Exponentials and logarithms practice problems A level Maths, Topic 4
The laws of logarithms, solving exponential equations, e and natural logarithms, and exponential growth and decay models. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for exponentials and logarithms
| log a + log b = log(ab) | product law |
|---|---|
| log an = n log a | power law |
| ax = b ⇔ x = loga b | definition |
| y = Aekt | exponential growth (k > 0) or decay (k < 0) |
A formula booklet is provided in the exam. Answers within 2.5% are marked correct, and you can type fractions such as 1/6. What the booklet gives you, and what it leaves out, is in our A level Maths formula booklet guide.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
2 to what power gives 32?
Worked solution
- 25 = 32.
- log2 32 = 5.
Hint
Take logs of both sides.
Worked solution
- x = ln 20/ln 3.
- x = 2.73.
Hint
The product law.
Worked solution
- Adding logs multiplies the arguments: log a + log b = log(ab).
- The answer is B.
Exam-style questions
Questions at the level of the A level Maths exam. The ones marked Challenge combine more than one idea.
Hint
Undo ln with e.
Worked solution
- x = e2.
- x = 7.39.
Hint
log3 9 = 2.
Worked solution
- log3 x = 4 − 2 = 2.
- x = 32 = 9.
Hint
Substitute t = 5.
Worked solution
- N = 500e−1 = 500 × 0.3679.
- N = 184.
Hint
Set e−0.2t = ½.
Worked solution
- −0.2t = ln ½, so t = ln 2/0.2.
- t = 3.47.
Hint
Let y = 2x, so 22x = y2.
Worked solution
- y2 − 6y + 8 = 0 gives y = 2 or y = 4.
- 2x = 2 gives x = 1; 2x = 4 gives x = 2.
- Larger solution: 2.
Exam question practice
One original exam-style question in the style of an A level paper. Answer it in full, with working, before you open the mark scheme: M marks are for method, A marks for accurate answers that follow from it, and B marks for correct statements on their own.
Exam-style question style · 8 points · about 8 minutes
The value of a machine, £V, t years after it is bought is modelled by V = 20 000e−kt. After 2 years it is worth £15 000.
- (a) Show that k = ½ ln(4/3).
- (b) Find the value after 5 years, to the nearest pound.
- (c) Find when the value first falls below £5000, to 1 decimal place.
- (d) Give one limitation of this model.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
15 000 = 20 000e−2k, so e−2k = 3/4 and −2k = ln(3/4), giving k = ½ ln(4/3).
Part (b)
k = 0.1438. V = 20 000e−0.719 = £9743.
Part (c)
e−kt < 1/4, so t > ln 4/k = 9.6 years.
Part (d)
It predicts the value never reaches zero, or it ignores factors such as condition and demand.
Your mark: 0 of 8
Common mistakes in exponentials and logarithms
- Writing log(a + b) = log a + log b, which is false.
- Forgetting that 22x = (2x)2 in hidden quadratics.
- Rounding k too early in a model, which throws later answers off.
- Using log base 10 where the model uses e.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: A level Maths past papers, with official scoring guidelines. · All A level Maths units