Practice › Number and Algebra › Exponential and Logarithmic Functions

Exponential and Logarithmic Functions: Paper 1 Challenge Long Questions

This is the Challenge tier of the Paper 1 set. The questions state a condition rather than naming a method: a change of base hidden inside an equation, roots that must be rejected, a logarithm whose base is below 1, an exponential equation with a parameter, and claims to prove or disprove. Work the standard Paper 1 set first. Detailed solutions can be requested by email.

10 questions · 44 marks · about 55 minutes · answers below each question

Jump to

Question 1 — a scaling condition fixes the power in y = kxⁿ
Question 2 — log₂ x + logₓ 2 = 5/2: change of base, then a quadratic
Question 3 — intersection of y = log₂ x and y = log₄(x + 12)
Question 4 — a logarithmic system that becomes a sum and a product of roots
Question 5 — log base 1/2: the inequality reverses, and the domain still applies
Question 6 — a vertical line across y = log₂ x and y = log₈ x: prove PR = 3QR
Question 7 — 4ˣ − k·2ˣ + 9 = 0 with two distinct real roots: both u-roots must be positive
Question 8 — is log₂ 3 rational? a proof by contradiction
Question 9 — can two reciprocal logarithms add up to 1? a negative discriminant
Question 10 — N = abᵗ from two data points; a claim; when one model overtakes another

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · Modelled on 2024 P1 Q15 — no slope or point on the line is given

New idea: multiplying x by 9 divides y by 27 — that ratio alone fixes the power n.

It is given that

lo g_{3} y

is a linear function of

lo g_{3} x

. When

x

is multiplied by

9

y

is divided by

27

. When

x = 1

y = 4

. Express

y

in terms of

x

. (4 marks)

Show answer

$y = 4 x^{- \frac{3}{2}}$

A linear relation between log₃ y and log₃ x means y = kxⁿ. Then k(9x)ⁿ = kxⁿ ÷ 27 gives 9ⁿ = 1/27.

Question 2 · New sub-type, no direct past-paper model

Technique not in the standard set: a change of base hidden inside the equation, with no substitution suggested.

Solve

lo g_{2} x + lo g_{x} 2 = \frac{5}{2}

. (4 marks)

Show answer

$x = 2$ or $x = 4$

logₓ 2 = 1 ÷ log₂ x. Put u = log₂ x to get 2u² − 5u + 2 = 0.

Question 3 · New sub-type — two logarithmic graphs with different bases

Technique not in the standard set: equating logarithms to different bases by converting log₄ to log₂; one root must be rejected.

Find the coordinates of the point of intersection of the graph of

y = lo g_{2} x

and the graph of

y = lo g_{4} (x + 12)

. (4 marks)

Show answer

$(4, 2)$

log₄(x + 12) = ½ log₂(x + 12). Then 2 log₂ x = log₂(x + 12), so x² = x + 12.

Question 4 · Modelled on 2025 P1 Q16 — the same kind of system, with the substitution and the quadratic no longer given

New idea: nothing tells you to set u = log₂ x and v = log₂ y; spotting it is the question.

Let

x

and

y

be real numbers such that

0 < x < y

lo g_{2} (x y) = 7

and

(lo g_{2} x) (lo g_{2} y) = 12

. Find

x

and

y

. (4 marks)

Show answer

$x = 8, y = 16$

With u = log₂ x and v = log₂ y: u + v = 7 and uv = 12, so u and v are 3 and 4.

Question 5 · New sub-type — an inequality in a logarithm whose base is less than 1

Trap: the base $\frac{1}{2}$ makes $y = lo g_{\frac{1}{2}} x$ a decreasing function, so the inequality reverses; and $x - 1$ must stay positive.

(a)Solve $lo g_{\frac{1}{2}} (x - 1) > - 2$ . (3 marks)
(b)Find the number of integers $x$ satisfying both $lo g_{\frac{1}{2}} (x - 1) > - 2$ and $lo g_{2} x > 1$ . (1 mark)

Show answer

(a) $1 < x < 5$ (b) 2 (namely $x = 3$ and $x = 4$ )

$- 2 = lo g_{\frac{1}{2}} 4$ . Because the base is below 1, $lo g_{\frac{1}{2}} (x - 1) > lo g_{\frac{1}{2}} 4$ means $x - 1 < 4$ .

Question 6 · Modelled on 2018 P2 Q32 — turned into a long question to prove, then solve

Technique not in the standard set: comparing two logarithmic graphs at the same x by a change of base.

Let

k > 1

. The vertical line

x = k

cuts the graph of

y = lo g_{2} x

, the graph of

y = lo g_{8} x

and the

x

-axis at the points

P

Q

and

R

respectively (see the figure).

Graphs of y = log 2 of x and y = log 8 of x cut by the vertical line x = k at P and Q, and the x-axis at R

(a)Prove that $P R = 3 QR$ . (2 marks)
(b)If $P Q = 4$ , find $k$ . (3 marks)

Show answer

(a) $P R = lo g_{2} k$ and $QR = \frac{1}{3} lo g_{2} k$ (b) $k = 64$

log₈ k = log₂ k ÷ log₂ 8 = ⅓ log₂ k. Then PQ = PR − QR = ⅔ log₂ k.

Question 7 · New sub-type — an exponential equation with a parameter

Trap: a positive discriminant is not enough. Each root in u = 2ˣ must be positive, which rules out k < −6.

Let

k

be a real constant. Find the range of values of

k

such that the equation

4^{x} - k \cdot 2^{x} + 9 = 0

has two distinct real roots. (5 marks)

Show answer

$k > 6$

Put u = 2ˣ > 0. Each positive root u gives exactly one x, and a negative root gives none — so u² − ku + 9 = 0 needs two distinct positive roots.

Question 8 · New sub-type — a claim about log₂ 3 to prove or disprove

New idea: a proof by contradiction using odd and even numbers; no amount of calculation settles it.

Someone claims that

lo g_{2} 3

is a rational number, that is,

lo g_{2} 3 = \frac{p}{q}

for some positive integers

p

and

q

. Do you agree? Explain your answer. (3 marks)

Show answer

Disagree. $lo g_{2} 3 = \frac{p}{q}$ would give $2^{p} = 3^{q}$ , but $2^{p}$ is even and $3^{q}$ is odd.

Remove the logarithm: 2^(p/q) = 3, then raise both sides to the power q.

Question 9 · New sub-type — a claim to disprove with the discriminant

Technique not in the standard set: $lo g_{b} a = \frac{1}{l o g _{a} b}$ turns the claim into a quadratic with no real root.

Let

a

and

b

be positive numbers, neither of which is equal to

1

. Someone claims that it is possible to choose

a

and

b

such that

lo g_{a} b + lo g_{b} a = 1

. Do you agree? Explain your answer. (4 marks)

Show answer

Disagree. With $u = lo g_{a} b$ , the equation becomes $u^{2} - u + 1 = 0$ , whose discriminant is $- 3 < 0$ .

$lo g_{b} a = \frac{1}{l o g _{a} b}$ , and $u = lo g_{a} b$ is a non-zero real number.

Question 10 · Modelled on 2026 P2 Q36 — the y = abˣ model built from two data points, then used to test a claim and compare two models

New idea: (c) compares two exponential models; dividing one by the other leaves a single power to solve.

The number of users of an app

t

months after its launch is

N

, where

N = a b^{t}

and

a

b

are positive constants. When

t = 2

N = 1800

and when

t = 5

N = 6075

(a)Find $a$ and $b$ . (3 marks)
(b)The developer claims that the number of users will exceed 100 000 within the first 12 months after the launch. Do you agree? Explain your answer. (2 marks)
(c)The number of users of another app $t$ months after its launch is $M$ , where $M = 5000 (1.2)^{t}$ . Find the least positive integer $t$ such that $N > M$ . (2 marks)

Show answer

(a) $a = 800, b = 1.5$ (b) Agree: $N \approx 103797 > 100000$ when $t = 12$ (c) $t = 9$

(a) Divide: b³ = 6075 ÷ 1800 = 27/8. (c) 800(1.5)ᵗ > 5000(1.2)ᵗ gives 1.25ᵗ > 6.25.

Continue practising this topic: If you have not yet worked the Paper 1 long questions, start there. Then move to Paper 2 multiple-choice — also available as Paper 2 Challenge MC.

Where this sits in the syllabus

Exponential and logarithmic functions sit in the senior Number and Algebra strand. Paper 2 has set at least one question on them in Section B of every paper from 2012 to 2026, and Paper 1 has used them in short Section B questions, most recently in 2024 and 2025, as well as inside longer sequence questions. Most marks are lost in two places: a root that makes a logarithm undefined is kept, and an inequality is not reversed after dividing by a negative logarithm such as log 0.8.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

Start with a free discovery session

No cost, no obligation. We look at where your child actually is, and show you what a term of real progress would look like.

Book a free discovery session See the programmes