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
1IALM-P2-5-B-01P2 · ASEasy6 marks
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)
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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.
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 9½ = 3
B1
(iii) −2, since 8−2 = 1⁄64
B1
Part (d)
16y = 8 ⇒ 24y = 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.
Figure 1 shows part of the curve with equation y = k × ax, 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 × 3x = 100 ⇒ 3x = 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.
Figure 1 shows four curves, labelled A, B, C and D. Their equations, in no particular order, are (i) y = 2x (ii) y = 4x (iii) y = (⅓)x (iv) y = 2x + 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, 4x > 2x, 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)
4x = 2x + 2; with u = 2x: u2 − u − 2 = 0
M1
(u − 2)(u + 1) = 0; u = −1 is impossible because 2x > 0
A1
2x = 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.
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 = 100.72 = 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.
Figure 1 shows the curve with equation y = ax + 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)
ax → 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)
3x = 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.
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 × 1.15t, 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)
1.15t = 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.
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.
Find the value of log7 50, giving your answer to 3 significant figures.
(2)
(d)
Solve logx 8 = 1.5, where x > 0 and x ≠ 1.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
log4x = log2x ÷ log2 4
M1
log2 4 = 2, so log4x = ½ log2x
A1
Part (b)
log2x − ½ log2x = 3
M1
log2x = 6
A1
x = 64
A1
Part (c)
log 50 ÷ log 7
M1
= 2.01 (2.0103…)
A1
Part (d)
x1.5 = 8
M1
x = 82/3 = 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.
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 × rt, 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).
In Figure 1, the logarithmic curve y = loga(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)
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Answer
Mark
Notes
Part (a)
loga(x + b) is defined only for x + b > 0, so the asymptote is x = −b; −b = −3 gives b = 3
B1
Part (b)
loga 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.
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)
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 = 108.5 ÷ 106 = 102.5
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 ≤ 101.3 = 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.
Figure 1 shows the curves y = log2x and y = log8x, 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 log8x = ⅓ log2x 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)
log8x = log2x ÷ 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 = log2k − ⅓ log2k = ⅔ log2k
M1
⅔ log2k = 4 ⇒ log2k = 6
A1
k = 64
A1
Part (d)
T: log2x = 1 ⇒ x = 2; S: log8x = 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.
Figure 1 shows the curves y = ax and y = a−x, where a is a constant, a > 1. The curves cross at R. The line y = 5 meets y = ax 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 = ax 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, ax = 5 ⇒ x = loga 5
M1
by symmetry Q has x = −loga 5, so PQ = 2 loga 5 = 4
M1
loga 5 = 2 ⇒ a2 = 5, and a > 1 so a = √5
A1
Part (c)
2 log√5c = 6 ⇒ log√5c = 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.
The curves C1 and C2 have equations C1: y = 32x and C2: y = 4 × 3x + 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 = 3x.
(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)
32x = (3x)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
3x > 0, so u = −5 is impossible: exactly one point
B1
3x = 9 ⇒ x = 2, y = 81: (2, 81)
A1
Part (d)
4 × 3x = 55 ⇒ 3x = 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.
This question involves two pairs of simultaneous equations.
(a)
Given that 2a × 8b = 64 and 9a ÷ 3b = 243, find the values of a and b.
(4)
(b)
The positive numbers x and y satisfy log3x + log3y = 3 and (log3x)(log3y) = 2. By letting p = log3x and q = log3y, 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)
2a + 3b = 26 ⇒ a + 3b = 6
M1
32a − 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 = 3p, y = 3q
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.
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 × 0.9t, and the mass of B, in grams, by 30 × 0.97t, 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)
0.9t = 0.5
M1
t = log 0.5 ÷ log 0.9
M1
= 6.58 days (6.5788…)
A1
Part (c)
80 × 0.9t = 30 × 0.97t ⇒ (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 × 0.913.09… = 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.
Figure 1 shows the curve with equation y = 4x − 10 × 2x + 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 = 2x, 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 → −∞, 4x → 0 and 2x → 0, so y → 16
B1
Part (c)
4x = (2x)2 = u2
M1
u2 − 10u + 16 = 0
A1
(u − 2)(u − 8) = 0
M1
2x = 2 ⇒ x = 1; 2x = 8 ⇒ x = 3
A1
Part (d)
u2 − 10u + 16 = 16 ⇒ u(u − 10) = 0
M1
u = 0 is impossible since 2x > 0, so 2x = 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.
Figure 1 shows the curves C1 and C2 with equations C1: y = 2 log3x, 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 log3x = log3x2 (power law)
M1
log3x2 − 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 log3x 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.
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.
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, hn m, is modelled by hn = 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….
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