Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 1 Number and algebra · 1.5 Laws of exponents and introduction to logarithms
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Use the laws of exponents with integer exponents, including zero and negative exponents | SL, HL | "Simplify (3x²)³ × x⁻⁴" (2 to 3 marks) |
| Rewrite a negative exponent as a fraction and back, without mixing up what the exponent applies to | SL, HL | "Write 5x⁻² without negative exponents" (1 mark) |
| Write numbers as powers of the same base in order to simplify | SL, HL | "Write 4³ × 8⁻¹ as a power of 2" (2 marks) |
| Know that aˣ = b is equivalent to logₐb = x, with a > 0, a ≠ 1 and b > 0 | SL, HL | "Write down the value of log₂ 32"; "find x if log₃ x = −2" (1 to 2 marks) |
| Know that log x means base 10 and ln x means base e | SL, HL | Any question using either |
| Evaluate logarithms with technology, and use the equivalence to solve simple equations | SL, HL | Paper 2: "Solve 3e²ˣ = 51" (3 marks) |
| Use logarithmic scales such as pH and decibels | SL, HL | Paper 2 context question, 4 to 6 marks in parts |
Before you start
You need powers as repeated multiplication (2⁵ = 2 × 2 × 2 × 2 × 2), the order of operations, and fractions. Scientific notation (1.1) already used two of the laws: multiplying powers of ten adds the exponents, and dividing subtracts them. This page explains why, and extends the idea to any base.
1The idea in one paragraph
A power such as a⁵ is shorthand for a multiplied by itself five times: a is the base and 5 is the exponent (or index). The laws of exponents are nothing more than counting how many times the base appears, and they let you simplify long products in one line. Following the pattern downwards past a¹ tells you what a⁰ and a negative exponent must mean. Then comes the reverse question. If 2 to some power is 32, what is the power? That power has a name, the logarithm: log₂ 32 = 5. A logarithm is an exponent, found backwards. The two you meet most are base 10, written log, and base e, written ln, and on Paper 2 your GDC evaluates both.
2The laws of exponents, and where they come from
Every law comes from counting factors. Figure 1 counts them for the three main laws.
- a³ × a² is three a's times two a's, so five a's: a⁵. Multiplying adds the exponents.
- a⁵ ÷ a² cancels two a's from top and bottom, leaving three: a³. Dividing subtracts the exponents.
- (a²)³ is three groups of two a's, so six: a⁶. A power of a power multiplies the exponents.
For integers m and n (and a, b ≠ 0 wherever something is divided or raised to a negative power):
| law | in words | example |
|---|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | same base, multiply: add the exponents | 7² × 7⁻⁵ = 7⁻³ |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | same base, divide: subtract the exponents | 9⁵ ÷ 9⁴ = 9¹ = 9 |
| (aᵐ)ⁿ = aᵐⁿ | power of a power: multiply the exponents | (3⁻²)³ = 3⁻⁶ |
| (ab)ⁿ = aⁿbⁿ | a power of a product: every factor gets the power | (3x)³ = 27x³ |
| (a/b)ⁿ = aⁿ/bⁿ | a power of a fraction: top and bottom both get it | (x/2)⁴ = x⁴/16 |
| a⁰ = 1 | any non-zero base to the power 0 | 12⁰ = 1 |
| a⁻ⁿ = 1/aⁿ | a negative exponent means "one over" | 5x⁻² = 5/x² |
The first two laws need the same base. 2³ × 3⁴ cannot be combined into a single power, because the factors are different numbers.
Two brackets rules that catch people. In (3x)³ the bracket holds both the 3 and the x, so both are cubed: 3³x³ = 27x³, not 3x³. In 5x⁻² there is no bracket, so the exponent belongs to the x alone: 5x⁻² = 5/x², not 1/(5x²) and not 1/(25x²). The same care applies to signs: (−2)⁴ = 16, but −2⁴ = −(2⁴) = −16.
3Zero and negative exponents
Why should a⁰ be 1, and why should a negative exponent mean a fraction? Because nothing else keeps the pattern. Figure 2 starts at 2³ = 8 and lowers the exponent one step at a time. Each step divides by 2.
From 2¹ = 2, one more division by 2 gives 1, so 2⁰ = 1. Another gives 1/2, so 2⁻¹ = 1/2. Then 2⁻² = 1/4 and 2⁻³ = 1/8. The same thing happens with any non-zero base, which is why
a⁰ = 1 and a⁻ⁿ = 1/aⁿ (a ≠ 0)
The division law agrees: a³ ÷ a³ = a⁰ by subtracting exponents, and it is also 1 because anything divided by itself is 1. And a² ÷ a⁵ = a⁻³, while cancelling gives 1/a³. The laws and the definitions fit together exactly.
A negative exponent does not make the number negative. 2⁻³ is 1/8, a small positive number. The minus sign in the exponent means "divide", not "negative".
A fraction with a negative exponent turns over: (2/3)⁻² = (3/2)² = 9/4.
4Simplifying expressions
Deal with numbers and each letter separately, and use the laws one at a time.
When a question says "give your answer with positive exponents" or "in the form…", the last line is where the final mark is.
Writing numbers as powers of the same base. Different bases can often be rewritten as one. Spot the powers: 4 = 2², 8 = 2³, 16 = 2⁴, 9 = 3², 27 = 3³, 25 = 5², 125 = 5³.
This is also how the simplest equations with an unknown exponent are solved: once both sides are powers of the same base, the exponents must be equal. For example 2ˣ⁺¹ = 32 = 2⁵ gives x + 1 = 5, so x = 4. Solving exponential equations in general, including with logarithms, is developed in 1.7.
5Logarithms: the exponent, found backwards
Suppose you know the base and the answer, and want the exponent. 2 to what power gives 32? The answer, 5, is called the logarithm of 32 to base 2, written log₂ 32 = 5. Figure 3 shows the two ways of writing the same fact.
aˣ = b is equivalent to logₐ b = x, where a > 0, a ≠ 1 and b > 0 (in the formula booklet)
Read logₐ b aloud as "the power you put on a to get b". The base stays the base in both forms; the exponent in one form is the logarithm in the other.
| exponential form | logarithmic form |
|---|---|
| 2⁵ = 32 | log₂ 32 = 5 |
| 10³ = 1000 | log₁₀ 1000 = 3 |
| 10⁻² = 0.01 | log₁₀ 0.01 = −2 |
| 3⁻² = 1/9 | log₃ (1/9) = −2 |
| 5⁰ = 1 | log₅ 1 = 0 |
| 7¹ = 7 | log₇ 7 = 1 |
The last two rows hold for every base: the log of 1 is 0, because a⁰ = 1, and the log of the base itself is 1, because a¹ = a.
Why the conditions? The base a must be positive and not 1: a negative base gives values that jump between positive and negative, and 1 to any power is just 1, so neither can reach every positive number. And b must be positive, because a positive base raised to any power is always positive. So log₂ (−8) and log₁₀ 0 do not exist. Your GDC will give an error, and in an equation a value that would need the log of a negative number or zero must be rejected.
Paper 1: evaluating by recognising powers. Rewrite the question as "base to what power gives this?"
6Base 10 and base e
Two bases are used so often that they have their own keys and their own notation.
Base 10. log x, with no base written, means log₁₀ x. It is called the common logarithm, and it answers "ten to what power?". Figure 4 shows what that means for a number that is not a whole power of ten.
250 is between 100 = 10² and 1000 = 10³, so log 250 is between 2 and 3. The GDC gives 2.397 94…, which is 2.40 to 3 s.f. Estimating first is a quick check on any calculator answer. Numbers between 0 and 1 have negative logarithms: log 0.5 = −0.301….
Base e. The number e = 2.718 28… is a constant like π. It turns up naturally in growth (1.4 shows it appearing from compound interest compounded more and more often) and it will be the base of the exponential function in calculus. The logarithm to base e is the natural logarithm, written ln x:
ln x means logₑ x, so eˣ = b is equivalent to x = ln b.
Figure 5 draws both pairs. The exponential graph and the logarithm graph are reflections of each other in the line y = x, because each undoes the other. Every exponential graph y = aˣ passes through (0, 1), because a⁰ = 1, and every log graph passes through (1, 0), because the log of 1 is 0. The log graph exists only for x > 0. Topic 2 (2.9) studies these graphs in detail.
Evaluating with technology. On Paper 2, use the LOG key for base 10 and the LN key for base e. For any other base, most GDCs have a template with a space for the base, so log₂ 20 can be entered directly: log₂ 20 = 4.32 (3 s.f.). (Changing the base by hand is part of 1.7.)
7Using the equivalence to solve equations
When the unknown is in the exponent, and the base is 10 or e, switch to logarithmic form. On Paper 1 leave the answer exact, as a logarithm; on Paper 2 evaluate it.
Get the power on its own first, then convert.
It works the other way too. If the logarithm is the unknown's partner, switch to exponential form.
An exact answer such as x = ln 12 or x = log 45 is the final answer on Paper 1. Do not try to "simplify" it into a decimal without a calculator.
8Logarithmic scales: pH, decibels and the Richter scale
Some quantities range over so many powers of ten that it is easier to record the exponent than the number. That is a logarithmic scale, and each step of 1 on it is a factor of 10 in the quantity.
pH in chemistry. The acidity of a solution depends on its hydrogen ion concentration, [H⁺], measured in mol dm⁻³, which can be anything from about 1 down to 10⁻¹⁴. The pH is defined by pH = −log[H⁺]. Figure 6 shows the scale.
The second line uses the equivalence: −log[H⁺] = 2.5 means log[H⁺] = −2.5, so [H⁺] is 10 to the power −2.5. A drop of 1 in pH means ten times the [H⁺]; a drop of 2 means a hundred times.
Decibels. Sound level in decibels is L = 10 log(I ÷ I₀), where I is the sound intensity and I₀ = 10⁻¹² W m⁻² is a fixed reference intensity. A sound a million times more intense than the reference has L = 10 log(10⁶) = 60 dB. Every 10 dB more is ten times the intensity.
The Richter scale. The Richter magnitude of an earthquake is also logarithmic: each step of 1 in magnitude is a tenfold increase in the amplitude of the ground motion that is measured.
On all three scales the trap is the same: a pH of 3 is not "a bit more acidic" than a pH of 4; it has ten times the concentration of hydrogen ions.
9Where marks are lost
Multiplying the exponents when you should add. a³ × a⁴ = a⁷, not a¹². Multiply exponents only for a power of a power, (a³)⁴ = a¹².
Combining different bases. 2³ × 3² is not 6⁵. The first two laws need the same base; rewrite as one base first if you can.
Forgetting to raise the number in a bracket. (2x)⁴ = 16x⁴, not 2x⁴. Everything inside the bracket gets the power.
Applying a negative exponent to the wrong thing. 3x⁻² = 3/x², because only the x is raised to −2.
Thinking a negative exponent makes a negative number. 4⁻² = 1/16, which is positive.
Writing a⁰ = 0. Any non-zero number to the power 0 is 1.
Swapping the base and the answer in a logarithm. log₂ 32 = 5 comes from 2⁵ = 32. Writing 5² = 32 or 32⁵ shows the base has moved. The base stays the base.
Taking the log of zero or a negative number. It does not exist. A value that needs one must be rejected, and you must say so.
Giving a decimal on Paper 1 when an exact answer is expected. x = ln 12 is complete. 2.48 needs a calculator you do not have.
10Work it right
- Multiplying the same base: add exponents. Dividing: subtract. Power of a power: multiply.
- Look at every bracket: what exactly is inside it, and so what gets the power?
- Handle numbers and each letter separately, then combine.
- Rewrite negative exponents as fractions when the question asks for positive exponents.
- If the bases differ, try writing them as powers of one base (4 = 2², 27 = 3³…).
- To evaluate or solve with a logarithm, write the equivalence: aˣ = b means logₐ b = x. The base stays the base.
- Check the argument of every logarithm is positive.
- Paper 1: leave log 45 or ln 12 exact. Paper 2: evaluate and give 3 s.f. unless told otherwise.
11Try it
Marks in brackets. Questions 1 to 3 are Paper 1 style: no calculator. Questions 4 and 5 are Paper 2 style.
Q1. Simplify fully, giving answers with positive exponents where needed.
(a) 4³ × 4⁻⁵ 1 mark
(b) (3x²)³ 2 marks
(c) 10a⁴b ÷ (2a⁶b³) 2 marks
Q2.
(a) Write 2⁷ = 128 in logarithmic form. 1 mark
(b) Find x, given that log₃ x = −2. 2 marks
(c) Find the value of log₅ 125 + log 0.01. 2 marks
Q3.
(a) Write 4³ × 8⁻¹ as a single power of 2. 2 marks
(b) Hence find x, given that 2ˣ = 4³ × 8⁻¹. 1 mark
Q4. Solve each equation, giving your answers correct to 3 significant figures.
(a) 10ˣ = 72 2 marks
(b) 3e²ˣ = 51 3 marks
(c) ln(x − 1) = 2 2 marks
Q5. The pH of a solution is given by pH = −log[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³.
(a) A sample of seawater (invented reading) has [H⁺] = 4.0 × 10⁻⁹ mol dm⁻³. Find its pH. 2 marks
(b) A sample of lemon juice (invented reading) has pH 2.3. Find its [H⁺], giving your answer in the form a × 10ᵏ, where 1 ≤ a < 10 and k ∈ ℤ. 2 marks
(c) Find how many times greater the [H⁺] of the lemon juice is than that of the seawater. 2 marks
12In one breath
A power aⁿ multiplies a by itself n times, and every law is counting: multiplying the same base adds exponents, dividing subtracts them, a power of a power multiplies them, and a power of a bracket reaches everything inside it. Lowering the exponent one step divides by the base, so a⁰ = 1 and a⁻ⁿ = 1/aⁿ; a negative exponent means "one over", not "negative". Watch what each exponent applies to: (2x)⁴ = 16x⁴ but 2x⁻³ = 2/x³. A logarithm is an exponent found backwards: aˣ = b means logₐ b = x, with a > 0, a ≠ 1 and b > 0, so the log of 1 is 0, the log of the base is 1, and the log of zero or a negative does not exist. log means base 10 and ln means base e ≈ 2.718; the GDC evaluates both, and switching between the two forms solves 10ˣ = 45 or eˣ = 12 in one line, left exact on Paper 1. pH, decibels and the Richter scale are logarithmic: one step is a factor of ten.
Answers
Q1. (a) 4³⁻⁵ = 4⁻² (or 1/16). (b) 3³(x²)³ = 27x⁶. (c) (10 ÷ 2)a⁴⁻⁶b¹⁻³ = 5a⁻²b⁻² = 5/(a²b²). (a) A1. (b) M1 for cubing both the 3 and x², A1 for 27x⁶ (3x⁶ scores M0 A0). (c) M1 for subtracting the exponents of a and of b, A1 for 5/(a²b²); 5a⁻²b⁻² is correct in value but does not use positive exponents, so it scores M1 A0.
Q2. (a) log₂ 128 = 7. (b) log₃ x = −2 means x = 3⁻² = 1/9. (c) log₅ 125 = 3 because 5³ = 125, and log 0.01 = −2 because 10⁻² = 0.01, so the sum is 1. (a) A1. (b) M1 for rewriting in exponential form, x = 3⁻², A1 for 1/9. (c) A1 for both values 3 and −2, A1 for 1.
Q3. (a)
(b) 2ˣ = 2³, so x = 3. (a) M1 for writing 4 and 8 as powers of 2, A1 for 2³. (b) A1, follow through from (a). A correct x found by evaluating 64 ÷ 8 = 8 = 2³ is equally valid.
Q4. (a) x = log 72 = 1.857 33… = 1.86. (b)
(c) ln(x − 1) = 2 means x − 1 = e², so x = 1 + e² = 8.389 05… = 8.39. Check: x − 1 = 7.389… > 0, so the logarithm exists. (a) M1 for x = log 72, A1. (b) M1 for dividing by 3, M1 for taking ln (or using the equivalence), A1 for 1.42. (c) M1 for x − 1 = e², A1 for 8.39. In (b), ln 51 ÷ 3 or similar, taking the log before isolating the power, scores M0.
Q5.
The lemon juice has about 1.25 × 10⁶ times the hydrogen ion concentration of the seawater. (a) M1 for substituting into −log, A1 for 8.40. (b) M1 for [H⁺ as 10 to the power −2.3, A1 for 5.01 × 10⁻³ in the form asked. (c) M1 for dividing the concentrations (or subtracting the pH values and raising 10 to that power), A1 for 1.25 × 10⁶ (accept 1.26 × 10⁶ from the rounded pH values). An answer of 3.65 from dividing one pH by the other scores M0 A0.]
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.5 Laws of exponents and introduction to logarithms. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.