2log₄(a)+log₄(b) = 2(3)+(−1/2) = 6−1/2 = 11/2. AO1 — power and product rules

A1 (a) 11/2. AO1 — correct value

A1 (b) log₄(a/√b) = log₄(a)−(1/2)log₄(b) = 3−(1/2)(−1/2) = 3+1/4 = 13/4. AO1 — quotient and power rules

A1 Check (b): a/√b = 4³/4^1/4 = 4^13/4 ✓. AO2 — verification using base-4 definition

Q11. Solve: 5^2x+1 = 8. Give your answer to 3 s.f.

Mid-tier

[4 marks]

M1 (2x+1) ln 5 = ln 8. AO1 — taking ln both sides

A1 2x+1 = ln8/ln5 = 2.0794/1.6094 ≈ 1.292. AO1 — computing the ratio

A1 2x = 0.292 ⇒ x ≈ 0.146 (3 s.f.). AO1 — solving for x

A1 Check: 5^1.292 ≈ 8.0 ✓. AO2 — verification

Q12. Express 2log(x) − 3log(y) + log(z) as a single logarithm.

Mid-tier

[3 marks]

M1 Power rule: 2log(x)=log(x²); 3log(y)=log(y³). AO1 — clearing coefficients first

A1 log(x²) − log(y³) + log(z) = log(x²z/y³). AO1 — applying quotient and product

A1 Single log: log(x²z/y³). AO1 — final form

Q13. Solve: e^2x − 5e^x + 6 = 0.

Mid-tier

[4 marks]

M1 u=e^x: u²−5u+6=0 ⇒ (u−2)(u−3)=0. AO1 — substitution u=e^x

A1 u=2 or u=3. e^x=2 ⇒ x=ln2; e^x=3 ⇒ x=ln3. AO1 — solving for x

A1 x = ln 2 or x = ln 3. Both valid (e^x>0 always). AO2 — exact answers and domain note

A1 Numerically: x ≈ 0.693 or x ≈ 1.099 (but give exact unless told otherwise). AO2 — numerical equivalents for checking

Q14. A population P grows by P = 500e^0.03t (t in years). (a) Initial population? (b) Time to double?

Mid-tier

[5 marks]

M1 (a) P(0) = 500e⁰ = 500. AO1 — initial value at t=0

M1 (b) 500e^0.03t=1000 ⇒ e^0.03t=2 ⇒ 0.03t=ln2. AO1 — setting up doubling equation

A1 t = ln2/0.03 = 23.1 years (to 3 s.f.). AO1 — solving for t

A1 To nearest month: 0.1 years ×12 ≈ 1.2 months ⇒ approximately 23 years and 1 month. AO2 — converting to months

A1 Check: P(23.1) = 500e^0.693 = 500×2 = 1000 ✓. AO2 — verification

Q15. Solve: log(x+8) − log(x) = 1 (base 10).

Mid-tier

[4 marks]

M1 Quotient rule: log((x+8)/x) = 1 ⇒ (x+8)/x = 10. AO1 — combining logs and converting

A1 x+8 = 10x ⇒ 8 = 9x ⇒ x = 8/9. AO1 — solving the linear equation

A1 Domain: x>0 ⇒ 8/9>0 ✓. Answer: x = 8/9. AO2 — domain check

A1 KEY: log(x+8)−log(x) = log((x+8)/x) by quotient rule. NOT log(8) — this is the most common error (2023 AQA ER). AO2 — law clarification — most common error

Q16. Solve: 7^x+1 = 3^2x−1. Give your answer to 3 d.p.

Mid-tier

[5 marks]

M1 ln(7^x+1) = ln(3^2x−1) ⇒ (x+1)ln7 = (2x−1)ln3. AO1 — taking ln and applying power rule

M1 x ln7 + ln7 = 2x ln3 − ln3. Collect x: x(ln7−2ln3) = −ln3−ln7 = −ln21. AO2 — collecting x terms

A1 ln7−2ln3 = ln7−ln9 = ln(7/9) = −0.2513... AO1 — computing the coefficient of x

A1 x = −ln21/(ln(7/9)) = −3.0445/(−0.2513) = 12.114 (3 d.p.). AO1 — final value of x

A1 Check: 7^13.114 and 3^23.228. Both ≈ same large number. ✓ AO2 — sanity check

Stretch — discriminators, twist-traps, top-band signals

Q17. Solve: log₂(x) + log₄(x) = 3.

Stretch

[6 marks]

M1 Change base: log₄(x) = log₂(x)/log₂(4) = log₂(x)/2. AO2 — applying change of base formula

M1 Equation: log₂(x) + log₂(x)/2 = 3 ⇒ (3/2)log₂(x) = 3. AO1 — combining like terms

A1 log₂(x) = 2 ⇒ x = 4. AO1 — solving for x

A1 x = 4. Check: log₂(4)+log₄(4) = 2+1 = 3 ✓. AO2 — verification

A1 General: logᵃ(x) = logᵇ(x)/logᵇ(a) for any valid base b. This is the change-of-base formula. AO2 — general formula

A1 This question type (mixed bases) requires the change-of-base tool and is a reliable stretch discriminator. AO3 — exam relevance

Q18. Solve: 4^x − 5⋅2^x + 4 = 0.

Stretch

[5 marks]

M1 4^x=(2²)^x=(2^x)². Let u=2^x: u²−5u+4=0. AO1 — recognising 4^x=(2^x)² Common mistake: Missing this step: 4^x and 2^x share base 2, which is the key.

A1 (u−1)(u−4)=0 ⇒ u=1 or u=4. AO1 — solving quadratic in u

A1 2^x=1 ⇒ x=0; 2^x=4=2² ⇒ x=2. AO1 — solving for x

A1 x = 0 or x = 2. Check: 1−5+4=0 ✓; 16−20+4=0 ✓. AO2 — verification

A1 Both u values >0, so both are valid (no domain rejection for 2^x). AO2 — confirming domain

Q19. The graph of log₁₀(y) against x is a straight line through (1, 2.3) and (3, 3.1). Find constants A and b such that y = Ab^x.

Stretch

[7 marks]

M1 log y = log A + x log b. Linear in x: gradient = log b. AO2 — linearising the exponential model

M1 Gradient = (3.1−2.3)/(3−1) = 0.4. So log₁₀(b) = 0.4. AO1 — finding gradient

A1 b = 10^0.4 ≈ 2.51 (3 s.f.). AO1 — value of b

M1 Use point (1, 2.3): 2.3 = log A + 1(0.4) ⇒ log A = 1.9. AO2 — finding log A

A1 A = 10^1.9 ≈ 79.4 (3 s.f.). AO1 — value of A

A1 Model: y = 79.4 × 2.51^x. Check: at x=1, log y = 1.9+0.4=2.3 ✓; at x=3, 1.9+1.2=3.1 ✓. AO2 — both points verified

A1 This technique (log-linear graphs ⇒ exponential models) appears regularly and earns top-band marks for method clarity. AO2 — exam relevance

Q20. Solve: log₂(x²−x−6) − log₂(x+2) = 3.

Stretch

[7 marks]

M1 Quotient rule: log₂((x²−x−6)/(x+2)) = 3. AO1 — combining logs

M1 Factorise numerator: x²−x−6=(x−3)(x+2). Cancel (x+2) for x≠−2: simplifies to x−3. AO2 — algebraic simplification before exponential form

A1 log₂(x−3) = 3 ⇒ x−3 = 8 ⇒ x = 11. AO1 — exponential form and solving

A1 Domain: x+2≠0 (x≠−2) ✓; x²−x−6>0 at x=11: 104>0 ✓; x−3>0: 8>0 ✓. AO2 — full domain check

A1 x = 11. AO1 — final answer

A1 Key insight: factorising the numerator before converting gave a much simpler log equation to solve. AO3 — strategic algebraic step

A1 Verify: log₂(121−11−6)−log₂(13) = log₂(104/13) = log₂(8) = 3 ✓. AO2 — full verification

Stuck on a question?

Exponentials & Logs — Primer

Pure Mathematics — Paper 2 · AQA AS Mathematics (sample)

What this primer is for. Focused on AQA AS Exponentials & Logs: the laws that score marks and the errors that lose them. The 2023 examiner report specifically cited log law errors as very common.

What this topic is

Logarithms and exponentials are inverse operations: a^x=b ⇔ logᵃ(b)=x. The natural log (ln) uses base e ≈2.718. Three skills assessed: Log Law Simplification; Solving equations (converting between forms); and Hidden Quadratics (substituting u=e^x).

Why it matters in this paper

A high-frequency topic, typically 8–12 marks. The 2023 AQA ER named splitting log(a+b) into log(a)+log(b) as a very common error costing 2–3 marks on one question. Fix this one habit and you reliably recover marks.

How to tackle this topic

Match the question to the method:

Log Laws Simplification — Three laws: log a + log b = log(ab); log a − log b = log(a/b); n log a = log(a^n). BANNED: log(a+b) ≠ log(a)+log(b). Apply to products/quotients inside logs, never to sums. Use when: simplifying, combining multiple log terms into one, proving identities.
Convert And Solve — Switch forms: logᵃ(b)=c ⇔ a^c=b. For a^x=b: x=ln(b)/ln(a) (change of base). Natural log identities: ln(e^x)=x; e^{ln x}=x; ln1=0. Use when: unknown is in the exponent, or a log equation reduces via exponential conversion.
Substitution For Hidden Quadratics — e^2x−5e^x+6=0 is a quadratic in e^x. Let u=e^x (always >0), solve the quadratic, then take ln. For 4^x−5⋅2^x+4=0: write 4^x=(2^x)² first. Use when: two exponential terms with one being the square of the other.

The key facts you must know

Product: log a + log b = log(ab).
Quotient: log a − log b = log(a/b).
Power: n log a = log(a^n).
FORBIDDEN: log(a+b) ≠ log a + log b — 2023 AQA ER top error.
Change of base: logᵃ(x) = ln(x)/ln(a) = log(x)/log(a).
a^x=b ⇒ x=ln(b)/ln(a).
ln(e^x)=x; e^{ln x}=x; ln1=0; ln e=1.
Domain: log(f(x)) requires f(x)>0; always check after solving.

Step-by-step worked example

Question: Solve e^2x−4e^x+3=0.

Step 1. Let u=e^x (note e^2x=(e^x)²=u²).
Step 2. u²−4u+3=0 ⇒ (u−1)(u−3)=0 ⇒ u=1 or u=3.
Step 3. e^x=1 ⇒ x=0; e^x=3 ⇒ x=ln3.
Step 4. Both valid (e^x>0 always). x=0 or x=ln3.

Why this lands top-band: naming the substitution u=e^x, giving exact answers (ln3 not 1.099), and checking domain all contribute to full marks.

Common mistakes (from examiner reports)

log(x+8) written as log(x)+log(8). — “Cost candidates 2–3 marks on a single question in 2023.”
Solving 3^x=7 as x=7/3 (dividing instead of taking logs).
Not checking domain after solving log equations (negative arguments).
Forgetting to cancel e^2x=(e^x)² and failing to spot the hidden quadratic.

Quick self-check

Simplify 2ln(x)+ln(3)−ln(x) as one log without looking.
Solve 5^x=12 exactly, then to 3 d.p.
What substitution turns e^2x−3e^x+2=0 into a quadratic?

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Exponentials & Logarithms High-frequency topic

The scaffold to use (LCSE Framework)

Q1. Write as a single logarithm: (a) log 5 + log 4; (b) log 20 − log 4; (c) 3 log 2.

Q2. Solve: log₂(x) = 5.

Q3. Solve: 3^x = 20. Give your answer to 3 significant figures.

Q4. Simplify: log 12 − log 4 + log 3.

Q5. Find: (a) ln(e³); (b) eln 7; (c) ln(1).

Q6. Solve: 2e^x = 14. Give your answer exactly.

Q7. Solve: log(3x + 4) = 2 (base 10).

Q8. Solve simultaneously: y = e2x and y = 3e^x − 2.

Q9. Solve: log₂(x+3) + log₂(x−1) = 5.

Q10. Given log₄(a)=3 and log₄(b)=−1/2, find: (a) log₄(a²b); (b) log₄(a/√b).

Q11. Solve: 52x+1 = 8. Give your answer to 3 s.f.

Q12. Express 2log(x) − 3log(y) + log(z) as a single logarithm.

Q13. Solve: e2x − 5e^x + 6 = 0.

Q14. A population P grows by P = 500e0.03t (t in years). (a) Initial population? (b) Time to double?

Q15. Solve: log(x+8) − log(x) = 1 (base 10).

Q16. Solve: 7x+1 = 32x−1. Give your answer to 3 d.p.

Q17. Solve: log₂(x) + log₄(x) = 3.

Q18. Solve: 4^x − 5⋅2^x + 4 = 0.

Q19. The graph of log₁₀(y) against x is a straight line through (1, 2.3) and (3, 3.1). Find constants A and b such that y = Ab^x.

Q20. Solve: log₂(x²−x−6) − log₂(x+2) = 3.

Q5. Find: (a) ln(e³); (b) e^{ln 7}; (c) ln(1).

Q8. Solve simultaneously: y = e^2x and y = 3e^x − 2.

Q11. Solve: 5^2x+1 = 8. Give your answer to 3 s.f.

Q13. Solve: e^2x − 5e^x + 6 = 0.

Q14. A population P grows by P = 500e^0.03t (t in years). (a) Initial population? (b) Time to double?

Q16. Solve: 7^x+1 = 3^2x−1. Give your answer to 3 d.p.