OxfordAQA International AS and A-level Mathematics: Exponentials, logarithms and solving a^x = b (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section PP1.3 (Exponential and logarithms) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International AS, Unit PSM1 (3.2, PP1), for International AS exams from May/June 2018 and International A-level exams from May/June 2019. Give decimal answers to 3 significant figures unless a question says otherwise, and do not round until the final line.

Related pages: the study guide, the revision notes, the OxfordAQA A-level Mathematics course page and the printable checklist.

Questions

1. The curve C has equation y = 5^x.

(a) State the coordinates of the point where C crosses the y-axis. [1] (b) Write down the equation of the asymptote of C. [1] (c) Find the value of y when x = −2. [1]

2. (a) Write 2⁻³ = 1/8 in the form log_a b = c. [1] (b) Find the value of x for which log_x 81 = 4. [2]

3. Evaluate, showing your working:

(a) 2 log₅15 − log₅9 [3] (b) log₉27 [2]

4. Suppose log_a 2 = p and log_a 3 = q. Write each of the following in terms of p and q:

(a) log_a 12 [2] (b) log_a (9/8) [2]

5. Solve 16^x = 8^(x−1). [4]

6. Solve 6^x = 100, giving your answer to 3 significant figures. [3]

7. Solve 3^x = 4^(x−1), giving your answer to 3 significant figures. [5]

8. Solve log₂(x − 1) + log₂(x + 3) = 5. [5]

9. A sum of £5000 is invested at 6% compound interest per year, so after n years it is worth 5000 × 1.06^n pounds.

(a) Show that the investment is worth at least £8000 when 1.06^n ≥ 1.6. [1] (b) Find the least whole number of years for this to happen. [3] (c) Verify your answer to part (b) by evaluating the investment after n − 1 years and after n years. [2]

10. The mass of a sample, in grams, is modelled by M = 500 × 0.92^t, where t is the time in days.

(a) Find the values of t for which M < 120, giving the boundary value to 3 significant figures. [4] (b) Find the least whole number of days after which the mass is below 120 grams, and justify your answer with two values of M. [2]

11. The curve y = a^x, where a > 0, passes through the point (−2, 1/49).

(a) Find the value of a. [2] (b) Solve 7^(2x − 1) = 30, giving your answer to 3 significant figures. [4] (c) Describe the single transformation that maps the curve y = 7^x onto the curve y = 7^(−x). [1] (d) Solve 7^(−x) = 0.4, giving your answer to 3 significant figures. [2]

12. (a) Show that 2 log₅x − log₅(x − 4) can be written as log₅[x² / (x − 4)]. [2] (b) Hence solve 2 log₅x − log₅(x − 4) = 2. [6]

Answers

1. (a) a⁰ = 1 for any base, so C crosses at (0, 1) [1]. (b) The curve approaches but never reaches the x-axis: y = 0 [1]. (c) 5⁻² = 1/5² = 1/25 (or 0.04) [1]. [3] Examiner insight: Give the intercept as a coordinate pair, not just “1”, and write the asymptote as an equation. Answers to short recall parts are easy to lose by giving only half of what is asked.

2. (a) The base 2 stays the base, and the index −3 becomes the logarithm: log₂(1/8) = −3 [1]. (b) x⁴ = 81 [1], so x = 3, since a base must be positive [1]. [3] Examiner insight: In part (b), stating that x = −3 is rejected because a logarithm base must be positive shows the reasoning and protects you if you wrote both values.

3. (a) 2 log₅15 = log₅225 [1]. Then log₅225 − log₅9 = log₅25 [1]. Since 5² = 25, the value is 2 [1]. (b) Let log₉27 = k, so 9^k = 27, which is 3^(2k) = 3³ [1]. So 2k = 3 and k = 3/2 [1]. [5] Examiner insight: “Showing your working” means the law you use should be visible. A bare answer with no log step gives the reader nothing to follow if it is wrong.

4. (a) 12 = 2² × 3, so log_a 12 = 2 log_a 2 + log_a 3 [1] = 2p + q [1]. (b) log_a(9/8) = log_a 9 − log_a 8 = 2 log_a 3 − 3 log_a 2 [1] = 2q − 3p [1]. [4] Examiner insight: Factorise into powers of 2 and 3 first, then apply the laws. Turning log_a 12 into log_a 4 + log_a 3 and stopping is an incomplete answer, because log_a 4 is not in terms of p.

5. 16 = 2⁴ and 8 = 2³, so 2^(4x) = 2^(3(x−1)) [2]. Equating indices: 4x = 3x − 3 [1]. So x = −3 [1]. [4] Examiner insight: Write each side as a power of the same base before equating indices. Check by substitution: 16⁻³ and 8⁻⁴ are both 1/4096.

6. Take logarithms of both sides: x log 6 = log 100 [1]. So x = log 100 / log 6 [1] = 2.57 [1]. [3] Examiner insight: log 100 is 2 if you use base 10, but any base gives the same quotient. Quote the exact fraction before the decimal, and give the decimal to 3 significant figures as the question asks.

7. Take logarithms: x ln 3 = (x − 1) ln 4 [1]. Expand: x ln 3 = x ln 4 − ln 4 [1]. Collect: ln 4 = x(ln 4 − ln 3) [1]. So x = ln 4 / (ln 4 − ln 3) [1] = 4.82 [1]. [5] Examiner insight: Unknowns on both sides need collecting and factorising before you divide. Do not divide log 4 by log 3 and treat the exponent x − 1 as if it were x.

8. Combine: log₂[(x − 1)(x + 3)] = 5 [1]. Convert: (x − 1)(x + 3) = 2⁵ = 32 [1]. Expand: x² + 2x − 35 = 0 [1]. Factorise: (x − 5)(x + 7) = 0 [1]. For x = −7 the term log₂(x − 1) is not defined, so reject it, leaving x = 5 [1]. [5] Examiner insight: A quadratic gives two roots, but they may not both be solutions. A log equation needs a final check that every argument is positive, and the rejection should be stated.

9. (a) 5000 × 1.06^n ≥ 8000 gives 1.06^n ≥ 8000/5000 = 1.6 [1]. (b) n ln 1.06 ≥ ln 1.6 [1]. Since ln 1.06 > 0, the inequality keeps its direction: n ≥ ln 1.6 / ln 1.06 = 8.07 [1]. The least whole number is n = 9 [1]. (c) After 8 years: 5000 × 1.06⁸ = £7969.24, which is below £8000 [1]. After 9 years: 5000 × 1.06⁹ = £8447.39, which is above £8000 [1]. [6] Examiner insight: Round the boundary 8.07 up to 9, not down to 8. For a “least whole number” question, reading the context carefully is what earns the final credit.

10. (a) 500 × 0.92^t < 120, so 0.92^t < 0.24 [1]. Take logarithms: t ln 0.92 < ln 0.24 [1]. ln 0.92 is negative, so dividing reverses the inequality: t > ln 0.24 / ln 0.92 [1], that is t > 17.1 [1]. (b) The least whole number is 18 days [1]. Check: M(17) = 121.16, which is above 120, and M(18) = 111.47, which is below 120 [1]. [6] Examiner insight: The sign reversal in part (a) is the usual place to lose credit. If you divide by ln 0.92 and keep the original sign, you get t < 17.1, which contradicts the fact that the mass is falling.

11. (a) a⁻² = 1/49, so a² = 49 [1]. Since a > 0, a = 7 [1]. (b) (2x − 1) ln 7 = ln 30 [1]. So 2x − 1 = ln 30 / ln 7 = 1.748 [1]. Then 2x = 2.748 [1] and x = 1.37 [1]. (c) A reflection in the y-axis [1]. (d) −x ln 7 = ln 0.4 [1], so −x = −0.471 and x = 0.471 [1]. [9] Examiner insight: A transformation answer needs the type and the line: “reflection” alone leaves out half of what is asked, and a wrong line such as the x-axis changes the curve. In part (d), taking logarithms of 0.4 gives a negative value, so watch the sign when you change −x to x.

12. (a) 2 log₅x = log₅(x²) [1]. Then log₅(x²) − log₅(x − 4) = log₅[x² / (x − 4)] [1]. (b) log₅[x² / (x − 4)] = 2, so x² / (x − 4) = 5² = 25 [1]. Then x² = 25(x − 4) = 25x − 100 [1]. Rearrange: x² − 25x + 100 = 0 [1]. Factorise: (x − 5)(x − 20) = 0 [1], so x = 5 or x = 20 [1]. Both are greater than 4, so every logarithm is defined and both are valid. Check x = 20: 2 log₅20 − log₅16 = log₅25 = 2 [1]. [8] Examiner insight: On a “show that” part, state which law you use at each line. Do not skip the middle step 2 log₅x = log₅(x²), because it is the whole of the demonstration.

Where marks are usually lost

• Adding or subtracting logarithms with different bases without converting first.
• Splitting log(x + y) into log x + log y, or log(x − y) into log x − log y.
• Dividing by a negative logarithm in an inequality and keeping the original inequality sign.
• Rounding intermediate values, such as 1.748 or 8.07, and then giving a final answer that differs in the third figure.
• Failing to check log equations, so that a root making an argument negative stays in the answer.
• Rounding “least whole number” answers to the nearest whole number instead of rounding up.
• Taking logarithms of both sides before isolating the power, for example in 5 × 3^x = 40.
• Describing a transformation by type only, such as “reflection”, without the mirror line.
• Giving the intercept as a number only, or leaving the asymptote off a sketch of y = a^x.

Next steps

• Reread the revision notes for the methods behind any question you lost marks on.
• Go back to the study guide for full explanations.
• Open the OxfordAQA A-level Mathematics course page and tick off topics on the printable checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class at /trial/.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2: section PP1.3, Exponential and logarithms. Published by OxfordAQA for International AS exams from May/June 2018 and International A-level exams from May/June 2019.