These twelve questions are original and run from easy to harder. They cover converting between index and log form, the log laws, equations with a domain restriction, straight-line graphs from logs and growth models. Attempt each on paper first, then open the worked answer.
Use a calculator only where a decimal answer is requested. Check the notation and calculator rules for your exam year on the Cambridge syllabus page. Start from the module overview if you need the study order.
Warm-up: converting and evaluating
1. Write 2⁶ = 64 as a logarithmic statement, and write log₇ 49 = 2 as an index statement.
Show answer
Base 2, index 6, number 64: log₂ 64 = 6.
Base 7, log value 2, number 49: 7² = 49.
2. Evaluate (a) log₅ 125, (b) log₄ 2, (c) log₃ (1/27).
Show answer
(a) 5³ = 125, so 3.
(b) 41/2 = 2, so 1/2.
(c) 3⁻³ = 1/27, so −3.
3. Solve logx 125 = 3/2.
Show answer
x3/2 = 125. Raise both sides to the power 2/3: x = 1252/3 = (∛125)² = 5² = 25.
Check: 253/2 = (√25)³ = 125. So x = 25.
4. Solve 3ˣ = 50, giving x to 3 significant figures.
Show answer
x = log₃ 50 = ln 50 ÷ ln 3 = 3.912 ÷ 1.0986 = 3.561.
Check: 3³ = 27 and 3⁴ = 81, so x lies between 3 and 4. So x = 3.56.
Building: laws and equations
5. Show that 3 lg 2 + lg 25 − lg 2 = 2.
Show answer
3 lg 2 = lg 8. So the expression is lg 8 + lg 25 − lg 2 = lg (8 × 25 ÷ 2) = lg 100.
Since 10² = 100, lg 100 = 2. ✓
6. Solve log₂ x + log₂ (x − 3) = 2.
Show answer
Domain: x > 0 and x − 3 > 0, so x > 3. Combine: x(x − 3) = 2² = 4, so x² − 3x − 4 = 0, which gives (x − 4)(x + 1) = 0.
x = −1 is rejected because it is not greater than 3. Check x = 4: log₂ 4 + log₂ 1 = 2 + 0 = 2. So x = 4.
7. Solve 22x − 5 × 2ˣ + 4 = 0.
Show answer
Let u = 2ˣ. Then 22x = u², so u² − 5u + 4 = 0, which gives (u − 1)(u − 4) = 0. So u = 1 or u = 4.
2ˣ = 1 gives x = 0, and 2ˣ = 4 gives x = 2. Check x = 2: 16 − 20 + 4 = 0. ✓ So x = 0 or x = 2. Both are valid because 2ˣ is always positive.
8. Solve lg (13x − 4) = 1 + lg x.
Show answer
Domain: x > 0 and 13x − 4 > 0, so x > 4/13. Write 1 = lg 10, so the right side is lg 10 + lg x = lg 10x. Equate: 13x − 4 = 10x, so 3x = 4 and x = 4/3.
Since 4/3 > 4/13, it is valid. Check: 13 × 4/3 − 4 = 40/3 and 10 × 4/3 = 40/3. So x = 4/3.
Stretch: graphs and models
9. The variables satisfy y = abˣ. A graph of lg y against x is a straight line through (0, 0.5) and (4, 2.1). Find a and b to 3 significant figures.
Show answer
Intercept = 0.5, so lg a = 0.5 and a = 100.5 = 3.16. Gradient = (2.1 − 0.5) ÷ 4 = 0.4, so lg b = 0.4 and b = 100.4 = 2.51.
So y = 3.16 × 2.51ˣ. Check at x = 4: lg y = 0.5 + 1.6 = 2.1. ✓
10. A graph of lg y against lg x is a straight line with gradient 2 passing through (1, 3). Find the relationship between y and x.
Show answer
Let y = axⁿ, so lg y = n lg x + lg a. The gradient gives n = 2. At (1, 3): 3 = 2 × 1 + lg a, so lg a = 1 and a = 10.
So y = 10x². Check: x = 10 gives lg x = 1 and y = 1000, with lg y = 3. ✓
11. A population is modelled by N = 800e0.03t, where t is in years. Find (a) the percentage growth per year to 3 significant figures, (b) the population after 20 years to the nearest whole number.
Show answer
(a) The yearly factor is e0.03 = 1.0305, so the growth is 3.05% per year, not 3%.
(b) N = 800e0.6 = 800 × 1.8221 = 1457.7. To the nearest whole number, 1458.
12. A culture of 2000 bacteria grows by 15% each hour. Find, to 3 significant figures, the time for the number to reach 10 000.
Show answer
N = 2000 × 1.15t. Set this equal to 10 000: 1.15t = 5. Take logs: t = ln 5 ÷ ln 1.15 = 1.6094 ÷ 0.13976 = 11.52.
Check: 1.1511.5 ≈ 5, since e0.13976 × 11.5 = e1.607 ≈ 4.99. So 11.5 hours.
If you got these wrong
Match each slip to the lesson that repairs it.
| What went wrong | Go back to |
|---|---|
| Swapped the base and the answer, or could not start a conversion (questions 1 to 4) | Convert between exponential and logarithmic statements |
| Doubled instead of squaring, or split log (a + b) (questions 5 and 8) | Combine logarithms with correct coefficients |
| Gave an extra root that the domain rules out (questions 6 and 8) | Solve an equation with a log-domain restriction |
| Copied the gradient and intercept as the constants (questions 9 and 10) | Linearise an exponential relationship |
| Read k as a percentage, or forgot the growth factor (questions 11 and 12) | Distinguish a growth constant from percentage growth |
Log each error in the mistake log and retest queue and try a fresh question on the same skill a few days later. The non-calculator working trainer and the quadratic structure explorer are useful for checking exact values and quadratic roots after you have tried by hand.
If the same slip returns, a teacher in online one-to-one Additional Mathematics tuition can go through your written solutions and find where the habit starts.