AQA A-Level Mathematics: F: Exponentials and logarithms (7357)

This guide teaches Section F: Exponentials and logarithms (content references F1 to F7) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Section F is listed under Paper 1, and Papers 2 and 3 can each assess any Paper 1 content, so these skills can appear on all three papers. A calculator is required in every 7357 paper, but many questions here ask for exact answers, so you still need to work with ln, e and log notation by hand.

Condensed recall is in the revision notes, and original questions with worked answers are in the practice set. Course hub: AQA A-Level Mathematics. Printable checklist: AQA A-Level Mathematics checklist. To find your weak spots first, try a free 10-minute diagnostic.

What this section covers

Ref	What you must be able to do
F1	Know and use the function aˣ (a positive) and its graph; know and use eˣ and its graph
F2	Know that the gradient of eᵏˣ is keᵏˣ, and why this makes exponential models suitable
F3	Know log_a x as the inverse of aˣ; know ln x, its graph, and ln x as the inverse of eˣ
F4	Use the laws of logarithms, including k log_a x ≡ log_a xᵏ for k = −1 and k = −½
F5	Solve equations of the form aˣ = b
F6	Use logarithmic graphs to estimate a and n in y = axⁿ, and k and b in y = kbˣ
F7	Use exponential growth and decay in modelling, and judge limitations and refinements

Appendix B of the specification lists the laws of logarithms, and the link x = aⁿ ⇔ n = log_a x (for a > 0 and x > 0), among the formulae you must be able to use without them being provided. Learn them.

F1: The graphs of aˣ and eˣ

For any positive a, the graph of y = aˣ:

• passes through (0, 1), because a⁰ = 1
• lies entirely above the x-axis: aˣ > 0 for every x
• has the x-axis (y = 0) as an asymptote
• is increasing when a > 1 and decreasing when 0 < a < 1 (since (1/2)ˣ = 2⁻ˣ, the graph of (1/a)ˣ is the reflection of aˣ in the y-axis)

The number e ≈ 2.718 gives the exponential function y = eˣ. Its graph has the same shape as 2ˣ or 3ˣ, and sits between them.

Worked example 1. Describe the graph of y = 3 + 2e⁻ˣ.

y-intercept:  x = 0  ->  y = 3 + 2e⁰ = 3 + 2 = 5        so (0, 5)
As x -> ∞,    e⁻ˣ -> 0   ->  y -> 3                     asymptote y = 3
e⁻ˣ > 0 for all x, so y > 3: no x-intercept
e⁻ˣ decreases as x increases, so the curve is decreasing

This is y = e⁻ˣ stretched by scale factor 2 parallel to the y-axis, then translated 3 up. The range is y > 3.

F2: The gradient of eᵏˣ

The defining property of e is that the gradient of eᵏˣ is keᵏˣ:

y = eᵏˣ   ->   dy/dx = k eᵏˣ = k y

So the rate of change of the quantity is proportional to the quantity itself. That is why exponential models suit so many situations: a population that grows in proportion to its size, a radioactive sample that decays in proportion to the amount left, money earning interest on its current value. If k > 0 you get growth; if k < 0, decay. (Differentiating eᵏˣ, aᵏˣ and ln x in general is Section G, covered in the differentiation study guide.)

Worked example 2. A culture has N = 200e^(0.3t) cells after t hours. Find the rate of growth when t = 5 and show that it is 0.3N.

dN/dt = 200 × 0.3 e^(0.3t) = 60 e^(0.3t)
t = 5:  dN/dt = 60 e^1.5 = 268.9...  ≈ 269 cells per hour
N(5) = 200 e^1.5 = 896.3...,   0.3 × 896.3... = 268.9...   ✓

F3: Logarithms as inverses

Definition: for a > 0 and x > 0,

y = log_a x   ⇔   aʸ = x

The logarithm answers “what power of a gives x?” For example, log₂ 32 = 5 because 2⁵ = 32. Two results follow at once: log_a 1 = 0 and log_a a = 1.

ln x means log_e x, the natural logarithm. Because ln x and eˣ are inverse functions:

e^(ln x) = x  (x > 0)        ln(eˣ) = x  (all x)

The graph of y = ln x is the reflection of y = eˣ in the line y = x. It:

• passes through (1, 0)
• has the y-axis (x = 0) as an asymptote
• exists only for x > 0, and is increasing, but ever more slowly

Worked example 3. Solve (a) e^(2x − 1) = 7 and (b) ln(3x) = 2. Give exact answers, then values to 3 significant figures.

(a) take ln of both sides:  2x − 1 = ln 7
    x = (1 + ln 7)/2 = 1.47 (3 s.f.)

(b) apply e to both sides:  3x = e²
    x = e²/3 = 2.46 (3 s.f.)

F4: Laws of logarithms

For a > 0 and x, y > 0:

Law	Statement
Multiplication	log_a x + log_a y ≡ log_a xy
Division	log_a x − log_a y ≡ log_a (x/y)
Power	k log_a x ≡ log_a xᵏ

The specification names two power cases you must handle:

• k = −1: log_a (1/x) = −log_a x
• k = −½: log_a (1/√x) = −½ log_a x

Worked example 4. (a) Write 2 log₃ 6 − log₃ 4 as an integer. (b) Solve log₂(x + 2) + log₂ x = 3.

(a) 2 log₃ 6 = log₃ 36
    log₃ 36 − log₃ 4 = log₃ (36/4) = log₃ 9 = 2

(b) log₂ [x(x + 2)] = 3
    x(x + 2) = 2³ = 8
    x² + 2x − 8 = 0  ->  (x + 4)(x − 2) = 0  ->  x = 2 or x = −4
    log₂ x needs x > 0, so reject x = −4.   x = 2

Always check solutions against the original equation: a logarithm of a negative number or zero is undefined.

F5: Solving aˣ = b

Take logs (base 10 or base e) of both sides, then use the power law:

aˣ = b   ->   x ln a = ln b   ->   x = ln b / ln a

Worked example 5. (a) Solve 5ˣ = 40 to 3 s.f. (b) Solve 2²ˣ − 6 × 2ˣ + 8 = 0.

(a) x = ln 40 / ln 5 = 2.29 (3 s.f.)

(b) 2²ˣ = (2ˣ)², so let u = 2ˣ:
    u² − 6u + 8 = 0  ->  (u − 2)(u − 4) = 0  ->  u = 2 or u = 4
    2ˣ = 2  ->  x = 1          2ˣ = 4  ->  x = 2

In (b), if one root for u had been negative, you would reject it: 2ˣ is never negative. This “hidden quadratic” pattern also appears with e²ˣ and eˣ.

F6: Using logarithmic graphs

Taking logs turns two non-linear relationships into straight lines. You can use log base 10 or ln, as long as you are consistent.

Model	Take logs	Plot	Gradient	Intercept
y = axⁿ	log y = log a + n log x	log y against log x	n	log a
y = kbˣ	log y = log k + x log b	log y against x	log b	log k

Spot which one by the horizontal axis: log x for a power law, plain x for an exponential.

Worked example 6. (a) Data for x and y give a line of best fit for ln y against ln x passing through (0.5, 1.85) and (2.5, 4.85). Estimate a and n in y = axⁿ. (b) Data give a line of best fit for log₁₀ y against x with intercept 0.70 and passing through (10, 2.10). Estimate k and b in y = kbˣ.

(a) gradient n = (4.85 − 1.85)/(2.5 − 0.5) = 3/2 = 1.5
    ln a = 1.85 − 1.5 × 0.5 = 1.10  ->  a = e^1.10 = 3.00 (3 s.f.)
    y ≈ 3.00 x^1.5

(b) gradient log₁₀ b = (2.10 − 0.70)/10 = 0.14  ->  b = 10^0.14 = 1.38 (3 s.f.)
    log₁₀ k = 0.70  ->  k = 10^0.70 = 5.01 (3 s.f.)
    y ≈ 5.01 × 1.38ˣ

Because the values come from a line of best fit, these are estimates.

F7: Exponential growth and decay in modelling

A general model is y = Aeᵏᵗ: A is the value at t = 0, k > 0 means growth and k < 0 means decay. Contexts named in the specification include continuous compound interest, radioactive decay, drug concentration decay and population growth.

• Continuous compound interest: £P invested at rate r per year compounded continuously grows to Pe^(rt) after t years.
• Half-life: for y = Ae^(−kt), the time to halve satisfies e^(−kt) = ½, so t = ln 2 / k.

Worked example 7. (a) £2000 is invested at 4% per year compounded continuously. Find its value after 5 years and the time for it to double. (b) A drug’s concentration is C = 15e^(−0.2t) mg/L, t hours after a dose. Find when it first falls to 2 mg/L.

(a) V = 2000 e^(0.04 × 5) = 2000 e^0.2 = £2442.81
    double: e^(0.04t) = 2  ->  t = ln 2 / 0.04 = 17.3 years (3 s.f.)

(b) 15 e^(−0.2t) = 2  ->  e^(−0.2t) = 2/15
    −0.2t = ln(2/15)  ->  t = 5 ln 7.5 = 10.1 hours (3 s.f.)

Limitations and refinements

You must be able to criticise a model, not just use it.

• Unlimited growth: y = Aeᵏᵗ with k > 0 grows without bound. Real populations are limited by food, space and disease, so long-term predictions overestimate. A refinement is a model whose growth slows and levels off at a maximum value.
• Constant rate assumed: interest rates, birth rates and how fast the body clears a drug may change over time.
• Continuous values: a model can give 896.3 cells; real counts are whole numbers.
• Extrapolation: predictions far outside the data range are unreliable.
• Decay never reaches zero: Ae^(−kt) > 0 for every t, but in reality the quantity will eventually be negligible or gone. A refinement might add a constant, as in C = Ae^(−kt) + c, if the quantity settles at a non-zero level.

Common errors

• Writing log(x + y) = log x + log y. There is no law for the log of a sum.
• Writing ln x / ln y = ln(x/y). The division law is about a difference of logs, not a quotient of logs.
• Forgetting that ln x, and log_a x, need x > 0, so keeping an invalid root.
• In the hidden quadratic, forgetting to go back from u to x, or keeping 2ˣ = −3.
• Mixing up the two log graphs: plotting log y against x for a power law.
• Taking the intercept of a log graph as a or k, instead of log a or log k.
• Rounding ln values early. Keep full calculator values until the final answer.

Next steps

Work through the practice questions, then use the revision notes for final review. The overarching themes guide covers modelling language (assumptions, refining a model) that F7 relies on, and the exam preparation guide covers the three papers. Check gaps with a free diagnostic.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018 (A-level exams June 2018 onwards), published by AQA. Section 3.7 F: Exponentials and logarithms, content F1 to F7.