Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 1 Number and algebra · 1.7 Laws of exponents and logarithms
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Evaluate and simplify powers with rational exponents, including negative ones | SL, HL | "Find the exact value of 27^(−2/3)" (2 marks, Paper 1) |
| Switch between y = aˣ and x = logₐ y, and use logₐ a = 1 and logₐ 1 = 0 | SL, HL | "Write down the value of log₃ 81" (1 mark) |
| Use the three laws of logarithms to combine or expand | SL, HL | "Write 2 log 3 + log 4 − log 6 as a single logarithm" or "express in terms of p and q" (3 to 5 marks) |
| Change the base of a logarithm, exactly and to evaluate | SL, HL | "Show that log₉ x = ½ log₃ x" and "hence solve" (4 to 6 marks) |
| Solve exponential equations by matching bases | SL, HL | "Solve 25ˣ = 125¹⁻ˣ" (3 marks, Paper 1) |
| Solve exponential equations using logarithms, exactly or to 3 s.f. | SL, HL | "Solve 3 × 2ˣ = 5ˣ⁻¹, giving x in the form ln a ÷ ln b" (4 to 5 marks) |
| Solve exponential equations graphically with technology | SL, HL | Paper 2: find where two exponential graphs meet |
| Solve equations that are quadratics in disguise, and reject impossible roots | SL, HL | "Solve 4ˣ − 2ˣ⁺³ + 15 = 0" (5 marks) |
Before you start
You need the laws of exponents with whole-number exponents from 1.5 (such as a³ × a⁴ = a⁷ and a⁻² = 1/a²), and the meaning of a logarithm from 1.5: logₐ b is the power you raise a to in order to get b, with ln meaning logₑ. You also need to factorise a quadratic and to solve a linear equation with the unknown on both sides.
1The idea in one paragraph
Exponents and logarithms are the same fact read in two directions. 2⁵ = 32 says "2 to the power 5 is 32"; log₂ 32 = 5 says "the power that turns 2 into 32 is 5". So every law of exponents has a partner law of logarithms. Rational exponents extend powers to fractions, so a^(1/2) is a square root. The laws of logarithms turn multiplying into adding, and change of base moves between log₂, log₁₀ and ln. The point of it all: when the unknown is in the exponent, as in 3ˣ⁺¹ = 20, taking logarithms is the algebraic way to bring it down.
2Rational exponents
The laws of exponents from 1.5 still hold when the exponent is a fraction, and that fixes what a fractional exponent has to mean. Take a^(1/2). By the power law, (a^(1/2))² = a¹ = a. So a^(1/2) is the number whose square is a: it is √a. In the same way, (a^(1/3))³ = a, so a^(1/3) = ∛a.
a^(1/m) = ᵐ√a, and a^(n/m) = (ᵐ√a)ⁿ. The denominator is the root, the numerator is the power. If m is even, the root means the positive root.
The laws themselves are not in the formula booklet, so learn them.
| Law | Example with fractions |
|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | x^(1/2) × x^(3/2) = x² |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | x ÷ x^(1/3) = x^(2/3) |
| (aᵐ)ⁿ = aᵐⁿ | (8^(2/3))^(3/2) = 8¹ = 8 |
| a⁻ⁿ = 1/aⁿ | 9^(−1/2) = 1/√9 = 1/3 |
| (ab)ⁿ = aⁿbⁿ | (4x⁶)^(1/2) = 2x³, for x > 0 |
Figure 1 shows the order to work in. Take the root first, then the power, then deal with any minus sign by taking one over.
Worked example 1. Paper 1. Find the exact values of (a) 27^(2/3), (b) 32^(−3/5), (c) (9/4)^(3/2).
The "positive root" rule matters in (c). √(9/4) means 3/2, not ±3/2. The ± belongs to solving an equation such as x² = 9/4, which is a different question.
Rewriting roots as powers. Calculus in Topic 5 needs single powers, so practise the reverse too: √x ÷ x² = x^(1/2 − 2) = x^(−3/2).
3What a logarithm is, and the two values you always know
From 1.5: for a > 0, a ≠ 1 and y > 0,
y = aˣ ⇔ x = logₐ y
The base a must be positive and not 1 (1ˣ is always 1, so no power of 1 gives 5). The number y must be positive, because a positive base to any power is positive. That is why the log of zero or of a negative number does not exist, and it is the reason for every "reject this root" in section 7.
Two values follow straight from the definition, and the guide lists them:
- logₐ a = 1, because a¹ = a.
- logₐ 1 = 0, because a⁰ = 1.
Two more say that logₐ and "a to the power" undo each other: logₐ(aˣ) = x and a^(logₐ x) = x. Figure 2 shows what "undo each other" looks like. The graph of y = logₐ x is the reflection of y = aˣ in the line y = x, because every point (p, q) on one is the point (q, p) on the other. You will study these graphs in 2.9.
Worked example 2. Paper 1. Write down the value of (a) log₂ 32, (b) log₃ (1/9), (c) log₄ 8, (d) log₂₅ 5.
Ask "what power?" each time. When it is not obvious, write both numbers as powers of one number: 4 = 2² and 8 = 2³.
4The three laws of logarithms
For a, x, y > 0 (and a ≠ 1), these are in the formula booklet:
| Law | In words |
|---|---|
| logₐ xy = logₐ x + logₐ y | the log of a product is the sum of the logs |
| logₐ (x/y) = logₐ x − logₐ y | the log of a quotient is the difference of the logs |
| logₐ xᵐ = m logₐ x | a power comes down in front as a multiplier |
They are not new facts. Each is a law of exponents read backwards, and Figure 3 lines them up. Here is the first one proved, which is also a model of section 1.6's layout. Let m = logₐ x and n = logₐ y, so x = aᵐ and y = aⁿ. Then xy = aᵐ × aⁿ = aᵐ⁺ⁿ, so logₐ xy = m + n = logₐ x + logₐ y.
Worked example 3. Paper 1. Write 2 log 3 + log 4 − log 6 as a single logarithm, in its simplest form.
The order matters. Move every coefficient up into an exponent before you combine, because the product and quotient laws only work on logs with a coefficient of 1.
Worked example 4. Paper 1. Find the exact value of (a) log₆ 4 + log₆ 9, (b) log₂ 40 − log₂ 5.
Neither log₆ 4 nor log₂ 40 is neat on its own; the combined log is, which is the signal to combine.
Worked example 5. Paper 1. Given that p = logₐ 2 and q = logₐ 3, write in terms of p and q: (a) logₐ 12, (b) logₐ 4.5, (c) logₐ (a²/3).
Break each number into 2s and 3s, the only logs you were given. In (c), logₐ a = 1 does the rest.
Three things the laws do not say, each a favourite wrong answer:
- log(x + y) is not log x + log y. There is no law for the log of a sum.
- (log x)² is not 2 log x. The power law applies to log(x²), where the square is inside.
- log x ÷ log y is not log(x/y). A quotient of two logs is a change of base (section 5), not a log of a quotient.
5Change of base
For a, b, x > 0 (with a, b ≠ 1), in the formula booklet:
To change base, divide: logₐ x = ln x ÷ ln a. The same is true with log₁₀, or with any other base, as long as it is the same base on top and underneath.
Why it is true:
It has three uses.
Evaluating a log in any base. On Paper 2 your GDC has ln and log₁₀ keys, and some have a logₐ template too. With either key, log₅ 70 = ln 70 ÷ ln 5 = 2.64 (3 s.f.). Check: 5² = 25 and 5³ = 125, so the answer must lie between 2 and 3.
Finding exact values when both numbers are powers of something else. On Paper 1:
Putting logs with different bases on the same base, so the laws in section 4 can be used. Figure 4 shows why this always works. The graph of y = log₄ x is exactly half as tall as the graph of y = log₂ x at every x, because log₄ x = log₂ x ÷ log₂ 4 = log₂ x ÷ 2. Changing the base does nothing but multiply by a constant.
Worked example 6. Paper 1. Solve log₄ x + log₂ x = 6.
6Solving exponential equations
An exponential equation has the unknown in an exponent. There are three methods, and choosing the right one is half the question.
Method A: match the bases (Paper 1). If both sides can be written as powers of the same number, the exponents must be equal, because y = aˣ never takes the same value twice.
Worked example 7. Solve (a) 4ˣ = 8ˣ⁻¹, (b) (1/9)ˣ = 27ˣ⁺².
Method B: take logarithms of both sides. When the bases will not match, take ln (or log) of both sides and use the power law to bring the unknown down. This is the workhorse, and it gives an exact answer on Paper 1 and a decimal on Paper 2.
Worked example 8. Solve 3ˣ⁺¹ = 20, giving (a) an exact answer, (b) x to 3 s.f.
Worked example 9. Solve 5ˣ = 2ˣ⁺³. Give x exactly, then to 3 s.f.
Bring down, expand, collect, factorise: the same four moves every time the unknown is in two exponents. ln 2 and ln 5 are just numbers; treat them like the numbers in 5x = 2(x + 3).
Method C: graphs (Paper 2). Draw each side as a graph on the GDC and find where they meet. Figure 5 is worked example 9 solved that way. The intersection is at x ≈ 2.27, which agrees with the algebra.
On Paper 2, graph it even when you do the algebra: the graph shows how many solutions to expect.
Quadratics in disguise. 4ˣ is (2ˣ)², so an equation with 4ˣ and 2ˣ in it is a quadratic in 2ˣ. Substitute y = 2ˣ, solve the quadratic, then undo the substitution.
Worked example 10. Paper 1. Solve 4ˣ − 6 × 2ˣ + 8 = 0.
Figure 6 graphs the left side and confirms the two roots.
Sometimes one root of the quadratic is impossible. In 9ˣ − 3ˣ⁺¹ − 10 = 0, put y = 3ˣ, so 9ˣ = y² and 3ˣ⁺¹ = 3y. Then y² − 3y − 10 = 0, so (y − 5)(y + 2) = 0 and y = 5 or y = −2. But 3ˣ is always positive, so 3ˣ = −2 has no solution. Say so in words, then x = log₃ 5 is the only answer. Writing "reject 3ˣ = −2 since 3ˣ > 0" is usually worth an R1 on its own.
An exponential model. A town has 2400 people and grows by 3.5% a year, so after t years its population is P = 2400 × 1.035ᵗ. When does it reach 5000?
Figure 7 shows the same answer as the time at which the growth curve crosses the line P = 5000.
7Equations with logarithms in them
The laws in section 4 also solve equations that start with logs. Combine the logs into one, then turn the log statement into a power statement with the definition.
Worked example 11. Paper 1. Solve log₃ x + log₃ (x − 8) = 2.
The quadratic had two roots; the original equation has one. Combining the logs let in x-values where both brackets are negative: their product is positive, but neither log exists. So every solution goes back into the original equation, and any that needs the log of zero or a negative number is rejected, in words.
8Where marks are lost
Inventing a law for log(x + y). There is none. If a sum sits inside a log, it stays there.
Combining before bringing up coefficients. 2 log 3 + log 4 is log 36, not log 12. Coefficients go up as exponents first.
Reading a^(2/3) upside down. The denominator is the root. 8^(2/3) is (∛8)² = 4, not (√8)³.
Giving ± for an even root. 16^(1/2) is 4. It is not ±4: the symbol means the positive root.
Taking the log of each term instead of each side. From 2ˣ + 5 = 13 you cannot write x ln 2 + ln 5 = ln 13. Isolate the power first: 2ˣ = 8, so x = 3.
Not rejecting impossible roots. 3ˣ = −2 has no solution, and log₃(−1) does not exist. Leaving either in, or dropping it without a reason, loses a mark.
Giving a decimal on Paper 1 when an exact answer is asked for. "x = 2.27" when the question says "in the form ln a ÷ ln b" loses the final A mark even though it is correct.
9Work it right
- Before anything else, decide the method: match bases if both sides are powers of one number; take logs if not; substitute if you see aˣ and a²ˣ together; graph if it is Paper 2 and the algebra is ugly.
- Isolate the power (aˣ = number) before you take logs.
- Write the step "x ln 5 = (x + 3) ln 2" in full. The M1 is for bringing the exponent down, and it must be seen.
- With the unknown in two exponents: bring down, expand, collect, factorise out x, divide.
- To combine logs: coefficients up first, then add for multiply and subtract for divide.
- Reject roots that make aˣ negative or a log's argument zero or negative, and write the reason.
- Read the form asked for: exact (log₃ 5, 3 ln 2 ÷ ln 2.5) on Paper 1; 3 s.f. on Paper 2 unless told otherwise.
- Check: substitute the answer into the original equation, or look at the graph on the GDC.
10Try it
Marks in brackets. Q1 to Q3 and Q5 to Q6 are Paper 1 style (no calculator). Q4 is Paper 2 style.
Q1. Find the exact value of
(a) 81^(−3/4) 2 marks
(b) log₈ 4 2 marks
Q2. Given that p = log 2 and q = log 3, express log 18 − log 4 − 2 log 3 in terms of p and q, and hence find its value in terms of p only. 4 marks
Q3. Solve 25ˣ = 125¹⁻ˣ. 3 marks
Q4. Paper 2. Solve 3 × 2ˣ = 5ˣ⁻¹, giving your answer (a) in the form ln a ÷ ln b, where a, b ∈ ℚ, (b) correct to 3 significant figures. 5 marks
Q5. Solve 4ˣ − 2ˣ⁺³ + 15 = 0, giving your answers in the form log₂ k. 5 marks
Q6. (a) Show that log₉ x = ½ log₃ x. 2 marks
(b) Hence solve log₃ x + log₉ x = 6. 3 marks
Q7. Solve log₂(x + 2) + log₂(x − 2) = 5. 5 marks
11In one breath
A fractional exponent is a root and a power: a^(n/m) is the mth root of a, raised to the power n, root first, with a minus sign meaning one over and an even root meaning the positive one. A logarithm is an exponent: y = aˣ means x = logₐ y, for a > 0, a ≠ 1 and y > 0, so logₐ a = 1 and logₐ 1 = 0. The three laws (the log of a product is a sum, of a quotient a difference, of a power a multiple) are the laws of exponents read backwards; bring coefficients up before combining, and never invent a law for log(x + y). Change of base, logₐ x = ln x ÷ ln a (or any base, the same top and bottom), evaluates any log and puts mixed bases on one base. To solve an exponential equation, match the bases if you can, otherwise isolate the power and take logs so the exponent comes down, substitute y = aˣ when you see a quadratic in disguise, and graph it on Paper 2; then reject anything that needs aˣ to be negative or a log of a non-positive number, and give the form the question asks for.
Answers
Q1. (a) 81^(−3/4) = 1 ÷ (⁴√81)³ = 1 ÷ 3³ = 1/27. (b) 8 = 2³ and 4 = 2², so 8^(2/3) = (∛8)² = 4, so log₈ 4 = 2/3. Alternatively log₈ 4 = log₂ 4 ÷ log₂ 8 = 2/3. (a) M1 for ⁴√81 = 3 or 81 = 3⁴, A1 for 1/27. (b) M1 for writing both as powers of 2 or a change of base to base 2, A1 for 2/3. A decimal such as 0.037 or 0.667 scores M0 A0 on Paper 1.
Q2. log 18 − log 4 − 2 log 3 = log 18 − log 4 − log 9 = log (18 ÷ (4 × 9)) = log (1/2) = −log 2 = −p. Or term by term: log 18 = p + 2q, log 4 = 2p, 2 log 3 = 2q, so the expression is p + 2q − 2p − 2q = −p. M1 for 2 log 3 = log 9 or for log 18 = log 2 + 2 log 3, M1 for correct use of the quotient law (or for writing each term in p and q), A1 for an expression in p and q, A1 for −p.
Q3. 5²ˣ = 5³⁽¹⁻ˣ⁾ = 5³⁻³ˣ, so 2x = 3 − 3x, so 5x = 3, so x = 3/5. M1 for writing both sides as powers of 5, M1 for equating the exponents, A1 for 3/5. Check: 250.6 = 1250.4 ≈ 6.90.
Q4. Taking ln: ln 3 + x ln 2 = (x − 1) ln 5 = x ln 5 − ln 5. Collect: ln 3 + ln 5 = x ln 5 − x ln 2 = x(ln 5 − ln 2). So x = (ln 3 + ln 5) ÷ (ln 5 − ln 2) = ln 15 ÷ ln 2.5, and (b) x = 2.96 (3 s.f.). M1 for taking logs of both sides with the power law used correctly on both, M1 for expanding and collecting the x terms, M1 for factorising x, A1 for ln 15 ÷ ln 2.5 (or ln 15 ÷ ln (5/2)), A1 for 2.96. A GDC intersection giving 2.96 alone earns the final A1 only, because (a) asks for an exact form.
Q5. Let y = 2ˣ, so 4ˣ = y² and 2ˣ⁺³ = 8y. Then y² − 8y + 15 = 0, so (y − 3)(y − 5) = 0 and y = 3 or 5. So 2ˣ = 3 or 2ˣ = 5, giving x = log₂ 3 or x = log₂ 5. M1 for 2ˣ⁺³ = 8 × 2ˣ, M1 for the quadratic in y, A1 for y = 3 and 5, A1A1 for each value of x.
Q6. (a) log₉ x = log₃ x ÷ log₃ 9 = log₃ x ÷ 2 = ½ log₃ x, as required. (b) log₃ x + ½ log₃ x = 6, so (3/2) log₃ x = 6, so log₃ x = 4, so x = 3⁴ = 81. (a) M1 for change of base to base 3, A1 for log₃ 9 = 2 and the result shown. (b) M1 for using (a) to reach a single log equation (this is "hence", so it must use (a)), A1 for log₃ x = 4, A1 for 81.
Q7. log₂((x + 2)(x − 2)) = 5, so x² − 4 = 2⁵ = 32, so x² = 36 and x = ±6. x = −6 makes x + 2 = −4 < 0, so its log does not exist: reject it. x = 6. Check: log₂ 8 + log₂ 4 = 3 + 2 = 5. M1 for the product law, M1 for converting to x² − 4 = 32, A1 for x = ±6, R1 for rejecting −6 with a reason, A1 for x = 6 alone. Giving x = ±6 as the final answer loses the last two marks.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.7 Laws of exponents and logarithms. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.