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

These notes condense 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. For full explanations and longer worked examples, read the study guide.

Also useful: the practice questions, the OxfordAQA A-level Mathematics course page, the printable checklist and the free 10-minute diagnostics.

Key facts at a glance

Topic	What to remember
Graph of y = a^x	Passes through (0, 1); always above the x-axis; asymptote y = 0
a > 1	Increasing, steeper as x grows
0 < a < 1	Decreasing; y = (1/a)^x is y = a^x reflected in the y-axis
Equivalence	y = a^x ⇔ x = logₐy
Special values	logₐ1 = 0 and logₐa = 1
Domain of a log	logₐx is defined only for x > 0
Solving a^x = b	Take logarithms of both sides, then x = log b / log a

Laws of indices

a^m × a^n = a^(m+n)       a^m ÷ a^n = a^(m−n)
(a^m)^n = a^(mn)          a^(−n) = 1/a^n
a^0 = 1                   a^(1/n) = n-th root of a

Laws of logarithms

Same base throughout.

logₐ x + logₐ y = logₐ(xy)
logₐ x − logₐ y = logₐ(x/y)
k logₐ x = logₐ(x^k)

Must-know distinctions:

Looks similar	Actually
log x + log y	= log(xy), not log(x + y)
log x − log y	= log(x/y), not log(x − y)
log(x²)	= 2 log x
(log x)²	is not 2 log x
log x / log y	is not log(x/y)

Method: convert and evaluate

Steps to evaluate logₐb without a calculator:

1. Write logₐb = k.
2. Convert: a^k = b.
3. Write a and b as powers of one base.
4. Equate the powers and solve for k.

Worked reminder: log₂(1/8) = k means 2^k = 2⁻³, so k = −3. And log₂₅5 = k means 25^k = 5, so 5^(2k) = 5¹, giving k = 1/2.

Worked reminder: log₁₀5 + log₁₀20 = log₁₀100 = 2, and log₂48 − log₂3 = log₂16 = 4.

Method: solve a^x = b

1. Isolate the power so the equation reads a^(expression) = number.
2. If both sides can be written with one base, equate the indices.
3. Otherwise take logarithms of both sides (same base each side).
4. Use k log a = log(a^k) to bring the index down.
5. Solve the resulting linear equation. Do not round until the final line.
6. Round to 3 significant figures (unless told otherwise) and check by substitution.

Worked reminders:

2^x = 11               x = log 11 / log 2 = 3.46
4^(3x) = 9             3x = log 9 / log 4 = 1.585, so x = 0.528
5^(x−2) = 12           x − 2 = log 12 / log 5 = 1.544, so x = 3.54
3^(x+1) = 2^(x+4)      (x+1) log 3 = (x+4) log 2
                       x (log 3 − log 2) = 4 log 2 − log 3
                       x = (4 log 2 − log 3) / (log 3 − log 2) = 4.13

If the equation has the unknown on both sides, gather the terms in x, factorise, then divide.

Method: log equations

1. Combine into one log using the laws.
2. Convert to index form.
3. Solve.
4. Reject any root that makes an argument of a log zero or negative.

Worked reminder: log₂(x + 1) + log₂(x − 1) = 3 gives x² − 1 = 8, so x = ±3. Only x = 3 works, since x = −3 makes x − 1 negative.

Growth and decay questions

For a model N = A × r^t, set N to the target and solve r^t = target / A by logarithms. If r < 1 then log r is negative. A division by it turns the inequality round.

Worked reminder: 3000 × 1.05^n > 4500 gives n > log 1.5 / log 1.05 = 8.31, so n = 9. Check: 3000 × 1.05⁸ = 4432.4 (too small) and 3000 × 1.05⁹ = 4654.0.

Converting between forms

Index form	Logarithm form
2⁵ = 32	log₂32 = 5
10³ = 1000	log₁₀1000 = 3
7⁰ = 1	log₇1 = 0
9^(1/2) = 3	log₉3 = 1/2
3⁻² = 1/9	log₃(1/9) = −2
a^x = y	logₐy = x

The base of the logarithm is the base of the power. The logarithm is the index.

Sketching y = a^x

Checklist for a sketch:

1. Mark the y-intercept (0, 1).
2. Draw the curve above the x-axis everywhere.
3. Write “y = 0” next to the asymptote and make the curve approach it without touching.
4. Show the right shape: rising for a > 1, falling for 0 < a < 1.
5. If asked, add one more labelled point such as (1, a).

For a table of values, remember that a⁻¹ = 1/a and a⁻² = 1/a². For y = 3^x this gives 1/9, 1/3, 1, 3, 9 at x = −2, −1, 0, 1, 2.

Two longer worked reminders

Laws example. Suppose logₐ2 = p and logₐ3 = q. Express logₐ(√12) in terms of p and q.

√12 = (2² × 3)^(1/2)
logₐ√12 = (1/2)(2logₐ2 + logₐ3) = p + q/2

Mixed example. Solve 2 log₃x − log₃(x − 2) = 2.

log₃[x² / (x − 2)] = 2
x² / (x − 2) = 9
x² = 9x − 18
x² − 9x + 18 = 0
(x − 3)(x − 6) = 0

So x = 3 or x = 6. Both make x and x − 2 positive, so both are valid. Check x = 3: 2log₃3 − log₃1 = 2 − 0 = 2.

Equation with a coefficient. Solve 4 × 3^x = 100.

3^x = 25
x = log 25 / log 3 = 2.93

Divide by 4 before taking logarithms. Taking logarithms of 4 × 3^x directly is allowed, but you then need log 4 + x log 3 = log 100, which has more places to slip.

Checking your answer

A minute spent checking is worth more than a minute spent on a new question.

• For a^x = b, substitute your decimal back. It should return b to about 3 significant figures.
• For a log equation, check every argument is positive.
• For an inequality, test the whole numbers either side of the boundary.
• Estimate first, because it catches key-press errors. If 2^x = 11, then x is between 3 and 4 because 2³ = 8 and 2⁴ = 16.

Quick self-test

1. Write 4³ = 64 in logarithm form.
2. Evaluate log₃81.
3. Evaluate log₂(1/16).
4. Evaluate log₅15 − log₅3 + log₅5.
5. Evaluate 3log₂2 + log₂5 − log₂10.
6. State the y-intercept and the asymptote of y = 6^x.
7. Solve 8^x = 32.
8. Solve 5^x = 40, giving 3 significant figures.
9. Solve 3^(x−1) = 20, giving 3 significant figures.
10. Solve 2^x = 5^(x−1), giving 3 significant figures.
11. Solve log₃(2x + 1) = 2.
12. Find the least whole number n for which 1.08^n > 3.

Answers

1. log₄64 = 3.
2. log₃81 = 4, because 3⁴ = 81.
3. −4, because 2⁻⁴ = 1/16.
4. log₅(15/3) + 1 = log₅5 + 1 = 1 + 1 = 2.
5. 3log₂2 = 3, and log₂5 − log₂10 = log₂(1/2) = −1, so the total is 2.
6. It crosses the y-axis at (0, 1), with asymptote y = 0.
7. 2^(3x) = 2⁵, so 3x = 5 and x = 5/3.
8. x = log 40 / log 5 = 2.29.
9. x − 1 = log 20 / log 3 = 2.727, so x = 3.73.
10. x log 2 = (x − 1) log 5, so x (log 5 − log 2) = log 5 and x = log 5 / (log 5 − log 2) = 1.76.
11. 2x + 1 = 9, so x = 4.
12. n > log 3 / log 1.08 = 14.27, so n = 15.

Where marks are usually lost

• Splitting log(x + y) into log x + log y, or reading log x / log y as log(x/y).
• Mixing bases when adding or subtracting logarithms.
• Not converting the log equation to index form before solving.
• Leaving an extraneous root that makes a log undefined.
• Rounding log values to 2 or 3 figures before dividing, which shifts the final answer.
• Taking logarithms before isolating the power, for example from 5 × 3^x = 40.
• Keeping the inequality sign after a division by a negative logarithm.
• Giving a decimal when the question asks for an exact value, such as log₂9.
• Omitting the y-intercept or the asymptote label on a sketch of y = a^x.

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.