Edexcel International A Level Maths · P3 · Pure Mathematics 3
Exponential and logarithmic functions
28 free, high-quality past-paper-style questions on Exponential and logarithmic functions 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 · 228 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-5-B-01P3 · A2Easy5 marks
The curve C has equation y = 3e2x − 5.
(a)
Find the coordinates of the point where C crosses the y-axis.
(1)
(b)
Find the exact coordinates of the point where C crosses the x-axis.
(2)
(c)
State the equation of the asymptote to C and the range of values of y.
(2)
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Answer
Mark
Notes
Part (a)
(0, −2), since 3e0 − 5 = −2
B1
Part (b)
e2x = 5/3 and takes ln: 2x = ln(5/3)
M1
(½ ln(5/3), 0)
A1
Part (c)
asymptote y = −5
B1
range y > −5
B1
Model sketch
Figure: C: y = 3e²ˣ − 5 with its asymptote
Common error: Taking ln before isolating: ln(3e2x − 5) is not ln 3 + 2x − ln 5. Make e2x the subject first.
Figure 1 shows the curve with equation y = Ae−x + B, where A and B are constants. The curve crosses the y-axis at C(0, 7) and the line y = 1 is an asymptote to the curve. The line y = 3 meets the curve at Q.
(a)
Use Figure 1 to write down the value of B, and find the value of A.
(2)
(b)
Find the exact x-coordinate of Q.
(2)
(c)
State the range of values of y on the curve for x ≥ 0.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
as x → ∞, e−x → 0, so B = 1
B1
7 = Ae0 + 1, so A = 6
B1
Part (b)
6e−x + 1 = 3 ⇒ e−x = ⅓
M1
−x = ln ⅓, so x = ln 3
A1
Part (c)
the greatest value is at C: y ≤ 7
B1
the curve never reaches its asymptote: y > 1, so 1 < y ≤ 7
B1
Common error: Writing 1 ≤ y ≤ 7. The curve gets ever closer to y = 1 but never reaches it, because e−x is never 0, so the lower end is a strict inequality.
The function f is defined by f(x) = ex + 2, x ∈ ℝ. Figure 1 shows the curves y = f(x) and y = f−1(x), together with the line y = x. The two straight dashed lines are asymptotes. The point A(0, 3) lies on y = f(x), and B is the image of A on y = f−1(x).
(a)
Write down the coordinates of B.
(1)
(b)
Find an expression for f−1(x).
(2)
(c)
State the domain of f−1 and the equation of the asymptote to y = f−1(x).
(2)
(d)
Solve the equation f−1(x) = 1, giving your answer in exact form.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
reflection in y = x swaps the coordinates: B(3, 0)
B1
Part (b)
y = ex + 2 ⇒ ex = y − 2 ⇒ x = ln(y − 2)
M1
f−1(x) = ln(x − 2)
A1
Part (c)
domain of f−1 = range of f: x > 2
B1
asymptote x = 2
B1
Part (d)
x − 2 = e, so x = 2 + e
B1
Exam tip: The domain of f−1 is the range of f. Since ex > 0, f(x) > 2, which is why ln(x − 2) only exists for x > 2.
Figure 1 shows part of the curve with equation y = a ln(x − 1) + b, x > 1, where a and b are constants. The curve passes through the points (2, 1) and (1 + e, 3). It crosses the x-axis at A and meets the line y = 5 at B.
(a)
Find the value of a and the value of b.
(3)
(b)
Write down the equation of the asymptote to the curve.
(1)
(c)
Find the exact x-coordinate of A.
(2)
(d)
Find the exact x-coordinate of B.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
(2, 1): a ln 1 + b = 1, and ln 1 = 0, so b = 1
B1
(1 + e, 3): a ln e + 1 = 3
M1
ln e = 1, so a = 2
A1
Part (b)
x = 1 (where x − 1 = 0)
B1
Part (c)
2 ln(x − 1) + 1 = 0 ⇒ ln(x − 1) = −½
M1
x = 1 + e−½ (or 1 + 1/√e)
A1
Part (d)
ln(x − 1) = 2 ⇒ x = 1 + e2
B1
Exam tip: Choose the points that make the logarithm simple: at x = 2 the bracket is 1 and ln 1 = 0, at x = 1 + e it is e and ln e = 1.
The air pressure, P kPa, at a height h km above sea level is modelled by P = P0e−kh, where P0 and k are positive constants. Figure 1 shows the graph of the model, which passes through the point (5, 55).
(a)
Use Figure 1 to write down the value of P0.
(1)
(b)
Show that k = ⅕ ln(20/11), and find k to 3 significant figures.
(2)
(c)
Find the height at which the model gives a pressure of 30 kPa.
(2)
(d)
Find dP/dh, and hence find the rate at which the pressure is decreasing with height when h = 2.
In an experiment, the pressure P and the volume V of a gas are thought to satisfy P = aVn, where a and n are constants. A graph of ln P against ln V is drawn. The line of best fit passes through the points (1.2, 3.86) and (2.7, 1.61).
Figure: line of best fit for ln P against ln V
(a)
Show that ln P = ln a + n ln V.
(2)
(b)
Find the value of n and the value of a, giving a to 3 significant figures.
(4)
(c)
Use the model to estimate P when V = 20.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes ln of both sides: ln P = ln(aVn)
M1
= ln a + ln Vn = ln a + n ln V, both laws seen
A1
Part (b)
gradient = (1.61 − 3.86)/(2.7 − 1.2)
M1
n = −1.5
A1
ln a = 3.86 + 1.5 × 1.2 = 5.66
M1
a = 287
A1
Part (c)
P = 287 × 20−1.5 = 3.21
B1
Exam tip: For P = aVn both axes are logged. The gradient is n itself; only the intercept needs undoing with e( ).
The mass, m grams, of a radioactive sample t days after it is prepared is modelled by m = kbt, where k and b are constants. A graph of log10m against t is a straight line which crosses the vertical axis at 1.85 and passes through the point (10, 0.65).
Figure: straight-line graph of log₁₀ m against t
(a)
Show that log10m = log10k + t log10b.
(2)
(b)
Find the values of k and b, giving your answers to 3 significant figures.
(3)
(c)
Interpret the value of b in the context of the model, and find the percentage of the mass lost each day.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes log10 of both sides
M1
log10k + log10bt = log10k + t log10b, as required
A1
Part (b)
gradient = (0.65 − 1.85)/10 = −0.12
M1
k = 101.85 = 70.8
A1
b = 10−0.12 = 0.759
A1
Part (c)
each day the mass is multiplied by 0.759 (it decays by a constant factor)
B1
about 24.1% is lost each day
B1
Common error: Undoing a base-10 log with e( ). The graph uses log10, so k = 101.85, not e1.85.
The concentration, C mg per litre, of a drug in a patient's blood t hours after an injection is modelled by C = 15e−0.2t, t ≥ 0.
(a)
State the concentration immediately after the injection.
(1)
(b)
Find the exact time at which the concentration has fallen to 3 mg per litre, and give this time to 3 significant figures.
(3)
(c)
Find dC/dt, and hence find the rate at which the concentration is decreasing when t = 4.
(2)
(d)
Show that dC/dt = −0.2C, and hence find the concentration at the instant when it is decreasing at 0.5 mg per litre per hour.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
15 mg per litre
B1
Part (b)
e−0.2t = 1/5
M1
−0.2t = ln(1/5) ⇒ t = 5 ln 5
A1
= 8.05 hours
A1
Part (c)
dC/dt = −3e−0.2t
B1
at t = 4: −3e−0.8 = −1.35, decreasing at 1.35 mg per litre per hour
B1
Part (d)
−3e−0.2t = −0.2 × 15e−0.2t = −0.2C
B1
−0.2C = −0.5 ⇒ C = 2.5 mg per litre
B1
Model sketch
Figure: concentration falls to 3 mg per litre at t = 5 ln 5 ≈ 8.05
Exam tip: For any model Aekt, the rate of change is k times the current value. That turns a question about a rate into a question about the value itself.
The curve C1 has equation y = 2ex + 1 and the curve C2 has equation y = e2x − 2.
(a)
State the equation of the asymptote of each curve.
(2)
(b)
Find the exact x-coordinate of the point where C2 crosses the x-axis.
(2)
(c)
Show that the x-coordinates of the points where the curves meet satisfy e2x − 2ex − 3 = 0, and hence find the exact coordinates of the point of intersection.
Exam tip: When a substitution u = ex gives a negative root, say explicitly why it is rejected: ex > 0 for all real x. That sentence can be worth a mark.
Light passes through n identical sheets of tinted glass. The intensity of the light that comes out, I units, is measured for n = 0, 1, 2, …, 8. Figure 1 shows log10I plotted against n, with a line of best fit through the points (0, 2.6) and (8, 1.0). It is suggested that I = kbn, where k and b are constants.
(a)
Show that log10I = log10k + n log10b, and explain how Figure 1 supports the suggested model.
(2)
(b)
Use the line of best fit to find the values of k and b, giving your answers to 3 significant figures.
(3)
(c)
Interpret the value of b in the context of the model.
(1)
(d)
Find the least number of sheets needed to reduce the intensity to less than 2% of its value with no sheets.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
log10I = log10k + log10bn = log10k + n log10b
B1
this is linear in n, and the points in Figure 1 lie close to a straight line
B1
Part (b)
gradient = (1.0 − 2.6) ÷ 8 = −0.2 = log10b
M1
k = 102.6 = 398
A1
b = 10−0.2 = 0.631
A1
Part (c)
each extra sheet multiplies the intensity by 0.631: each sheet lets through about 63% of the light that reaches it
B1
Part (d)
bn < 0.02 ⇒ −0.2n < log10 0.02, so n > 8.49…
M1
9 sheets
A1
Common error: Taking the gradient −0.2 as the value of b. The gradient is log10 b, so b = 10−0.2; a negative gradient means b is between 0 and 1, not that b is negative.
The rate of flow, Q litres per minute, of a liquid through a pipe of internal radius r cm is thought to satisfy Q = arn, where a and n are constants. Figure 1 shows ln Q plotted against ln r for seven pipes, with a line of best fit through the points (0.5, 1.2) and (2.0, 7.2).
(a)
Explain why the straight line in Figure 1 supports the suggested relationship.
(2)
(b)
Find the value of n and the value of a, giving a to 3 significant figures.
(3)
(c)
Use the model to find the radius of a pipe for which the rate of flow is 100 litres per minute.
(2)
(d)
State the effect on Q, according to the model, of doubling the radius of the pipe.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
taking ln: ln Q = ln a + n ln r
B1
this has the form Y = c + mX with Y = ln Q, X = ln r, so the graph should be a straight line with gradient n
B1
Part (b)
gradient = (7.2 − 1.2) ÷ (2.0 − 0.5)
M1
n = 4
A1
ln a = 1.2 − 4 × 0.5 = −0.8, so a = e−0.8 = 0.449
A1
Part (c)
r4 = 100 ÷ a = 100e0.8
M1
r = 3.86 cm (3.862…; a = 0.449 or 0.45 also gives 3.86)
A1
Part (d)
Q is multiplied by 24: it becomes 16 times as large
B1
Exam tip: On a ln–ln graph the gradient is the power n itself. Only the intercept needs undoing: a = eintercept, and the intercept is found from a point on the line, not read where the line meets the vertical axis of a cropped graph.
Figure 1 is a sketch, not drawn to scale, of the curves C1 with equation y = 4x and C2 with equation y = 8 × 2x. The curves meet at the point R. The point P on C2 and the point Q on C1 have the same x-coordinate, and the gradient of C2 at P equals the gradient of C1 at Q.
Figure 1 shows the curves C1 with equation y = 2 ln x, x > 0, and C2 with equation y = ln(x + 6), x > −6. C1 crosses the x-axis at A and C2 crosses the x-axis at B. The curves meet at P. The line y = −1 meets C1 at Q and C2 at R.
(a)
Write down the coordinates of A and of B.
(2)
(b)
Show that the x-coordinate of P satisfies x2 − x − 6 = 0, and hence find the exact coordinates of P.
(4)
(c)
Find the exact length of QR.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
A(1, 0)
B1
B(−5, 0)
B1
Part (b)
2 ln x = ln x2, so ln x2 = ln(x + 6)
M1
x2 = x + 6 ⇒ x2 − x − 6 = 0
A1
(x − 3)(x + 2) = 0; x = −2 is rejected as ln x is undefined for x ≤ 0
A1
P(3, 2 ln 3) or (3, ln 9)
A1
Part (c)
Q: x = e−½; R: x + 6 = e−1, so x = e−1 − 6
M1
QR = 6 + e−½ − e−1
A1
Common error: Keeping x = −2 as a second meeting point. It solves the quadratic, but 2 ln x is not defined there, and Figure 1 shows only one crossing.
A new sports club opens. The number of members, N, t months after it opens is modelled by N = A − Be−0.25t, where A and B are positive constants. Figure 1 shows the graph of the model. The line N = 800 is an asymptote.
(a)
Use Figure 1 to find the value of A and the value of B.
(2)
(b)
Find the time at which the model predicts 650 members, giving your answer in the form p ln q and to 3 significant figures.
(2)
(c)
Show that dN/dt = 0.25(800 − N).
(2)
(d)
Hence find the number of members at the moment when the membership is increasing at 40 members per month.
(1)
(e)
Give one limitation of the model.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
as t → ∞, e−0.25t → 0, so N → A: A = 800
B1
200 = 800 − Be0, so B = 600
B1
Part (b)
600e−0.25t = 150 ⇒ e−0.25t = ¼
M1
t = 4 ln 4 (or 8 ln 2) = 5.55 months
A1
Part (c)
dN/dt = 150e−0.25t
B1
600e−0.25t = 800 − N, so 150e−0.25t = 0.25(800 − N), as required
B1
Part (d)
0.25(800 − N) = 40 ⇒ N = 640
B1
Part (e)
e.g. it gives non-integer numbers of members; or it assumes membership never falls (members do leave); or the limit of 800 may change if the club expands
B1
Exam tip: For a model of the form A − Be−kt, the rate of growth is k times the gap to the limit A. That is why the curve in Figure 1 is steepest at the start and flattens as N approaches 800.
Figure 1 shows the curve with equation y = eax + b + c, where a, b and c are constants. The line y = −2 is an asymptote to the curve. The curve passes through A(1.5, −1) and B(2.5, e2 − 2), crosses the x-axis at P and crosses the y-axis at Q.
(a)
Write down the value of c.
(1)
(b)
Find the value of a and the value of b.
(3)
(c)
Find the exact x-coordinate of P.
(2)
(d)
Find the exact y-coordinate of Q.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
c = −2
B1
Part (b)
A: e1.5a + b = 1 ⇒ 1.5a + b = 0
M1
B: e2.5a + b = e2 ⇒ 2.5a + b = 2
M1
a = 2, b = −3
A1
Part (c)
e2x − 3 = 2 ⇒ 2x − 3 = ln 2
M1
x = (3 + ln 2)/2
A1
Part (d)
y = e−3 − 2
B1
Exam tip: Once c is known, move it across: eax + b = y + 2 at each point. Taking ln then gives two linear equations in a and b.
A biologist records the number of fish, P, in a lake t years after the lake is first stocked. She suggests the model P = kbt, where k and b are constants. On a graph of ln P against t, her line of best fit passes through the points (2, 6.14) and (8, 7.22).
Figure: line of best fit for ln P against t; vertical axis starts at 5.5
(a)
Show that ln P = ln k + t ln b, and explain why this supports the use of the graph.
(2)
(b)
Find the values of k and b, giving your answers to 3 significant figures.
(4)
(c)
Writing the model as P = ket ln b, find the rate at which the number of fish is increasing when t = 10.
(3)
(d)
The lake can support at most about 5000 fish. Find the value of t at which the model predicts 5000 fish, and use this to comment on the model, suggesting a refinement.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
ln P = ln k + ln bt = ln k + t ln b
B1
this is linear in t (form Y = c + mX), so ln P against t should be a straight line with gradient ln b and intercept ln k
B1
Part (b)
gradient = (7.22 − 6.14)/(8 − 2) = 0.18
M1
intercept = 6.14 − 2 × 0.18 = 5.78
M1
k = e5.78 = 324
A1
b = e0.18 = 1.20
A1
Part (c)
dP/dt = k ln b et ln b = 0.18 e5.78 + 0.18t
M1
substitutes t = 10: 0.18 e7.58
M1
≈ 353 fish per year (accept 350 to 370 from rounded k and b)
A1
Part (d)
5.78 + 0.18t = ln 5000
M1
t = 15.2 years
A1
after this the model predicts more fish than the lake can support, so it is unreliable for large t; a refined model should level off at about 5000, e.g. P = 5000 − Be−ct
B1
Exam tip: Use the equation of the line, ln P = 5.78 + 0.18t, for later parts. Working from rounded values of k and b can move the final answer noticeably.
A metal component is taken out of an oven and left to cool in a workshop. Its temperature, T °C, t minutes later is modelled by T = 22 + Ae−kt, where A and k are positive constants. When t = 0, T = 85, and when t = 5, T = 60.
(a)
Find the value of A.
(1)
(b)
Show that k = ⅕ ln(63/38), and find k to 3 significant figures.
(3)
(c)
Find the time taken for the temperature of the component to fall to 30 °C.
(3)
(d)
Find the rate at which the temperature is falling when t = 5.
(3)
(e)
State the temperature that the model predicts in the long term, and give one limitation of the model.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
85 = 22 + A ⇒ A = 63
B1
Part (b)
60 = 22 + 63e−5k ⇒ e−5k = 38/63
M1
−5k = ln(38/63) ⇒ k = ⅕ ln(63/38), as required
A1
k = 0.101
A1
Part (c)
63e−kt = 8
M1
t = ln(63/8)/k, using their k
M1
20.4 minutes
A1
Part (d)
dT/dt = −63ke−kt
M1
at t = 5: −63k × 38/63 = −38k
M1
falling at 3.84 °C per minute
A1
Part (e)
22 °C, as e−kt → 0
B1
e.g. it assumes the workshop stays at 22 °C; or the model never reaches 22 °C exactly, whereas the component will in practice reach room temperature
B1
Model sketch
Figure: cooling curve approaching 22 °C
Exam tip: At t = 5 you already know e−5k = 38/63 from part (b). Reusing exact values like this avoids rounding errors.
The curve C has equation y = 4 − 3e−2x and the curve D has equation y = 2 + e−2x.
(a)
For the curve C, find the coordinates of the point where it meets the y-axis, the exact x-coordinate of the point where it meets the x-axis, and the equation of its asymptote.
(4)
(b)
Show that C and D meet at exactly one point, and find the exact coordinates of this point, P.
(3)
(c)
Show that the gradient of C at P is 3, and hence find an equation of the tangent to C at P, in the form y = 3x + c, where c is an exact constant.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
(0, 1)
B1
3e−2x = 4 ⇒ −2x = ln(4/3)
M1
x = ½ ln(3/4) or −½ ln(4/3)
A1
asymptote y = 4
B1
Part (b)
4 − 3e−2x = 2 + e−2x ⇒ 4e−2x = 2
M1
e−2x = ½ has exactly one solution, x = ½ ln 2
A1
P(½ ln 2, 5/2)
A1
Part (c)
dy/dx = 6e−2x
B1
at P: 6 × ½ = 3, as required
B1
y − 5/2 = 3(x − ½ ln 2)
M1
y = 3x + 5/2 − (3/2) ln 2
A1
Model sketch
Figure: C and D meet at one point P(½ ln 2, 5/2)
Exam tip: Once you know e−2x = ½ at P, substitute that value directly into the derivative and the equation of C; there is no need to work with ln 2 inside the exponential.
A car costs £18 000 when new. Two models are proposed for its value, £V, t years after it is bought: Model A: V = 18 000e−0.15t Model B: V = 18 000 × 0.84t
(a)
Find the value of the car after 4 years according to Model A, to the nearest £10.
(1)
(b)
Find, according to Model B, the time at which the value of the car first falls below £9000.
(3)
(c)
Show that Model B can be written as V = 18 000e−ct, stating the value of c to 3 significant figures. Hence state, with a reason, which model predicts the faster rate of depreciation.
(3)
(d)
Find dV/dt for Model B, and hence find the rate at which the value is decreasing when t = 2, according to this model.
(3)
(e)
The car will always be worth at least £1500 as scrap. Explain why neither model takes this into account, and suggest a refined model of the form V = p + qe−0.15t that does.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
18 000e−0.6 = £9880
B1
Part (b)
0.84t = 0.5
M1
t = ln 0.5 / ln 0.84
M1
3.98 years
A1
Part (c)
0.84t = et ln 0.84, so c = −ln 0.84
M1
c = 0.174
A1
Model B, since 0.174 > 0.15, so its value falls by a larger proportion each year
B1
Part (d)
dV/dt = 18 000 × 0.84t ln 0.84
M1
at t = 2: 18 000 × 0.7056 × ln 0.84 = −2214
A1
decreasing at about £2210 per year
A1
Part (e)
both models predict values tending to £0 as t → ∞, so for large t they fall below the scrap value
B1
V = 1500 + 16 500e−0.15t (still £18 000 at t = 0 and tends to £1500)
B1
Model sketch
Figure: Model B (c ≈ 0.174) falls faster than Model A (c = 0.15)
Exam tip: To compare two exponential models, write both with base e. The constant in the exponent is the proportional rate of decay: the larger it is, the faster the value falls.
Figure 1 shows the curves C1 with equation y = e2x and C2 with equation y = 5ex − 4. The curves meet at the points A and B, and C2 crosses the x-axis at D. The vertical line x = k, where k lies between the x-coordinates of A and B, meets C2 at P and C1 at Q.
(a)
Write down the equation of the asymptote to C2, and find the exact x-coordinate of D.
(2)
(b)
Show that the x-coordinates of A and B satisfy e2x − 5ex + 4 = 0, and hence find the exact coordinates of A and B.
(4)
(c)
Write down an expression for the length PQ in terms of k.
(1)
(d)
Find the exact value of k for which PQ is greatest, show that this gives a maximum, and find the greatest length of PQ.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = −4
B1
5ex = 4 ⇒ x = ln(4/5)
B1
Part (b)
e2x = 5ex − 4 ⇒ e2x − 5ex + 4 = 0
B1
with u = ex: (u − 4)(u − 1) = 0
M1
x = 0 or x = ln 4
A1
A(0, 1) and B(ln 4, 16)
A1
Part (c)
C2 is above C1 between A and B: PQ = 5ek − 4 − e2k
B1
Part (d)
d(PQ)/dk = 5ek − 2e2k
M1
= 0 ⇒ ek = 5/2, k = ln(5/2) (ek = 0 is impossible)
A1
second derivative 5ek − 4e2k = 12.5 − 25 < 0, so a maximum
B1
greatest PQ = 25/2 − 4 − 25/4 = 9/4
A1
Exam tip: Treat e2k as (ek)2 throughout. Setting 5ek − 2e2k = 0 then factorises as ek(5 − 2ek) = 0, and ek is never 0.
A biologist measures the length, L cm, and the mass, m grams, of seven fish of one species. The lengths are between 12 cm and 42 cm. To test two possible models, m = kbL and m = aLn, where k, b, a and n are constants, she draws the two graphs in Figure 1: (i) ln m against L and (ii) ln m against ln L. On graph (ii) she draws a line of best fit through the points (2.5, 3.08) and (3.5, 6.08).
(a)
Explain which model is supported by Figure 1.
(2)
(b)
Find the value of n and the value of a, giving a to 3 significant figures.
(4)
(c)
Use the model to estimate the mass of a fish of this species that is 60 cm long, and comment on the reliability of your estimate.
(2)
(d)
Find the length of a fish of mass 500 g, according to the model.
(2)
(e)
Describe what the value of n tells you about how the mass changes when the length of a fish doubles.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
for m = kbL, ln m = ln k + L ln b, so graph (i) would be a straight line, but its points lie on a curve
B1
for m = aLn, ln m = ln a + n ln L; graph (ii) is close to a straight line, so m = aLn is supported
B1
Part (b)
gradient = (6.08 − 3.08) ÷ (3.5 − 2.5)
M1
n = 3
A1
ln a = 3.08 − 3 × 2.5 = −4.42
M1
a = e−4.42 = 0.0120
A1
Part (c)
0.0120 × 603 ≈ 2600 g (allow 2590 to 2600)
B1
60 cm is outside the range of lengths measured (12 to 42 cm), so this is an extrapolation and may be unreliable
B1
Part (d)
L3 = 500 ÷ a, or ln L = (ln 500 + 4.42) ÷ 3
M1
L = 34.6 cm (allow 34.6 to 34.7)
A1
Part (e)
m is proportional to L3, so doubling the length multiplies the mass by 23 = 8
B1
Exam tip: Match the graph to the model before calculating: log only the left-hand side for kbL (ln m against L), log both sides for aLn (ln m against ln L). Whichever graph is straight tells you the model.
A sample contains two radioactive isotopes, X and Y. The masses of X and Y, x grams and y grams, t days after measurements begin are modelled by x = 80e−pt and y = 50e−qt, where p and q are positive constants. Figure 1 shows the graphs of both models. The graph for X passes through (10, 40), the graph for Y passes through (25, 25), and the graphs cross at E.
(a)
Show that p = (ln 2)/10, and find the exact value of q.
(3)
(b)
Find the value of t at E, giving your answer to 3 significant figures.
(3)
(c)
Show that the masses of X and Y are decreasing at the same rate when t = 100/3.
(3)
(d)
Describe what the models predict about the make-up of the sample in the long term.
x/y = 1.6e−(p − q)t → 0 since p > q: both masses tend to 0, but almost all the remaining mass is Y
B1
Common error: Dividing 80e−pt = 50e−qt and writing e−pt + qt = 1.6. Collect the exponentials on the side that keeps the power positive: e(p − q)t = 80/50.
A weed is spreading across the surface of a pond of area 500 m2. The area covered, A m2, t weeks after it is first measured is recorded every 2 weeks. A first model is Model 1: A = 4e0.3t. Figure 1 shows the recorded areas as crosses, together with the curve of Model 1. The recorded area when t = 12 is 114 m2.
(a)
State the area covered when t = 0 according to Model 1, and find the time at which Model 1 predicts that the whole pond is covered.
(3)
(b)
Show that, for Model 1, dA/dt = 0.3A, and hence find the rate at which the area is increasing when A = 100.
(2)
(c)
Use Figure 1 to describe how the recorded areas compare with Model 1, and suggest a reason for this.
(1)
(d)
A refined model is Model 2: A = 500 ÷ (1 + 124e−0.3t). Show that Model 2 gives the same area as Model 1 when t = 0, and state the area that Model 2 predicts in the long term.
(2)
(e)
Find the time at which Model 2 predicts that 400 m2 of the pond is covered.
(2)
(f)
Compare the predictions of the two models when t = 12 with the recorded area, and state which model fits better.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
4 m2
B1
4e0.3t = 500 ⇒ t = ln 125 ÷ 0.3
M1
t = 16.1 weeks
A1
Part (b)
dA/dt = 1.2e0.3t = 0.3 × 4e0.3t = 0.3A
B1
30 m2 per week
B1
Part (c)
from about t = 10 the recorded areas are increasingly below Model 1, because growth slows as the weed runs out of space (or light or nutrients) on the pond
B1
Part (d)
t = 0: 500 ÷ (1 + 124) = 4 m2, the same as Model 1
B1
e−0.3t → 0, so A → 500 m2, the area of the pond
B1
Part (e)
1 + 124e−0.3t = 1.25 ⇒ e−0.3t = 1/496
M1
t = ln 496 ÷ 0.3 = 20.7 weeks
A1
Part (f)
Model 1: 4e3.6 ≈ 146 m2; Model 2: 500 ÷ (1 + 124e−3.6) ≈ 114 m2; the recorded area is 114 m2, so Model 2 fits better
B1
Exam tip: When asked to refine an exponential growth model, check two things: the refined model must agree with the starting value, and its long-term value must respect the physical limit (here the 500 m2 of the pond).
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