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Edexcel International A Level Maths · P2 · Pure Mathematics 2

Exponentials and logarithms

28 free, high-quality past-paper-style questions on Exponentials and logarithms for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 236 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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The curve C has equation y = (½)^x + 3, for all real values of x.

(a)

Write down the coordinates of the point where C crosses the y-axis.

(1)

(b)

Write down the equation of the asymptote to C.

(1)

(c)

Explain why C does not meet the x-axis.

(1)

(d)

State whether y increases or decreases as x increases, giving a reason.

(1)

(e)

Find the value of x for which y = 11.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
(½)⁰ + 3 = 4, so (0, 4)	B1
Part (b)
y = 3	B1
Part (c)
(½)^x > 0 for every x, so y > 3 and y is never zero	B1
Part (d)
decreases: the base ½ lies between 0 and 1 (equivalently (½)^x = 2^−x)	B1
Part (e)
(½)^x = 8, i.e. 2^−x = 2³	M1
x = −3	A1

Model sketch

Figure: model sketch of C with its asymptote y = 3

Common error: Putting the asymptote of y = (½)x + 3 on the x-axis. Adding 3 lifts the whole curve, asymptote included, by 3 units.

Theory: P2.5 Exponentials and logarithms

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This question is about the meaning of a logarithm. Do not use a calculator.

(a)

Write 5⁴ = 625 in logarithmic form.

(1)

(b)

Write log₂ 0.5 = −1 in index form.

(1)

(c)

Find the value of (i) log₂ 32 (ii) log₉ 3 (iii) log₈(¹⁄₆₄).

(3)

(d)

Find the exact value of log₁₆ 8.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
log₅ 625 = 4	B1
Part (b)
2⁻¹ = 0.5	B1
Part (c)
(i) 5	B1
(ii) ½, since 9^½ = 3	B1
(iii) −2, since 8⁻² = 1⁄64	B1
Part (d)
16^y = 8 ⇒ 2^4y = 2³	M1
y = ¾	A1

Exam tip: For a log whose base and argument are powers of the same number, write both as powers of that number and compare indices — the answer comes out exactly.

Theory: P2.5 Exponentials and logarithms

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In this question a is a positive constant, a ≠ 1.

(a)

Write 3 log_a 2 + log_a 5 − log_a 10 as a single logarithm, simplifying your answer.

(2)

(b)

Show that log₆ 4 + log₆ 9 = 2.

(2)

(c)

Find the exact value of log_a(a⁵) − log_a(¹⁄_a).

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
log_a 8 + log_a 5 − log_a 10, using the power law	M1
= log_a(40 ÷ 10) = log_a 4	A1
Part (b)
log₆ 4 + log₆ 9 = log₆ 36 (product law)	M1
6² = 36, so log₆ 36 = 2	A1
Part (c)
5 log_a a + log_a a, using the power and reciprocal laws	M1
= 5 + 1 = 6	A1

Common error: Turning 3 log 2 into log 6. A number in front of a log becomes a power inside it: 3 log 2 = log 23 = log 8.

Theory: P2.5 Exponentials and logarithms

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Solve each equation, giving your answers to 3 significant figures.

(a)

7^x = 30

(2)

(b)

2^3x = 50

(2)

(c)

0.6^x = 0.05

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
x = log 30 ÷ log 7	M1
= 1.75 (1.7478…)	A1
Part (b)
3x log 2 = log 50	M1
x = log 50 ÷ (3 log 2) = 1.88 (1.8812…)	A1
Part (c)
x = log 0.05 ÷ log 0.6	M1
= 5.86 (5.8644…)	A1

Exam tip: Write the exact form, such as log 30 ÷ log 7, before the decimal. If the rounding slips, the method is still visible.

Theory: P2.5 Exponentials and logarithms

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Figure 1 shows part of the curve with equation y = k × a^x, where k and a are positive constants. The curve passes through the points A(1, 6) and B(3, 54), crosses the y-axis at C and meets the line y = 100 at D.

(a)

Use the coordinates of A and B to show that a = 3, and find the value of k.

(3)

(b)

Write down the coordinates of C.

(1)

(c)

Find the x-coordinate of D, giving your answer to 3 significant figures.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
ka = 6 and ka³ = 54	M1
dividing: a² = 9, and a > 0 so a = 3	A1
k = 2	A1
Part (b)
y = 2 × 3⁰ = 2, so C(0, 2)	B1
Part (c)
2 × 3^x = 100 ⇒ 3^x = 50, so x = log 50 ÷ log 3	M1
= 3.56 (3.5609…)	A1

Common error: Dividing 54 by 6 and writing a = 9. The x-coordinates of A and B differ by 2, so 9 is a2, not a.

Theory: P2.5 Exponentials and logarithms

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Figure 1 shows four curves, labelled A, B, C and D. Their equations, in no particular order, are
(i) y = 2^x (ii) y = 4^x (iii) y = (⅓)^x (iv) y = 2^x + 2.
Curves A and C meet at the point P.

(a)

Match each of the curves A, B, C and D to its equation, giving a reason for each choice.

(4)

(b)

Write down the equation of the asymptote to curve A.

(1)

(c)

Use algebra to find the coordinates of P.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
B is (iii): the only decreasing curve, as its base ⅓ lies between 0 and 1	B1
A is (iv): it crosses the y-axis at (0, 3), since 2⁰ + 2 = 3, not at (0, 1)	B1
C is (ii): for x > 0, 4^x > 2^x, so it is the steeper of the two increasing curves through (0, 1)	B1
D is (i): the remaining curve, through (0, 1), (1, 2) and (2, 4)	B1
Part (b)
y = 2	B1
Part (c)
4^x = 2^x + 2; with u = 2^x: u² − u − 2 = 0	M1
(u − 2)(u + 1) = 0; u = −1 is impossible because 2^x > 0	A1
2^x = 2 ⇒ P(1, 4)	A1

Exam tip: To tell exponential curves apart, check three things: where each crosses the y-axis, whether it rises or falls, and which is steeper for x > 0.

Theory: P2.5 Exponentials and logarithms

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Figure 1 shows a ruler with a logarithmic scale. The mark for the number N, where 1 ≤ N ≤ 10, is placed a distance 250 log N mm from the mark for 1, where log means log₁₀. The pointer P is 180 mm from the mark for 1.

(a)

Explain why the mark for 10 is 250 mm from the mark for 1.

(1)

(b)

Show that the distance from the mark for 1 to the mark for 6 is the sum of the distances to the marks for 2 and 3.

(2)

(c)

The mark for 2 is p mm from the mark for 1. Find, in terms of p, the distance from the mark for 1 to the mark for (i) 8 (ii) 5.

(3)

(d)

Find the number at the pointer P, giving your answer to 3 significant figures.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
log 10 = 1 (since 10¹ = 10), so the distance is 250 × 1 = 250 mm	B1
Part (b)
250 log 6 = 250 log(2 × 3)	M1
= 250 log 2 + 250 log 3 by the product law, which is the sum of the two distances	A1
Part (c)
(i) 250 log 2³ = 3 × 250 log 2 = 3p mm	B1
(ii) 250 log(10 ÷ 2) = 250(log 10 − log 2)	M1
= 250 − p mm	A1
Part (d)
250 log N = 180 ⇒ log N = 0.72	M1
N = 10^0.72 = 5.25 (5.248…)	A1

Exam tip: Read 250 log 6 as 250 log(2 × 3): splitting a number into factors is what lets the product law turn one distance into a sum of two.

Theory: P2.5 Exponentials and logarithms

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Figure 1 shows the curve with equation y = a^x + b, where a and b are constants and a > 0. The line y = −4 is an asymptote to the curve. The curve passes through the point (2, 5), crosses the x-axis at Q and crosses the y-axis at R.

(a)

Write down the value of b.

(1)

(b)

Find the value of a.

(2)

(c)

Write down the coordinates of R.

(1)

(d)

Find the x-coordinate of Q, giving your answer to 3 significant figures.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
a^x → 0 as x → −∞, so the asymptote is y = b: b = −4	B1
Part (b)
a² − 4 = 5 ⇒ a² = 9	M1
a = 3 (−3 rejected since a > 0)	A1
Part (c)
3⁰ − 4 = −3, so R(0, −3)	B1
Part (d)
3^x = 4 ⇒ x = log 4 ÷ log 3	M1
= 1.26 (1.2618…)	A1

Common error: Giving a = ±3. The base of an exponential function must be positive, so only a = 3 is possible.

Theory: P2.5 Exponentials and logarithms

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A laboratory culture of yeast cells is monitored. The number of cells, N, at time t hours after monitoring begins is modelled by N = 400 × 1.15^t, t ≥ 0.

(a)

Write down the number of cells when monitoring begins.

(1)

(b)

Find the number of cells predicted by the model after 6 hours.

(1)

(c)

Find the value of t for which the model predicts 5000 cells, giving your answer to 3 significant figures.

(3)

(d)

Hence write down the number of complete hours after which the model first predicts more than 5000 cells.

(1)

(e)

Show that, according to the model, the number of cells approximately doubles every 5 hours.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
400	B1
Part (b)
400 × 1.15⁶ = 925 (925.2…)	B1
Part (c)
1.15^t = 12.5	M1
t = log 12.5 ÷ log 1.15	M1
= 18.1 (18.07…)	A1
Part (d)
19 (since 18.07 is not a whole number of hours and N is increasing)	B1
Part (e)
N(t + 5) ÷ N(t) = 1.15⁵, the same for every t	M1
1.15⁵ = 2.01 ≈ 2, so the number approximately doubles every 5 hours	A1

Exam tip: When a question asks for a whole number of hours, days or years, solve the equation first, then round in the direction the context needs — here up, because the number must have passed 5000.

Theory: P2.5 Exponentials and logarithms

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Given that log₃ x = p and log₃ y = q, where x and y are positive,

(a)

express log₃(9x) in terms of p.

(2)

(b)

express log₃(^x²⁄_y) in terms of p and q.

(1)

(c)

express log₃(^√x⁄_27y³) in terms of p and q.

(2)

(d)

Given also that log₃(xy) = 4 and log₃(^x⁄_y) = 1, find the values of p and q, and hence find the exact values of x and y, writing each in the form k√3.

(4)

Show mark scheme

Answer	Mark	Notes
Part (a)
log₃ 9 + log₃ x	M1
= 2 + p	A1
Part (b)
2p − q	B1
Part (c)
½log₃ x − log₃ 27 − 3log₃ y	M1
= ½p − 3 − 3q	A1
Part (d)
p + q = 4 and p − q = 1	M1
p = 2.5, q = 1.5	A1
x = 3^2.5 = 9√3	A1
y = 3^1.5 = 3√3	A1

Exam tip: Break the argument into prime factors and powers before using the laws: 27y3 = 33 × y3, so its log is 3 + 3q.

Theory: P2.5 Exponentials and logarithms

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This question is about equations in which the unknown appears in two different powers.

(a)

Show that the solution of 3^{x + 2} = 5^x can be written as x = log 9 ÷ (log 5 − log 3).

(3)

(b)

Hence find x to 3 significant figures.

(1)

(c)

Solve 2^{2x − 1} = 7^{1 − x}, giving your answer to 3 significant figures.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
(x + 2) log 3 = x log 5	M1
x log 3 + 2 log 3 = x log 5 ⇒ 2 log 3 = x(log 5 − log 3)	M1
2 log 3 = log 9, so x = log 9 ÷ (log 5 − log 3)	A1
Part (b)
4.30 (4.3013…)	B1
Part (c)
(2x − 1) log 2 = (1 − x) log 7	M1
x(2 log 2 + log 7) = log 2 + log 7, i.e. x = log 14 ÷ log 28	M1
x = 0.792 (0.7919…)	A1

Common error: Taking logs and then forgetting the brackets: (x + 2) log 3 must be expanded to x log 3 + 2 log 3 before the x terms are collected.

Theory: P2.5 Exponentials and logarithms

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The equation 5^2x − 7 × 5^x + 10 = 0 is to be solved.

(a)

Using the substitution u = 5^x, write the equation as a quadratic in u.

(1)

(b)

Hence solve the equation, giving any non-exact answer to 3 significant figures.

(4)

(c)

Explain why the equation 5^2x + 7 × 5^x + 10 = 0 has no real solutions.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
5^2x = (5^x)², so u² − 7u + 10 = 0	B1
Part (b)
(u − 2)(u − 5) = 0	M1
u = 2 or u = 5	A1
5^x = 5 ⇒ x = 1	A1
5^x = 2 ⇒ x = log 2 ÷ log 5 = 0.431	A1
Part (c)
u² + 7u + 10 = (u + 2)(u + 5) = 0 gives u = −2 or −5	M1
5^x > 0 for all real x, so neither value is possible: no real solutions	A1

Exam tip: After solving for u, go back to 5x = u for each value separately. A root of the quadratic is not yet a value of x.

Theory: P2.5 Exponentials and logarithms

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Solve the following logarithmic equations. Show each step of your working.

(a)

log₂(x + 4) + log₂(x − 2) = 4, where x > 2

(4)

(b)

log₅(2y + 1) − log₅(y − 1) = 1, where y > 1

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
log₂[(x + 4)(x − 2)] = 4	M1
(x + 4)(x − 2) = 2⁴ = 16	M1
x² + 2x − 24 = 0 ⇒ (x + 6)(x − 4) = 0	A1
x = 4; x = −6 rejected because x − 2 would be negative	A1
Part (b)
log₅[(2y + 1) ÷ (y − 1)] = 1	M1
2y + 1 = 5(y − 1)	M1
y = 2 (both arguments, 5 and 1, are positive)	A1

Exam tip: Check every root in the original logs, not in the quadratic. A root is rejected only because some log argument would be zero or negative — say which one.

Theory: P2.5 Exponentials and logarithms

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This question uses the change of base formula.

(a)

Show that log₄ x = ½ log₂ x for x > 0.

(2)

(b)

Hence solve log₂ x − log₄ x = 3.

(3)

(c)

Find the value of log₇ 50, giving your answer to 3 significant figures.

(2)

(d)

Solve log_x 8 = 1.5, where x > 0 and x ≠ 1.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
log₄ x = log₂ x ÷ log₂ 4	M1
log₂ 4 = 2, so log₄ x = ½ log₂ x	A1
Part (b)
log₂ x − ½ log₂ x = 3	M1
log₂ x = 6	A1
x = 64	A1
Part (c)
log 50 ÷ log 7	M1
= 2.01 (2.0103…)	A1
Part (d)
x^1.5 = 8	M1
x = 8^2/3 = 4	A1

Exam tip: In a “hence” part, write the result you have just shown as your first line. Here, replacing log4 x by ½ log2 x is the step that earns the method mark.

Theory: P2.5 Exponentials and logarithms

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A patient takes a dose of a medicine. The concentration of the medicine in the patient's blood, C mg l⁻¹, t hours after the dose is taken, is modelled by C = C₀ × r^t, where C₀ and r are positive constants. Figure 1 shows the graph of C against t, which passes through the points (0, 80) and (2, 51.2). The dashed line shows the level C = 20.

(a)

Write down the value of C₀.

(1)

(b)

Show that r = 0.8.

(2)

(c)

A second dose is taken when the concentration first falls to 20 mg l⁻¹. Find the time between the two doses, giving your answer in hours to 3 significant figures.

(3)

(d)

Find the time taken for the concentration to halve, and explain how this time is related to your answer to part (c).

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
at t = 0, r⁰ = 1, so C₀ = 80	B1
Part (b)
80r² = 51.2 ⇒ r² = 0.64	M1
r > 0, so r = 0.8	A1
Part (c)
80 × 0.8^t = 20 ⇒ 0.8^t = 0.25	M1
t = log 0.25 ÷ log 0.8	M1
= 6.21 hours (6.2126…)	A1
Part (d)
0.8^t = 0.5 ⇒ t = log 0.5 ÷ log 0.8 = 3.11 hours (3.1063…)	B1
20 = 80 × ½ × ½, so the time in (c) is exactly two halving times: 2 × 3.1063… = 6.21	B1

Exam tip: Both log 0.25 and log 0.8 are negative, so their quotient is positive. A negative time means the equation has been set up upside down.

Theory: P2.5 Exponentials and logarithms

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In Figure 1, the logarithmic curve y = log_a(x + b) is drawn for its whole domain; a > 1 and b are constants. It has the vertical asymptote x = −3 and goes through (5, 3). Its intercepts are labelled P, on the x-axis, and Q, on the y-axis.

(a)

Explain why b = 3.

(1)

(b)

Find the value of a.

(2)

(c)

Find the coordinates of P.

(2)

(d)

Find the y-coordinate of Q, giving your answer to 3 significant figures.

(2)

(e)

Find the value of x for which y = 6.

(1)

Show mark scheme

Answer	Mark	Notes
Part (a)
log_a(x + b) is defined only for x + b > 0, so the asymptote is x = −b; −b = −3 gives b = 3	B1
Part (b)
log_a 8 = 3 ⇒ a³ = 8	M1
a = 2	A1
Part (c)
log₂(x + 3) = 0 ⇒ x + 3 = 2⁰ = 1	M1
P(−2, 0)	A1
Part (d)
y = log₂ 3 = log 3 ÷ log 2	M1
= 1.58 (1.5849…)	A1
Part (e)
x + 3 = 2⁶ = 64 ⇒ x = 61	B1

Common error: Writing log2(x + 3) = 0 as x + 3 = 0. A logarithm is 0 when its argument is 1, because a0 = 1.

Theory: P2.5 Exponentials and logarithms

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A sum of money is invested in a savings account. No money is added or withdrawn, and interest at the same percentage rate is added at the end of each year. Figure 1 shows the balance, £B, at the start and after each of the first five complete years, with each bar labelled to the nearest pound.

(a)

Use the first three bars to show that the balance is multiplied by the same number each year, and state the annual percentage rate of interest.

(2)

(b)

Write down an expression for B after n complete years.

(1)

(c)

Find the number of complete years after which the balance first exceeds £4000.

(3)

(d)

A second account, also starting with £2500, would reach exactly £4000 after 8 complete years. Find its annual percentage rate of interest, giving your answer to 3 significant figures.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
2600 ÷ 2500 = 1.04 and 2704 ÷ 2600 = 1.04	M1
rate = 4% per year	A1
Part (b)
B = 2500 × 1.04ⁿ	B1
Part (c)
2500 × 1.04ⁿ > 4000 ⇒ 1.04ⁿ > 1.6	M1
n > log 1.6 ÷ log 1.04 = 11.98…	A1
12 years (after 11 years the balance is £3849; after 12 years it is £4003)	A1
Part (d)
(1 + p⁄100)⁸ = 1.6 ⇒ 1 + p⁄100 = 1.6^1/8 = 1.06050…	M1
p = 6.05 (6.0504…)	A1

Exam tip: When the inequality gives n > 11.98…, check the balance after 11 and after 12 years to confirm which whole year first passes £4000.

Theory: P2.5 Exponentials and logarithms

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The level of a sound, L decibels (dB), is given by L = 10 log(I ÷ I₀), where I is the intensity of the sound, I₀ is a fixed reference intensity and log means log₁₀. Figure 1 shows the levels of four sounds.

(a)

Find the value of I ÷ I₀ for the conversation.

(1)

(b)

Find how many times more intense the busy road is than the conversation, giving your answer to 3 significant figures.

(2)

(c)

Show that doubling the intensity of any sound increases its level by 3.01 dB, to 3 significant figures.

(2)

(d)

A workshop has n identical machines. One machine running alone gives a level of 72 dB, and the intensities of machines running together add. Find the greatest number of machines that can run together without the level exceeding that of the busy road in Figure 1.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
log(I ÷ I₀) = 6 ⇒ I ÷ I₀ = 10⁶	B1
Part (b)
ratio = 10^8.5 ÷ 10⁶ = 10^2.5	M1
= 316 (316.2…)	A1
Part (c)
10 log(2I ÷ I₀) = 10 log 2 + 10 log(I ÷ I₀)	M1
increase = 10 log 2 = 3.0103… = 3.01 dB	A1
Part (d)
level = 10 log(nI₁ ÷ I₀) = 10 log n + 72, where I₁ is the intensity of one machine	M1
10 log n + 72 ≤ 85 ⇒ n ≤ 10^1.3 = 19.95…	A1
n = 19	A1

Common error: Rounding 19.95 up to 20. Twenty machines give 10 log 20 + 72 = 85.01 dB, which is above 85 dB, so the greatest number is 19.

Theory: P2.5 Exponentials and logarithms

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Figure 1 shows the curves y = log₂ x and y = log₈ x, which meet at R. The vertical line x = k meets the curves at P and Q, where PQ = 4. The line y = 1 meets the curves at T and S.

(a)

Show that log₈ x = ⅓ log₂ x for x > 0.

(2)

(b)

Write down the coordinates of R.

(1)

(c)

Find the value of k.

(3)

(d)

Find the length TS.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
log₈ x = log₂ x ÷ log₂ 8 (change of base)	M1
log₂ 8 = 3, giving the result	A1
Part (b)
both logarithms are 0 when x = 1: R(1, 0)	B1
Part (c)
PQ = log₂ k − ⅓ log₂ k = ⅔ log₂ k	M1
⅔ log₂ k = 4 ⇒ log₂ k = 6	A1
k = 64	A1
Part (d)
T: log₂ x = 1 ⇒ x = 2; S: log₈ x = 1 ⇒ x = 8	M1
TS = 6	A1

Exam tip: A vertical distance between two curves is the difference of their y-values at the same x. Writing both logs in the same base first turns that difference into a single multiple of log2 k.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test20IALM-P2-5-FB-10 P2 · ASMedium8 marks

Figure 1 shows the curves y = a^x and y = a^−x, where a is a constant, a > 1. The curves cross at R. The line y = 5 meets y = a^x at P and y = a^−x at Q, and PQ = 4.

(a)

Write down the coordinates of R, and describe the single transformation that maps the curve y = a^x onto the curve y = a^−x.

(2)

(b)

Show that a = √5.

(3)

(c)

The line y = c, where c > 1, meets the two curves at points that are 6 units apart. Find the exact value of c, giving your answer in the form m√5.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
R(0, 1)	B1
reflection in the y-axis	B1
Part (b)
at P, a^x = 5 ⇒ x = log_a 5	M1
by symmetry Q has x = −log_a 5, so PQ = 2 log_a 5 = 4	M1
log_a 5 = 2 ⇒ a² = 5, and a > 1 so a = √5	A1
Part (c)
2 log_√5 c = 6 ⇒ log_√5 c = 3	M1
c = (√5)³	M1
= 5√5	A1

Common error: Using PQ = loga 5 instead of twice it. P and Q lie on opposite sides of the y-axis, each a distance loga 5 from it.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test21IALM-P2-5-B-11 P2 · ASHard12 marks

The curves C₁ and C₂ have equations C₁: y = 3^2x and C₂: y = 4 × 3^x + 45, for all real x.

(a)

For each curve, write down the coordinates of the point where it crosses the y-axis, and write down the equation of the asymptote to C₂.

(3)

(b)

Show that the x-coordinate of any point where the curves meet satisfies u² − 4u − 45 = 0, where u = 3^x.

(2)

(c)

Hence show that the curves meet at exactly one point, and find its coordinates.

(4)

(d)

Find the value of x at which C₂ has y = 100, giving your answer to 3 significant figures.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
C₁: (0, 1)	B1
C₂: (0, 49)	B1
asymptote to C₂: y = 45	B1
Part (b)
3^2x = (3^x)² = u²	M1
u² = 4u + 45 ⇒ u² − 4u − 45 = 0	A1
Part (c)
(u − 9)(u + 5) = 0	M1
u = 9 or u = −5	A1
3^x > 0, so u = −5 is impossible: exactly one point	B1
3^x = 9 ⇒ x = 2, y = 81: (2, 81)	A1
Part (d)
4 × 3^x = 55 ⇒ 3^x = 13.75	M1
x = log 13.75 ÷ log 3	M1
= 2.39 (2.3857…)	A1

Model sketch

Figure: model sketch of C₁ and C₂, meeting once at (2, 81)

Exam tip: Where two exponential curves meet, set the expressions equal and look for a quadratic in 3x. The sign of each root of that quadratic tells you how many intersection points there are.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test22IALM-P2-5-B-12 P2 · ASHard10 marks

This question involves two pairs of simultaneous equations.

(a)

Given that 2^a × 8^b = 64 and 9^a ÷ 3^b = 243, find the values of a and b.

(4)

(b)

The positive numbers x and y satisfy log₃ x + log₃ y = 3 and (log₃ x)(log₃ y) = 2. By letting p = log₃ x and q = log₃ y, find all possible pairs of values of x and y.

(5)

(c)

Write down the value of xy for these pairs.

(1)

Show mark scheme

Answer	Mark	Notes
Part (a)
2^{a + 3b} = 2⁶ ⇒ a + 3b = 6	M1
3^{2a − b} = 3⁵ ⇒ 2a − b = 5	M1
solving simultaneously	M1
a = 3, b = 1	A1
Part (b)
q = 3 − p, so p(3 − p) = 2 ⇒ p² − 3p + 2 = 0	M1
p = 1 or p = 2	A1
x = 3^p, y = 3^q	M1
x = 3, y = 9	A1
x = 9, y = 3	A1
Part (c)
log₃(xy) = 3, so xy = 27	B1

Common error: Treating (log3 x)(log3 y) as log3(xy). There is no law for a product of two logs; substitute p and q and solve the algebra instead.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test23IALM-P2-5-B-13 P2 · ASHard10 marks

Two samples of radioactive material, A and B, are stored together. The mass of A, in grams, t days after storage begins is modelled by 80 × 0.9^t, and the mass of B, in grams, by 30 × 0.97^t, t ≥ 0.

(a)

Find the total mass of the two samples when storage begins.

(1)

(b)

Find the time taken for the mass of A to halve, giving your answer in days to 3 significant figures.

(3)

(c)

Show that the two masses are equal when t = log(8⁄3) ÷ log(97⁄90).

(3)

(d)

Find this value of t, and the mass of each sample at this time, giving your answers to 3 significant figures.

(2)

(e)

Explain why, according to the model, the mass of A is less than the mass of B at all later times.

(1)

Show mark scheme

Answer	Mark	Notes
Part (a)
80 + 30 = 110 g	B1
Part (b)
0.9^t = 0.5	M1
t = log 0.5 ÷ log 0.9	M1
= 6.58 days (6.5788…)	A1
Part (c)
80 × 0.9^t = 30 × 0.97^t ⇒ (0.97 ÷ 0.9)^t = 80 ÷ 30	M1
(97⁄90)^t = 8⁄3	M1
taking logs: t log(97⁄90) = log(8⁄3), giving the result	A1
Part (d)
t = 13.1 days (13.09…)	B1
mass of each = 80 × 0.9^13.09… = 20.1 g	B1
Part (e)
the ratio (mass of A) ÷ (mass of B) = (8⁄3)(90⁄97)^t keeps decreasing because 90⁄97 < 1 (A loses a larger fraction each day), so once the masses are equal A stays smaller	B1

Model sketch

Figure: model sketch of the masses of A and B against time

Exam tip: Collect the two exponentials on one side before taking logs: (97⁄90)t = 8⁄3 has positive logs on both sides, so there is no sign to worry about.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test24IALM-P2-5-B-14 P2 · ASHard10 marks

In this question x > 0 and x ≠ 1. Let L = log₃ x.

(a)

Show that log_x 9 = 2 ÷ log₃ x.

(2)

(b)

Hence show that the equation log₃ x + 3 log_x 9 = 7 can be written as L² − 7L + 6 = 0.

(2)

(c)

Hence solve log₃ x + 3 log_x 9 = 7.

(3)

(d)

Solve log₄(x + 1) − log₄(x − 2) = ½, where x > 2.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
log_x 9 = log₃ 9 ÷ log₃ x (change of base)	M1
log₃ 9 = 2, giving the result	A1
Part (b)
L + 6 ÷ L = 7	M1
multiplying by L (non-zero as x ≠ 1): L² − 7L + 6 = 0	A1
Part (c)
(L − 1)(L − 6) = 0	M1
L = 1 ⇒ x = 3	A1
L = 6 ⇒ x = 729	A1
Part (d)
log₄[(x + 1) ÷ (x − 2)] = ½	M1
(x + 1) ÷ (x − 2) = 4^½ = 2	M1
x + 1 = 2x − 4 ⇒ x = 5	A1

Common error: Solving L2 − 7L + 6 = 0 and giving x = 1 and x = 6. Those are values of L = log3 x; the values of x are 31 and 36.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test25IALM-P2-5-FB-11 P2 · ASHard12 marks

Figure 1 shows the curve with equation y = 4^x − 10 × 2^x + 16. The curve crosses the x-axis at A and B and the y-axis at C. The line y = 16 is an asymptote to the curve, and the curve meets this line at the point D. The line y = −10 is also shown.

(a)

Write down the coordinates of C.

(1)

(b)

Explain why y = 16 is an asymptote to the curve.

(1)

(c)

Using the substitution u = 2^x, show that the x-coordinates of A and B are 1 and 3.

(4)

(d)

Find the x-coordinate of D, giving your answer to 3 significant figures.

(3)

(e)

Show that the curve does not meet the line y = −10.

(3)

Show mark scheme

Answer	Mark	Notes
Part (a)
1 − 10 + 16 = 7, so C(0, 7)	B1
Part (b)
as x → −∞, 4^x → 0 and 2^x → 0, so y → 16	B1
Part (c)
4^x = (2^x)² = u²	M1
u² − 10u + 16 = 0	A1
(u − 2)(u − 8) = 0	M1
2^x = 2 ⇒ x = 1; 2^x = 8 ⇒ x = 3	A1
Part (d)
u² − 10u + 16 = 16 ⇒ u(u − 10) = 0	M1
u = 0 is impossible since 2^x > 0, so 2^x = 10	A1
x = log 10 ÷ log 2 = 3.32 (3.3219…)	A1
Part (e)
u² − 10u + 16 = −10 ⇒ u² − 10u + 26 = 0	M1
discriminant = 10² − 4 × 26 = −4	A1
−4 < 0, so there is no real u, and the curve does not meet the line	A1

Exam tip: A root u = 0 or a negative u gives no point on the curve, because 2x is always strictly positive. Say so explicitly when you reject it.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test26IALM-P2-5-FB-12 P2 · ASHard11 marks

Figure 1 shows the curves C₁ and C₂ with equations
C₁: y = 2 log₃ x, x > 0 and C₂: y = 1 + log₃(x + 6), x > −6.
The curve C₂ crosses the x-axis at S, and the curves meet at the point P.

(a)

Write down the equation of the asymptote to each curve.

(2)

(b)

Find the exact x-coordinate of S.

(2)

(c)

Show that the x-coordinate of P satisfies x² − 3x − 18 = 0.

(3)

(d)

Hence find the coordinates of P, explaining why the curves meet only once. Give the y-coordinate in the form p + q log₃ 2, where p and q are integers.

(4)

Show mark scheme

Answer	Mark	Notes
Part (a)
C₁: x = 0	B1
C₂: x = −6	B1
Part (b)
1 + log₃(x + 6) = 0 ⇒ x + 6 = 3⁻¹	M1
x = −17⁄3	A1
Part (c)
2 log₃ x = log₃ x² (power law)	M1
log₃ x² − log₃(x + 6) = 1 ⇒ x² ÷ (x + 6) = 3	M1
x² = 3x + 18, giving x² − 3x − 18 = 0	A1
Part (d)
(x − 6)(x + 3) = 0	M1
x = −3 is rejected because log₃ x is undefined for x ≤ 0, so x = 6 only	A1
y = 2 log₃ 6 = 2 log₃ 3 + 2 log₃ 2	M1
P(6, 2 + 2 log₃ 2)	A1

Common error: Keeping x = −3 as a second meeting point. It solves the quadratic, but log3(−3) does not exist, so −3 is not on C1; the figure confirms the curves meet only once.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test27IALM-P2-5-FB-13 P2 · ASHard11 marks

A hot drink is left to cool in a room. Its temperature, T °C, t minutes after it is poured, is modelled by T = 20 + A × 2^−kt, where A and k are positive constants. Figure 1 shows the graph of T against t. It passes through the points (0, 90) and (2, 55), and the line T = 20 is an asymptote.

(a)

Find the value of A.

(2)

(b)

Show that k = ½.

(2)

(c)

Find the time taken for the drink to cool to 30 °C, giving your answer in minutes to 3 significant figures.

(3)

(d)

Find the length of time for which the temperature of the drink is between 60 °C and 70 °C, giving your answer in minutes to 3 significant figures.

(3)

(e)

State, with a reason, whether the model predicts that the drink will ever cool to 18 °C.

(1)

Show mark scheme

Answer	Mark	Notes
Part (a)
at t = 0: 20 + A × 2⁰ = 90	M1
A = 70	A1
Part (b)
20 + 70 × 2^−2k = 55 ⇒ 2^−2k = ½ = 2⁻¹	M1
−2k = −1, so k = ½	A1
Part (c)
70 × 2^−t/2 = 10 ⇒ 2^−t/2 = 1⁄7	M1
t⁄2 = log 7 ÷ log 2	M1
t = 5.61 minutes (5.6147…)	A1
Part (d)
T = 70: 2^−t/2 = 5⁄7 ⇒ t = 2 log 1.4 ÷ log 2 = 0.9709…	M1
T = 60: 2^−t/2 = 4⁄7 ⇒ t = 2 log 1.75 ÷ log 2 = 1.6147…	A1
time = 1.6147… − 0.9709… = 0.644 minutes	A1
Part (e)
no: 70 × 2^−t/2 > 0 for all t, so T > 20 at all times	B1

Exam tip: Isolate the power before taking logs: subtract 20, divide by 70, then take logs of 2−t/2 = 1⁄7. Taking logs of the sum 20 + 70 × 2−t/2 leads nowhere.

Theory: P2.5 Exponentials and logarithms

+ Add to test✓ In your test28IALM-P2-5-FB-14 P2 · ASHard12 marks

A ball is dropped from a height of 2.5 m onto a hard floor. Figure 1 shows its height, h m, against time, t s, with the heights of the drop and of the first two rebounds labelled. The height of the nth rebound, h_n m, is modelled by h_n = 2.5 × rⁿ, where r is a constant.

(a)

Use the heights in Figure 1 to show that the model with r = 0.64 fits the first two rebounds.

(2)

(b)

Find the height of the 3rd rebound, in metres to 3 significant figures.

(1)

(c)

Find the number of the first rebound whose height is less than 1 cm.

(4)

(d)

A second ball is dropped from a height of 2.0 m. The height of its nth rebound is modelled by 2.0 × 0.75ⁿ m. Show that the nth rebound of the second ball is higher than the nth rebound of the first ball when (75⁄64)ⁿ > 5⁄4.

(3)

(e)

Hence find the smallest value of n for which this happens.

(2)

Show mark scheme

Answer	Mark	Notes
Part (a)
1.6 ÷ 2.5 = 0.64	M1
1.024 ÷ 1.6 = 0.64, the same ratio, so r = 0.64	A1
Part (b)
2.5 × 0.64³ = 0.655 m (0.65536)	B1
Part (c)
2.5 × 0.64ⁿ < 0.01 ⇒ 0.64ⁿ < 0.004	M1
n log 0.64 < log 0.004	M1
log 0.64 is negative, so dividing by it reverses the inequality: n > 12.37…	A1
the 13th rebound	A1
Part (d)
2.0 × 0.75ⁿ > 2.5 × 0.64ⁿ	M1
(0.75 ÷ 0.64)ⁿ > 2.5 ÷ 2.0	M1
0.75 ÷ 0.64 = 75⁄64 and 2.5 ÷ 2.0 = 5⁄4, giving the result	A1
Part (e)
n > log 1.25 ÷ log(75⁄64) = 1.40…	M1
n = 2 (first rebounds: 1.5 m < 1.6 m; second rebounds: 1.125 m > 1.024 m)	A1

Common error: Forgetting to reverse the inequality when dividing by log 0.64. Since 0.64 < 1 its logarithm is negative, so the result is n > 12.37…, not n < 12.37….

Theory: P2.5 Exponentials and logarithms