Indices & Standard Form — Revision Notes

The question	Where it is answered	The tool
How do I combine powers without writing out all the factors?	1.1 Laws of Indices	the five laws — add, subtract or multiply the indices
What can an index be, if not a positive whole number?	1.2 Zero, Negative & Rational Indices	the three definitions — and the laws still hold
How do I write down, and calculate with, an enormous or a tiny number?	1.3 Standard Form	`A` × 10ⁿ, with 1 ≤ `A` < 10
What happens to money left in a bank, when the interest earns interest too?	1.4 Compound Interest	$A = P {(1 + \frac{r}{100})}^{n}$ — an index counting the periods

Index notation — the recap ⓘ Recap

Everything in this chapter rests on one piece of shorthand you already have.

The index notation of

a \times a \times \dots \times a

(n factors), where a is a real number and n is a positive integer, is

a^{n}

. The number a is the base and n is the index (plural: indices).

Fig. 1.1

5⁴ is read "5 to the power of 4". The base says what is being multiplied; the index says how many factors there are. Numbers with a base of 5, such as 5¹, 5² and 5³, are called powers of 5.

An index is not a multiplier. 2¹⁰ is 1024, not 20 — and that gap is what makes the notation worth having.

Where this comes from

The textbook opens the chapter with a donation that doubles every day. Thirty doublings of a single cent is 2³⁰ cents — 1 073 741 824 cents, about $10.7 million. Thirty times a cent would have been 30 cents, which is the same gap between 2¹⁰ and 20, drawn out far enough to see.

The five Laws of Indices

Each law is counting factors, written in shorthand. Here are all five, with the condition each one needs.

	In symbols	In words	What must match
Law 1	$a^{m} \times a^{n} = a^{m + n}$	multiplying → add the indices	the two bases
Law 2	$a^{m} \div a^{n} = a^{m - n}$ , or $\frac{a^{m}}{a^{n}} = a^{m - n}$	dividing → subtract the indices	the two bases
Law 3	${(a^{m})}^{n} = a^{m n}$	a power of a power → multiply the indices	—
Law 4	$a^{n} \times b^{n} = {(a \times b)}^{n}$	multiplying → multiply the bases, index unchanged	the two indices
Law 5	$a^{n} \div b^{n} = {(\frac{a}{b})}^{n}$	dividing → divide the bases, index unchanged	the two indices

The conditions, exactly as the textbook states them. In Laws 1–5 the indices m and n are positive integers. In Law 2 the base a is a real number with a ≠ 0 and m > n; in Law 5 the bases are real with b ≠ 0. Laws 1, 3 and 4 hold for any real bases. Section 1.2 lifts all five laws to zero, negative and rational indices — and tightens the conditions on the bases when it does.

Walkthrough 1 — Law 1 with coefficients Basic

Simplify 8m⁵n³ × 3m²n⁴, leaving your answer in index notation.

8m⁵n³ × 3m²n⁴ = 8 × 3 × m⁵ × m² × n³ × n⁴

Multiplication is commutative, so the factors may be reordered freely. Gathering the numbers together and each letter with its own kind puts every pair of matching bases side by side — the one condition Law 1 needs.
= 24m^{5 + 2}n^{3 + 4} (Law 1 of Indices)

8 × 3 = 24 is ordinary multiplication: the 24 is a coefficient, not a base, so it never collects an index. For the letters, Law 1 adds the indices because the bases match.
= 24m⁷n⁷

5 + 2 = 7 and 3 + 4 = 7. "In index notation" means this is the answer — writing the seven factors of m back out would be undoing the whole point of the notation.

Walkthrough 2 — Law 2 with coefficients Basic

Simplify 15p⁶q⁵ ÷ 5p⁴q², leaving your answer in index notation.

15p⁶q⁵ ÷ 5p⁴q² = $\frac{15 p^{6} q^{5}}{5 p^{4} q^{2}}$

Written inline, "÷ 5p⁴q²" is easy to read as dividing by the 5 only. As a single fraction, every factor of the divisor is visibly underneath, which is what the next step depends on.
= $\frac{15}{5}$ × $\frac{p^{6}}{p^{4}}$ × $\frac{q^{5}}{q^{2}}$

Split the one fraction into three, each holding a single base. A fraction may be split like this because multiplication and division travel together — and now each piece is exactly the shape Law 2 describes.
= 3 × p^{6 − 4} × q^{5 − 2} (Law 2 of Indices)

15 ÷ 5 = 3 for the coefficients. For each letter, Law 2 subtracts the index below from the index above — in that order, because subtraction is not symmetrical.
= 3p²q³

6 − 4 = 2 and 5 − 2 = 3. The answer is exact and in index notation, so nothing further is wanted.

Walkthrough 3 — Laws 1 to 3, with a letter in the index Intermediate

Simplify (6^k)⁴ × (6³)^k ÷ (6²)^k, leaving your answer in index notation.

(6^k)⁴ × (6³)^k ÷ (6²)^k = 6^{k × 4} × 6^{3 × k} ÷ 6^{2 × k} (Law 3 of Indices)

Law 3 clears every bracket by multiplying the two indices. Doing this first is what turns three differently-shaped terms into three plain powers of the same base, 6.
= 6^4k × 6^3k ÷ 6^2k

Tidy each index. A letter in an index behaves like a letter anywhere else: 4 × k is written 4k, and it is still a single index.
= 6^{4k + 3k − 2k} (Laws 1 & 2 of Indices)

Law 1 adds the indices of the two factors being multiplied and Law 2 subtracts the index of the divisor. The base 6 is written once and never changes — only the index is doing any work.
= 6^5k

4k + 3k − 2k = 5k, collecting like terms in the index. Index notation is the only possible answer here: with k unknown, there is no number to evaluate.

Walkthrough 4 — Laws 1 to 4 on a fraction Intermediate

Simplify

\frac{{(a b^{3})}^{2} \times {(−2 a^{3} b)}^{3}}{{(2 a^{2} b^{2})}^{2}}

, leaving your answer in index notation.

(ab³)² = a²(b³)², (−2a³b)³ = (−2)³(a³)³b³, (2a²b²)² = 2²(a²)²(b²)² (Law 4)

Law 4 hands the outside index to every factor inside the bracket — the number in front included, so the −2 and the 2 are raised as well.
= $\frac{a^{2} b^{6} \times (−8) a^{9} b^{3}}{4 a^{4} b^{4}}$ (Law 3)

Law 3 multiplies the indices of a power of a power: (b³)² = b⁶. The numbers are ordinary arithmetic, and an odd index keeps a minus sign: (−2)³ = −8.
= $\frac{−8 a^{11} b^{9}}{4 a^{4} b^{4}}$ (Law 1)

The numerator is now a product of matching bases, so Law 1 adds: 2 + 9 = 11 and 6 + 3 = 9. The minus sign travels with the coefficient, never with an index.
= $\frac{−8}{4}$ × a^{11 − 4} × b^{9 − 4} (Law 2)

Divide the coefficients as ordinary numbers and subtract the indices base by base — top index minus bottom index.
= −2a⁷b⁵

−8 ÷ 4 = −2, 11 − 4 = 7 and 9 − 4 = 5. The sign is written once, in front, and the answer is left in index notation.

Walkthrough 5 — all five laws together Advanced

Simplify

{(\frac{−2 m^{3}}{n^{2}})}^{5} \div \frac{8 m^{7}}{m n^{14}}

, leaving your answer in index notation.

${(\frac{−2 m^{3}}{n^{2}})}^{5} = \frac{{(−2 m^{3})}^{5}}{{(n^{2})}^{5}}$

Law 5, read as ${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$ : an index outside a fraction goes to the top and the bottom, not to one of them.
= $\frac{−32 m^{15}}{n^{10}}$ (Laws 4 & 3)

Law 4 spreads the index 5 over both factors of the numerator and Law 3 multiplies the indices: (−2)⁵ = −32 (odd index, sign kept), (m³)⁵ = m¹⁵ and (n²)⁵ = n¹⁰.
= $\frac{−32 m^{15}}{n^{10}} \times \frac{m n^{14}}{8 m^{7}}$

Dividing by a fraction is multiplying by its reciprocal — the second fraction is turned upside down. Doing this before any index work means every factor ends up unambiguously on the top or on the bottom of one product.
= $\frac{−32 m^{16} n^{14}}{8 m^{7} n^{10}}$ (Law 1)

Multiply the two numerators and the two denominators, then use Law 1 on the top: m¹⁵ × m = m^{15 + 1} = m¹⁶. A letter written with no index has an index of 1.
= −4m⁹n⁴

−32 ÷ 8 = −4, and Law 2 subtracts the indices: 16 − 7 = 9 and 14 − 10 = 4. Every index in the answer is a positive integer, so no further tidying is needed.

Where this comes from

Each law is nothing more than counting factors, which is why you can always rebuild one you have forgotten. Written out, 7² × 7⁴ is (7 × 7) × (7 × 7 × 7 × 7): two factors then four factors, so six factors altogether — and 2 + 4 = 6. That is Law 1.

Laws 4 and 5 are the two that read backwards as often as forwards, so learn both directions: ${(a b)}^{n} = a^{n} b^{n}$ and ${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$ . Read left to right they are how a bracket is opened; read right to left they are how two separate powers are gathered.

Past-paper questions

G3 N2022 · Paper 1 · Q2(a) 1 mark

Simplify 3y⁵ × 5y³. [1]

Worked solution

3y⁵ × 5y³ = (3 × 5) × (y⁵ × y³) = 15y^{5 + 3} (Law 1)

= 15y⁸

Why this works. The mark is for keeping the two halves of the term apart: 3 and 5 multiply as ordinary numbers, while Law 1 takes the matching letters and adds 5 + 3 = 8. Multiplying the indices instead gives 15y¹⁵. Writing the y^{5 + 3} line names the law you used.

The four errors this topic is built to catch

Each of these four is a law used without its condition. Read the right-hand column and the condition is always the thing that was missing.

Wrong	Right	What went wrong
2³ × 5⁴ = 10⁷	2³ × 5³ = 10³, but 2³ × 5⁴ will not combine at all	Law 1 needs the same base; Law 4 needs the same index. With neither matching, there is no law to use.
(`x`²)³ = `x`⁵	(`x`²)³ = `x`⁶	Adding the indices is Law 1 (a product of two powers). A power of a power is Law 3, and it multiplies.
(2`x`²)³ = 2`x`⁶	(2`x`²)³ = 8`x`⁶	Law 4 gives the index to every factor inside the bracket. The coefficient is a factor, so it is cubed too.
(`a` + `b`)² = `a`² + `b`²	(`a` + `b`)² = `a`² + 2`ab` + `b`²	There is no law of indices for a sum. Test it: (3 + 4)² = 49, but 3² + 4² = 25.

Brackets decide what the base is. In 2x⁴ only the x is raised: 2 × x⁴. In (2x)⁴ the whole of 2x is the base: 16x⁴.

An even index removes a minus sign; an odd index keeps it. (−2)⁴ = 16, the same as 2⁴, but (−2)⁵ = −32.

More detail

The same reading of brackets settles −3² against (−3)²: in −3² only the 3 is squared, so it is −9, while (−3)² squares the whole of −3 and gives 9.

Because an even index removes the sign, an answer can often be written two ways and both are right: (−2)⁴ and 2⁴ are both 16, so either may be left as the answer. An odd index is different — (−2)⁵ = −32 is not 2⁵.

Check yourself

Simplify 5a⁴b⁷ × 4a³b².
Answer
= 5 × 4 × a⁴ × a³ × b⁷ × b²
= 20a^{4 + 3}b^{7 + 2}
= 20a⁷b⁹ (Law 1).
Simplify (−12x⁷y⁵) ÷ (3xy²).
Answer
= $\frac{−12 x^{7} y^{5}}{3 x y^{2}}$ = −4 × x^{7 − 1} × y^{5 − 2}
= −4x⁶y³ (Law 2). Remember x counts as x¹.
Simplify (2c³d⁴)⁵.
Answer
= 2⁵(c³)⁵(d⁴)⁵ (Law 4)
= 32c¹⁵d²⁰ (Law 3)
= 32c¹⁵d²⁰. The coefficient 2 is raised to the fifth power as well: 2⁵ = 32, not 2.
Simplify $\frac{{(3 h^{2} k)}^{3} \times {(−2 h k^{4})}^{2}}{{(6 h^{3} k^{2})}^{2}}$ .
Answer
Numerator: (3h²k)³ = 27h⁶k³ and (−2hk⁴)² = 4h²k⁸, so the numerator is 108h⁸k¹¹.
Denominator: (6h³k²)² = 36h⁶k⁴.
108 ÷ 36 = 3, 8 − 6 = 2, 11 − 4 = 7, so = 3h²k⁷. Note (−2)² = +4: an even index removes the minus sign.
Given that y⁹ × (y²)ⁿ ÷ (y³)² = y¹⁷, find the value of n.
Answer
y⁹ × y²ⁿ ÷ y⁶ = y^{9 + 2n − 6} = y^{3 + 2n}
Same base, so the indices are equal: 3 + 2n = 17
2n = 14, n = 7.
Explain why a⁴ × b⁴ can be written as (ab)⁴, but a⁴ × b³ cannot be written as (ab)⁴ or as (ab)⁷.
Answer
Law 4 combines two powers with the same index by multiplying the bases, and 4 = 4, so a⁴ × b⁴ = (ab)⁴. Law 1 combines two powers with the same base by adding the indices, and here the bases differ. In a⁴ × b³ neither the bases nor the indices match, so no law applies and the expression is already in its simplest form. Check with numbers: 2⁴ × 3³ = 432, while 6⁴ = 1296 and 6⁷ = 279 936.

The zero index

You cannot multiply 3 by itself zero times, so the zero index is not discovered — it is defined, in the one way that keeps the five laws working.

Definition 1.

a^{0} = 1

, where the base a is a real number such that a ≠ 0.

Walkthrough 6 — the zero index Basic

Evaluate 3 × 7⁰ − (−5)⁰ + (2y)⁰ without using a calculator, given that y ≠ 0.

7⁰ = 1 and (−5)⁰ = 1

With no calculator allowed, the mark is for quoting Definition 1: $a^{0} = 1$ for every real base except 0. The bracket in (−5)⁰ makes −5 the base, and it is not zero, so the definition applies to it too.
(2y)⁰ = 1

The bracket makes the whole of 2y the base, so the term collapses to 1. Without it, 2y⁰ would mean 2 × y⁰ = 2. The condition y ≠ 0 is what lets Definition 1 apply here.
3 × 7⁰ − (−5)⁰ + (2y)⁰ = 3 × 1 − 1 + 1

Substitute the three values back in. Replacing every zero-index term before doing any arithmetic keeps all three uses of the definition visible to the marker.
= 3

3 − 1 + 1 = 3. The value is exact, so it is left as 3 — there is nothing here to round.

Where this comes from

"a multiplied by itself n times" stops making sense the moment n is not a positive whole number. Follow the pattern down a column of powers of 3 instead, dividing by 3 each time, and the value of 3⁰ is forced:

Index form	3⁴	3³	3²	3¹	3⁰
Value	81	27	9	3	1

The algebra says the same thing. For a ≠ 0, a number divided by itself is 1:

a^{n} \div a^{n} = 1, but Law 2 gives a^{n} \div a^{n} = a^{n - n} = a^{0}

And that is why a ≠ 0. The argument divides by aⁿ, which is only legal when a is not zero. 0⁰ is left undefined.

Negative indices

Dividing a power of 3 by 3 once more takes the index below zero. As with the zero index, what a negative index means is defined so that the five laws carry on working.

Definition 2.

a^{- n} = \frac{1}{a^{n}}

, where the base a is a real number such that a ≠ 0, and n is a positive integer.

A negative index does not make a number negative. It makes it a reciprocal. 2⁻³ is

\frac{1}{8}

, which is small and positive; −2³ is −8, which is negative. Read the minus sign as "one over", every time.

A negative index turns a fraction upside down.

{(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}

Not a sixth law: it is Definition 2 followed by Law 5, as Walkthrough 7 shows line by line.

Walkthrough 7 — a negative index on a fraction Basic

Evaluate

{(\frac{2}{5})}^{−3}

without using a calculator, giving your answer as a fraction.

${(\frac{2}{5})}^{−3} = \frac{1}{{(\frac{2}{5})}^{3}}$

Definition 2, $a^{- n} = \frac{1}{a^{n}}$ , applied with the whole fraction as the base. The minus sign has now been used up; the rest of the question is ordinary index work.
$= 1 \div \frac{2^{3}}{5^{3}}$ (Law 5 of Indices)

Law 5 sends the index 3 to the numerator and to the denominator separately, so the cube of the fraction becomes a fraction of two cubes.
$= 1 \times \frac{5^{3}}{2^{3}} = \frac{5^{3}}{2^{3}}$

Dividing by a fraction is multiplying by its reciprocal. This is the line that turns $\frac{2}{5}$ into $\frac{5}{2}$ — the "flip it over" shortcut, proved rather than assumed.
$= \frac{125}{8}$

5³ = 125 and 2³ = 8. "As a fraction" means the exact value $\frac{125}{8}$ (or 15 $\frac{5}{8}$ ) is wanted — not the decimal 15.625.

Walkthrough 8 — the laws with negative indices Intermediate

Simplify 27p⁻⁴q⁵ ÷ (3p³q)², expressing your answer in positive index form.

(3p³q)² = 3²(p³)²q² = 9p⁶q² (Laws 4 & 3)

Clear the bracket first. Law 4 gives the index 2 to every factor inside, the 3 included, and Law 3 doubles the index of p³.
= $\frac{27 p^{−4} q^{5}}{9 p^{6} q^{2}}$

Writing the division as one fraction puts the whole of the simplified divisor underneath, so nothing can be left out when the indices are subtracted.
= $\frac{27}{9}$ × p^{−4 − 6} × q^{5 − 2} (Law 2)

Law 2 still applies once zero and negative indices are defined — subtract the bottom index from the top one, signs and all.
= 3p⁻¹⁰q³

27 ÷ 9 = 3, −4 − 6 = −10 and 5 − 2 = 3. Watch that first subtraction: taking 6 away from −4 moves further down the number line, not back up towards zero.
$= \frac{3 q^{3}}{p^{10}}$

Definition 2 turns p⁻¹⁰ into $\frac{1}{p^{10}}$ . "Positive index form" means no index in the final answer may be negative, so the p has to move to the denominator — and only the p, because q³ is already positive.

Where this comes from

Carry the powers-of-3 pattern one step further, still dividing by 3 each time:

Index form	3²	3¹	3⁰	3⁻¹	3⁻²
Value	9	3	1	$\frac{1}{3}$	$\frac{1}{9}$

And again the algebra agrees. Cancelling factors, $a^{4} \div a^{7} = \frac{1}{a^{3}}$ , while Law 2 gives $a^{4} \div a^{7} = a^{4 - 7} = a^{−3}$ . The two must agree, and Definition 2 is what makes them agree.

Past-paper questions

G3 N2017 · Paper 1 · Q1 1 mark

Given that $\frac{1}{81}$ = 3^k, find k. [1]

Worked solution

81 = 3 × 3 × 3 × 3 = 3⁴, so $\frac{1}{81} = \frac{1}{3^{4}} = 3^{−4}$ (Definition 2)

3^k = 3⁻⁴, and the bases match, so k = −4.

Check: 3⁻⁴ = $\frac{1}{3^{4}}$ = $\frac{1}{81}$ , the given number.

Why this works. First write both sides to the same base: 81 = 3⁴. Then Definition 2, $\frac{1}{a^{n}} = a^{- n}$ , reads the reciprocal as a negative index: the 4 changes sign and nothing else. Check your answer against k = 4 and k = $\frac{1}{4}$ — the reciprocal ignored, or a negative index read as a root.

Rational indices

The last gap is an index that is not a whole number at all. What $3^{\frac{1}{4}}$ means comes from roots.

If a is a non-negative number such that a = bⁿ for some non-negative number b, then b is the positive nth root of a:

b = \sqrt[n]{a}

Definition 3.

a^{\frac{1}{n}} = \sqrt[n]{a}

where the base a is a real number such that a ≥ 0, and n is a positive integer.

And in general.

a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}

where the base a is a real number such that a ≥ 0, and m and n are positive integers.

Bottom is the root, top is the power. In

125^{\frac{4}{3}}

the 3 underneath is the cube root and the 4 on top is the power. Either order gives the same value; taking the root first keeps the numbers small.

Walkthrough 9 — evaluating a rational index Intermediate

Evaluate

125^{\frac{4}{3}}

without using a calculator.

$125^{\frac{4}{3}} = {(\sqrt[3]{125})}^{4}$

Use $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ : denominator underneath is the root, numerator on top is the power. With no calculator allowed, take the root first — the other order would mean cubing 125 by hand.
$\sqrt[3]{125} = 5$ , since 125 = 5 × 5 × 5

The cube root is found by prime factorisation: 125 splits into three equal factors of 5, and three equal factors is exactly what a cube root asks for.
= 5⁴

The 4 from the top of the index is an ordinary power, now applied to the root just found.
= 625

5⁴ = 625. A rational index of a perfect power gives an exact answer, so 625 is the whole answer — there is nothing to round.

Where this comes from

The positive square root of 9 is 3, because 3 × 3 = 9, and the positive 4th root of 81 is 3, because 3 × 3 × 3 × 3 = 81:

\sqrt{9} = 3 \sqrt[4]{81} = 3

Now let P stand for 5 raised to the index one third. Cubing it, and using Law 3 to multiply the two indices:

P^{3} = {(5^{\frac{1}{3}})}^{3} = 5^{\frac{1}{3} \times 3} = 5^{1} = 5

so P must be the cube root of 5. That is the only value the index $\frac{1}{3}$ can be given without breaking Law 3, which is what Definition 3 records. The textbook's memory hook for the general form is that the denominator is "below the ground", which is where roots live.

Why the base must be non-negative. If negative bases were allowed, both ${(- 1)}^{\frac{1}{2}}$ and Law 3 could be used to "prove" that 1 = −1. The conditions on the base are what keep the laws honest.

The laws, extended to every rational index

With the three definitions in place, all five laws survive — but the conditions on the bases tighten, because zero and negative indices need a ≠ 0 while rational indices need a ≥ 0. Combining both gives a > 0.

	Positive integer indices	Any rational indices
Law 1	$a^{m} \times a^{n} = a^{m + n}$ , any real base	same, if `a` > 0
Law 2	$a^{m} \div a^{n} = a^{m - n}$ , if `a` ≠ 0 and `m` > `n`	same, if `a` > 0 — and `m` > `n` is no longer needed
Law 3	${(a^{m})}^{n} = a^{m n}$ , any real base	same, if `a` > 0
Law 4	$a^{n} \times b^{n} = {(a b)}^{n}$ , any real bases	same, if `a` > 0 and `b` > 0
Law 5	$a^{n} \div b^{n} = {(\frac{a}{b})}^{n}$ , if `b` ≠ 0	same, if `a` > 0 and `b` > 0

Walkthrough 10 — a negative and fractional index Advanced

Simplify

{(\frac{16 x^{8}}{y^{12}})}^{- \frac{1}{4}}

, expressing your answer in positive index form.

${(\frac{16 x^{8}}{y^{12}})}^{- \frac{1}{4}} = {(\frac{y^{12}}{16 x^{8}})}^{\frac{1}{4}}$

A negative index on a fraction inverts the fraction: ${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}$ . Clearing the minus sign first means only one awkward thing — the quarter index — is left to deal with.
$= \frac{{(y^{12})}^{\frac{1}{4}}}{{(16 x^{8})}^{\frac{1}{4}}}$ (Law 5)

Law 5 sends the index to the numerator and the denominator separately. It is allowed here because both parts are positive.
$= \frac{{(y^{12})}^{\frac{1}{4}}}{16^{\frac{1}{4}} {(x^{8})}^{\frac{1}{4}}}$ (Law 4)

Law 4 hands the quarter index to every factor of the denominator — the 16 as well as the x⁸, exactly as a coefficient inside a bracket gets cubed.
$= \frac{y^{3}}{16^{\frac{1}{4}} x^{2}}$ (Law 3)

Law 3 multiplies the indices: 12 × $\frac{1}{4}$ = 3 and 8 × $\frac{1}{4}$ = 2. Fractional indices very often cancel down to whole numbers like this — that is how the question was built.
$= \frac{y^{3}}{2 x^{2}}$

By Definition 3, $16^{\frac{1}{4}} = \sqrt[4]{16} = 2$ , since 16 = 2 × 2 × 2 × 2. Every index in the answer is now a positive whole number, which is what "positive index form" asks for.

Why this works

Why m > n can be dropped from Law 2. With only positive indices allowed, $a^{3} \div a^{7}$ had no answer to give. Now it does: a⁻⁴. Extending the definitions is exactly what removes the restriction.

Past-paper questions

G3 N2018 · Paper 1 · Q2 2 marks

Simplify ${(\frac{x^{8}}{y^{6}})}^{- \frac{3}{2}}$ . [2]

Worked solution

${(\frac{x^{8}}{y^{6}})}^{- \frac{3}{2}} = {(\frac{y^{6}}{x^{8}})}^{\frac{3}{2}}$ (the negative index inverts the fraction)

$= \frac{{(y^{6})}^{\frac{3}{2}}}{{(x^{8})}^{\frac{3}{2}}}$ (Law 5, then Law 3 on each part)

$= \frac{y^{9}}{x^{12}}$ , since 6 × $\frac{3}{2}$ = 9 and 8 × $\frac{3}{2}$ = 12.

Why this works. Two marks, two decisions. Deal with the sign of the index first by turning the fraction upside down, ${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}$ ; then multiply each index by $\frac{3}{2}$ (Law 3). Both indices are even, so the answer has whole-number indices — a fraction left in an index means the multiplication went the other way.

G3 N2024 · Paper 2 · Q1(c) 2 marks

Simplify ${(\frac{a^{9}}{64 b^{12}})}^{- \frac{2}{3}}$ . [2]

Worked solution

${(\frac{a^{9}}{64 b^{12}})}^{- \frac{2}{3}} = {(\frac{64 b^{12}}{a^{9}})}^{\frac{2}{3}}$ (the negative index inverts the fraction)

$= \frac{64^{\frac{2}{3}} {(b^{12})}^{\frac{2}{3}}}{{(a^{9})}^{\frac{2}{3}}}$ (Laws 5 and 4, then Law 3)

$64^{\frac{2}{3}} = {(\sqrt[3]{64})}^{2} = 4^{2} = 16$ , 12 × $\frac{2}{3}$ = 8 and 9 × $\frac{2}{3}$ = 6, so

$= \frac{16 b^{8}}{a^{6}}$

Why this works. Same two decisions as the 2018 question, plus the 64: the index is handed to every factor of the denominator, the number included, so 64 × $\frac{2}{3}$ is not it. Read $64^{\frac{2}{3}}$ as "cube root first, then square" — the cube root of 64 is 4 and 4² = 16. Taking the root before the power keeps the arithmetic small.

Equations with the unknown in the index

To solve x² = 100 you take a square root, because the unknown is the base. In 2^x = 32 the unknown is the index, so the move instead is to make both sides powers of the same base.

If

a^{x} = a^{n}

then x = n, provided the base a is not 0, 1 or −1.

Walkthrough 11 — solving for an unknown index Intermediate

Solve 8^x =

\frac{1}{32}

.

8 = 2³ and 32 = 2⁵

The unknown is in the index, so the two sides can only be compared once they are powers of one common base. Both 8 and 32 are powers of 2, so 2 is that base.
(2³)^x = $\frac{1}{2^{5}}$

Rewrite each side using base 2. Nothing has changed in value — only the way the two numbers are written.
2^3x = 2⁻⁵ (Law 3, then Definition 2)

Law 3 multiplies the indices on the left. On the right, Definition 2 read backwards turns $\frac{1}{2^{5}}$ into 2⁻⁵. Both sides are now a single power of 2.
3x = −5

Same base, and that base is not 0, 1 or −1, so the indices must be equal. This is the line that removes the index altogether and leaves an ordinary linear equation.
$x = - \frac{5}{3}$

Divide both sides by 3. Leave it as the exact fraction − $\frac{5}{3}$ (or −1 $\frac{2}{3}$ ); −1.67 would only be an approximation to an answer that is exact.

Why this works

Why those three bases are barred. 1⁴ = 1⁹ without 4 being 9, and 0³ = 0⁷; (−1)² = (−1)⁴. For every other base, one value of the index gives one value of the power, so the indices can safely be equated.

Check yourself

Evaluate 5 × 9⁰ + (−3)⁰ − (4k)⁰ without using a calculator, given that k ≠ 0.
Answer
9⁰ = 1, (−3)⁰ = 1 and (4k)⁰ = 1 by Definition 1.
= 5 × 1 + 1 − 1
= 5.
Evaluate ${(\frac{3}{4})}^{−3}$ without using a calculator, giving your answer as a fraction.
Answer
A negative index inverts the fraction:
${(\frac{3}{4})}^{−3} = {(\frac{4}{3})}^{3} = \frac{4^{3}}{3^{3}}$
= $\frac{64}{27}$ (or 2 $\frac{10}{27}$ ).
Simplify 15m⁻³ × 2m⁵ ÷ (3m²)², expressing your answer in positive index form.
Answer
15m⁻³ × 2m⁵ = 30m^{−3 + 5} = 30m² (Law 1)
(3m²)² = 9m⁴ (Laws 4 & 3)
$\frac{30 m^{2}}{9 m^{4}}$ = $\frac{10}{3}$ m^{2 − 4} = $\frac{10}{3}$ m⁻² (Law 2)
= $\frac{10}{3 m^{2}}$ .
Evaluate $64^{- \frac{2}{3}}$ without using a calculator, giving your answer as a fraction.
Answer
$64^{- \frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}}$ (Definition 2)
$64^{\frac{2}{3}} = {(\sqrt[3]{64})}^{2} = 4^{2} = 16$
= $\frac{1}{16}$ .
Simplify $\sqrt[3]{c^{2}} \times \sqrt[6]{c}$ , expressing your answer in index form, given that c > 0.
Answer
Rewrite both factors as indices: $\sqrt[3]{c^{2}} = c^{\frac{2}{3}}$ and $\sqrt[6]{c} = c^{\frac{1}{6}}$ .
Law 1 adds the indices: $\frac{2}{3} + \frac{1}{6} = \frac{4}{6} + \frac{1}{6} = \frac{5}{6}$
= $c^{\frac{5}{6}}$ .
Solve 9^x = $\frac{1}{27}$ .
Answer
9 = 3² and 27 = 3³, so (3²)^x = $\frac{1}{3^{3}}$
3^2x = 3⁻³
2x = −3
x = − $\frac{3}{2}$ (or −1.5).

Past-paper questions

G3 N2019 · Paper 1 · Q3 2 marks

Simplify ${(x^{6})}^{\frac{2}{3}}$ . [1]
3^a = 3⁷ + 3⁷ + 3⁷
Find the value of a. [1]

Worked solution

(a) Law 3 multiplies the indices: 6 × $\frac{2}{3}$ = 4, so ${(x^{6})}^{\frac{2}{3}}$ = $x^{4}$ .

(b) The right-hand side is three copies of the same number:

3⁷ + 3⁷ + 3⁷ = 3 × 3⁷ = 3^{1 + 7} = 3⁸ (Law 1)

So 3^a = 3⁸, and since the bases match, a = 8.

Why this works. Part (b) is set because there is no index law for adding powers: the three terms are identical, so the sum is a multiplication, 3 × 3⁷, and multiplication does have a law. Once both sides are a single power of 3 the indices can be compared. Steer clear of a = 21, which reads the plus signs as times signs.

G3 N2022 · Paper 1 · Q23 2 marks

Given that 25^2x = 125⁷, find x. [2]

Worked solution

25 = 5² and 125 = 5³, so both sides can be written to base 5:

(5²)^2x = (5³)⁷

5^4x = 5²¹ (Law 3)

4x = 21, so x = 5.25 (or $\frac{21}{4}$ ).

Why this works. The unknown is in the index, so it needs a common base: 25 and 125 are both powers of 5, and writing each side to base 5 earns the first mark. Law 3 then makes each side a single power and the indices can be equated. 2 × 2x = 4x, and an index may be a fraction: 5.25 is right.

G3 N2023 · Paper 1 · Q3 3 marks

Simplify (2c³d²)³ [1]
16 × 5⁴ + 9 × 25² = 5^k
Use the laws of indices to find the value of k.
Show your working. [2]

Worked solution

(a) Law 4 hands the outside index to every factor inside the bracket:

(2c³d²)³ = 2³ × (c³)³ × (d²)³ = 8c⁹d⁶

(b) 25 = 5², so 25² = (5²)² = 5⁴ (Law 3). Both terms now carry the same power of 5:

16 × 5⁴ + 9 × 5⁴ = (16 + 9) × 5⁴ = 25 × 5⁴

= 5² × 5⁴ = 5^{2 + 4} = 5⁶ (Law 1)

So k = 6.

Check: 16 × 625 + 9 × 625 = 10 000 + 5625 = 15 625, and 5⁶ = 15 625.

Why this works. "Show your working" makes part (b) a three-line answer rather than a number. There is no law for adding powers, so give the two terms a common factor — which needs 25² rewritten as 5⁴ — and take it out. Hold on to the last step: 25 is 5², not a left-over coefficient, and folding it in makes one power.

What standard form is

Singapore's population in June 2020 was about 5 850 000; the mass of an oxygen atom is about 0.000 000 000 000 000 000 000 000 026 56 kg. Standard form is what removes the zero-counting from both.

A number is said to be expressed in standard form when it is written as

A \times 10^{n}

, where 1 ≤ A < 10 and n is an integer.

Written as	Standard form?	Why
6.3 × 10⁴	yes	1 ≤ 6.3 < 10, and 4 is an integer
63 × 10³	no	`A` = 63 is not less than 10 → 6.3 × 10⁴
0.72 × 10⁻⁵	no	`A` = 0.72 is less than 1 → 7.2 × 10⁻⁶
1.0 × 10¹²	yes	1 ≤ 1.0 < 10; an `A` of exactly 1 is allowed
8.4 × 10^2.5	no	the index must be an integer
4.5 × 2⁷	no	the base must be 10

A negative index means a small number, not a negative one. 5.3 × 10⁻¹¹ is 0.000 000 000 053 — positive, and tiny. A negative number in standard form carries its minus sign on the A: −5.3 × 10¹¹.

Why this works

Both conditions do real work. "1 ≤ A < 10" forces exactly one non-zero digit in front of the decimal point, so every number has exactly one standard form and two numbers can be compared at a glance. "n is an integer" is what lets Law 1 and Law 2 be used on the powers of ten.

Converting, both ways

Converting to standard form is counting: how many places must the decimal point move to leave exactly one non-zero digit in front of it?

Fig. 1.2

Count the places the decimal point has to travel to sit just after the first non-zero digit. Travelling left gives a positive index; travelling right gives a negative one. The number itself never changes — only the way it is written.

Check the sign of the index. A number bigger than 10 ends up with a positive index; a number smaller than 1 ends up with a negative one. So 0.000 061 is 6.1 × 10⁻⁵, never 6.1 × 10⁵.

Walkthrough 12 — converting both ways Basic

(a) Express 58 400 000 in standard form. (b) Express 9.06 × 10⁻⁵ in ordinary notation.

58 400 000 = 5.84 × 10 000 000

Standard form allows exactly one non-zero digit in front of the decimal point, so the point goes straight after the 5. Everything taken out of the number has to be given back as a multiplier — here, ten million.
= 5.84 × 10⁷

10 000 000 has 7 zeros, so it is 10⁷ — the same 7 as the places the point moved left. A = 5.84 is between 1 and 10 and the index is an integer, so this is standard form.
9.06 × 10⁻⁵ = 9.06 ÷ 10⁵

By Definition 2 a negative power of ten is a division: 10⁻⁵ = $\frac{1}{10^{5}}$ . Dividing is what makes the number small; the number itself is still positive.
= 0.000 090 6

Dividing by 10⁵ moves the decimal point 5 places to the left, so four zeros appear between the point and the 9. Ordinary notation means every digit written out with no power of ten left anywhere.

Why this works

Multiplying by 10 moves every digit one place up the place-value columns, which looks like the decimal point moving one place to the right; dividing by 10 moves it one place to the left. Each move is one factor of 10, so the number of moves is the index — which is why counting places and counting powers give the same answer.

Past-paper questions

G3 N2021 · Paper 2 · Q2(a) 1 mark

The value of fuels and lubricants exported from Singapore in 2017 was $90 182 million.
Write $90 182 million in standard form correct to 3 significant figures. [1]

Worked solution

"Million" is 10⁶, so

$90 182 million = $90 182 × 10⁶ = $9.0182 × 10⁴ × 10⁶ = $9.0182 × 10¹⁰ (Law 1)

= $9.02 × 10¹⁰ (3 s.f.)

Why this works. The word million is worth 10⁶ and has to become a power before anything is counted; then 90 182 = 9.0182 × 10⁴, and Law 1 adds 4 + 6 = 10. Round last and round only A: 9.0182 to 3 s.f. is 9.02, and the × 10¹⁰ is not part of the count. Keep the dollar sign.

Arithmetic in standard form

There are only two situations here, and they behave completely differently: × and ÷ on one side, + and − on the other.

Multiplying and dividing. Handle the A-numbers and the powers of ten separately:

(A \times 10^{m}) \times (B \times 10^{n}) = (A B) \times 10^{m + n}

The powers of ten obey Laws 1 and 2, so the indices are added when multiplying and subtracted when dividing. The powers need not match.

Adding and subtracting. There is no law of indices for a sum. The two terms must first be written with the same power of ten; then that power is a common factor:

A \times 10^{n} + B \times 10^{n} = (A + B) \times 10^{n}

Finish by re-checking A. 3.2 × 7.5 = 24, and 1.2 ÷ 6 = 0.2; neither is between 1 and 10. Fix A and move the index to match: 24 × 10³ = 2.4 × 10⁴.

Walkthrough 13 — multiplying in standard form Basic

Calculate (3.2 × 10⁵) × (7.5 × 10⁻²), giving your answer in standard form.

= (3.2 × 7.5) × (10⁵ × 10⁻²)

Multiplication may be reordered, so the two A-numbers go together and the two powers of ten go together. Each pair is then a separate and much easier calculation.
= 24 × 10^{5 + (−2)} (Law 1 of Indices)

3.2 × 7.5 = 24 for the numbers, and Law 1 adds the indices of the powers of ten because the base 10 is the same in both. Adding a negative index brings the power down.
= 24 × 10³

5 + (−2) = 3. The arithmetic on the number line is worth writing out: it is where sign slips happen.
= 2.4 × 10⁴

24 is not between 1 and 10, so the previous line is a correct value but not standard form. Writing 24 as 2.4 × 10¹ and using Law 1 again gives 2.4 × 10^{1 + 3} = 2.4 × 10⁴.

Walkthrough 14 — adding, with unequal powers Intermediate

Calculate 3.7 × 10⁶ + 4.5 × 10⁵, giving your answer in standard form.

4.5 × 10⁵ = 0.45 × 10⁶

The two powers of ten differ, and there is no law of indices for a sum. Raising the index from 10⁵ to 10⁶ multiplies by ten, so the A-number is divided by ten to keep the value the same.
3.7 × 10⁶ + 4.5 × 10⁵ = 3.7 × 10⁶ + 0.45 × 10⁶

Substitute the rewritten term. Both terms are now "so many lots of 10⁶", which is exactly the condition that makes them addable — the same condition as adding like terms in algebra.
= (3.7 + 0.45) × 10⁶

10⁶ is now a common factor of both terms, so it comes outside a bracket and only the A-numbers are added. The power of ten is not added to anything.
= 4.15 × 10⁶

3.7 + 0.45 = 4.15, and 1 ≤ 4.15 < 10, so this is already in standard form and no adjusting of the index is needed. (A quick check: the answer should be a little more than 3.7 million, and it is.)

Walkthrough 15 — subtracting, then dividing Intermediate

Evaluate

\frac{6.4 \times 10^{−3} - 5.2 \times 10^{−4}}{2.7 \times 10^{2}}

, giving your answer in standard form, correct to 3 significant figures.

5.2 × 10⁻⁴ = 0.52 × 10⁻³

The numerator is a subtraction, so both terms need the same power of ten before anything can be taken away. Raising the index from 10⁻⁴ to 10⁻³ is compensated by moving the decimal point of 5.2 one place to the left.
6.4 × 10⁻³ − 0.52 × 10⁻³ = (6.4 − 0.52) × 10⁻³ = 5.88 × 10⁻³

With a common power of ten, 10⁻³ is a common factor and only the A-numbers are subtracted. 5.88 is between 1 and 10, so the numerator is itself in standard form.
$\frac{5.88 \times 10^{−3}}{2.7 \times 10^{2}} = \frac{5.88}{2.7} \times \frac{10^{−3}}{10^{2}}$

Split the division into "numbers ÷ numbers" and "power ÷ power". The whole numerator had to be collapsed into one term first — dividing each of its two pieces by 2.7 × 10² separately would be a different, longer calculation.
= 2.1777… × 10⁻⁵ (Law 2: −3 − 2 = −5)

Law 2 subtracts the indices of the powers of ten, and 5.88 ÷ 2.7 does not terminate. Carry the extra decimals on this line: rounding here and then continuing is what pushes a final answer outside the accepted range.
= 2.18 × 10⁻⁵ (3 s.f.)

Round once, here: 2.1777… to 3 s.f. is 2.18. Count the significant figures on the A-number only — the × 10⁻⁵ is not part of them.

Why this works

Laws 1 and 2 are laws about products of powers, which is why multiplying and dividing can treat the A-numbers and the powers of ten as two separate jobs. A sum has no such law, so the only way through is to make the power of ten a common factor — the same move as collecting like terms in algebra, with 10ⁿ playing the part of the letter.

The final tidy-up goes both ways: 24 × 10³ = 2.4 × 10⁴ when A is too big, and 0.2 × 10³ = 2 × 10² when it is too small. Each time the digits move one way and the index moves the other, so the value never changes.

Past-paper questions

G3 N2015 · Paper 1 · Q4 2 marks

A field has an area of 1400 m².
An average of 60 dandelion plants grow on each square metre.
Each plant has an average of 20 flowers.
Joel estimates that these plants will produce 2 × 10⁸ seeds in total.
Calculate an estimate of the average number of seeds produced by each flower. [2]

Worked solution

Number of flowers = 1400 × 60 × 20 = 1 680 000 = 1.68 × 10⁶

Average number of seeds per flower $= \frac{2 \times 10^{8}}{1.68 \times 10^{6}} = \frac{2}{1.68} \times 10^{2}$ (Law 2: 8 − 6 = 2)

= 1.190 47… × 10² = 119.047…

= 119 (3 s.f.)

Why this works. An average is a total divided by a count, so the work is deciding what to divide by: the flowers, 1400 × 60 × 20. Writing that as 1.68 × 10⁶ splits the division in two — A-numbers divide, Law 2 subtracts the indices. The answer counts seeds, so it is not standard form; it is not exact, so 119.047… becomes 119.

G3 N2019 · Paper 2 · Q3(c) 6 marks

The table shows information about telecommunications in Singapore.

Year	2010	2015
Mobile phone subscriptions	7.29 × 10⁶	8.21 × 10⁶
SMS messages sent	2.77 × 10¹⁰	1.14 × 10¹⁰
Broadband subscriptions	7.85 × 10⁶	1.20 × 10⁷

Calculate how many more mobile phone subscriptions there were in 2015 than in 2010. Give your answer in standard form. [1]
Calculate the percentage increase in the number of broadband subscriptions from 2010 to 2015. [2]
The population of Singapore in 2015 was 5.54 × 10⁶.
Calculate the mean number of SMS messages sent per person per day in 2015. [3]

Worked solution

(a) Both figures already carry 10⁶, so the A-numbers subtract directly:

8.21 × 10⁶ − 7.29 × 10⁶ = 0.92 × 10⁶ = 9.2 × 10⁵

(b) Write 1.20 × 10⁷ as 12.0 × 10⁶ so that the two broadband figures share an index:

Increase = 12.0 × 10⁶ − 7.85 × 10⁶ = 4.15 × 10⁶

Percentage increase = $\frac{4.15 \times 10^{6}}{7.85 \times 10^{6}}$ × 100% = 52.866…%

= 52.9% (3 s.f.)

(c) A year is 365 days, so the number of person-days in 2015 is 5.54 × 10⁶ × 365 = 2.0221 × 10⁹.

Mean $= \frac{1.14 \times 10^{10}}{2.0221 \times 10^{9}} = 5.6377…$

= 5.64 messages per person per day (3 s.f.)

Why this works. Terms in standard form add or subtract only once their powers of ten match. In (a) they already do, so the work is at the end: 0.92 × 10⁶ becomes 9.2 × 10⁵. In (b) rewrite 1.20 × 10⁷ as 12.0 × 10⁶ first, measuring the increase against 2010; (c) needs the right row, the right year and per day.

G3 N2023 · Paper 2 · Q1(b) 2 marks

Population density is defined as the number of people per square kilometre.
In 2020, the population density of Singapore was 7810 people per square kilometre.
In 2020, the total population of Singapore was 5.69 million.
Calculate the area of Singapore. [2]

Worked solution

The definition says $density = \frac{population}{area}$ , so $area = \frac{population}{density}$ .

Population = 5.69 million = 5.69 × 10⁶

Area $= \frac{5.69 \times 10^{6}}{7810} = 728.55…$

= 729 km² (3 s.f.)

Why this works. The question hands you the definition, so the first mark is for turning it the right way round: people per km² means population over area, so area is population over density. The units are the check — people ÷ (people / km²) leaves km². "5.69 million" becomes 5.69 × 10⁶ before dividing, and the rounding happens once, at the end.

G3 N2024 · Paper 1 · Q18 3 marks

Write 263 in standard form. [1]
1. Write 3.4 × 10⁹⁹ in the form A × 10¹⁰⁰. [1]
2. Work out (4.7 × 10¹⁰⁰) + (3.4 × 10⁹⁹).
  Give your answer in standard form. [1]

Worked solution

(a) The point moves two places left, from 263. to 2.63, so the index is 2: 2.63 × 10²

(b)(i) The index is to go up by one, from 99 to 100, so the A-number must come down by a factor of ten: 0.34 × 10¹⁰⁰

(b)(ii) Using (b)(i), both terms now carry 10¹⁰⁰:

(4.7 × 10¹⁰⁰) + (3.4 × 10⁹⁹) = 4.7 × 10¹⁰⁰ + 0.34 × 10¹⁰⁰

= (4.7 + 0.34) × 10¹⁰⁰ = 5.04 × 10¹⁰⁰

Why this works. Part (b)(i) hands you an illegal intermediate on purpose: 0.34 is not a legal A, but matching the indices is the only way to add at 10¹⁰⁰. Watch the direction of the trade — index up means the number in front gets smaller (3.4 → 0.34); the other way gives 38.7 × 10¹⁰⁰. Then check A: 5.04 is legal.

Doing it on the calculator

Every scientific calculator has a dedicated standard-form key, usually marked ×10^x (older models: EXP or EE). One keystroke enters "× 10 to the power of", and the display then shows the answer in standard form.

To enter	Press	Not
3.2 × 10⁵	3 . 2 ×10^x 5	3.2 × 10 ×10^x 5, which is ten times too big
4.19 × 10⁻⁴	4 . 1 9 ×10^x (−) 4	the subtraction key −, which the calculator reads as an operation
a whole numerator over a whole denominator	put brackets round each, or use the fraction key	typing it straight across, which divides by the first term only

Write the working, not just the answer. Put the substitution line on the page, and write the unrounded value before you round it. A method that is right but for one digit still earns most of the marks.

Significant figures are counted on A only. So 6.817 03 × 10⁻⁶ becomes 6.82 × 10⁻⁶, and 0.000 06 × 10³ is not an answer at all until A is fixed.

More detail

The keys vary between models, so check the manual for yours. On nearly all of them the ×10^x key already contains the × 10, which is why pressing × 10 as well gives an answer ten times too big.

Past-paper questions

G3 N2020 · Paper 1 · Q1 1 mark

Calculate $\frac{- (−11) - \sqrt{{(−11)}^{2} - 4 \times 16 \times (−50)}}{2 \times 16}$ . [1]

Worked solution

Inside the root: (−11)² − 4 × 16 × (−50) = 121 + 3200 = 3321, and $\sqrt{3321} = 57.628…$

Numerator = 11 − 57.628… = −46.628…; denominator = 32

$= \frac{−46.628…}{32} = −1.4571…$ = −1.46 (3 s.f.)

Why this works. The single mark is for typing it as one expression: brackets round the whole numerator, brackets round 2 × 16, or the fraction key. Watch both signs — −(−11) is +11, and −4 × 16 × (−50) is +3200, so the number under the root is bigger than 121. Keep 57.628… unrounded on the way through, and round once, to 3 s.f.

G3 N2022 · Paper 1 · Q1 1 mark

Calculate $\sqrt[5]{{12.5}^{2} - \frac{6.8}{0.037}}$ . [1]

Worked solution

Inside the root: 12.5² = 156.25 and $\frac{6.8}{0.037} = 183.783…$ , so

12.5² − $\frac{6.8}{0.037}$ = 156.25 − 183.783… = −27.5337…

A fifth root is an odd root, so a negative number has one:

$\sqrt[5]{−27.5337…} = −1.9407…$ = −1.94 (3 s.f.)

Check: (−1.9407…)⁵ = −27.53…, the number under the root.

Why this works. The whole bracket sits under the root, so work out 12.5² − 6.8 ÷ 0.037 first and take the fifth root of that. The result is negative, and $\sqrt[5]{x}$ is $x^{\frac{1}{5}}$ with an odd denominator, so a negative number does have one. Some calculators refuse ${(−27.5…)}^{\frac{1}{5}}$ : take the root of 27.5337… and put the minus sign back.

Standard form in the real world

Once every quantity is written as A × 10ⁿ, the index alone tells you the magnitude — roughly how big something is — and two quantities can be compared without reading a single zero.

Fig. 1.3

The index is the magnitude. A person is more than 10¹⁰ times as long as a hydrogen atom's radius, and a light-year is about 10¹⁶ times as long as a person — both read straight off the scale, with no zeros counted.

SI prefixes are standard form with a name attached. Every one of them stands for a power of ten, so a measurement given with a prefix converts to standard form in one step.

Prefix	Symbol	Power of 10	Prefix	Symbol	Power of 10
tera-	T	10¹²	milli-	m	10⁻³
giga-	G	10⁹	micro-	µ	10⁻⁶
mega-	M	10⁶	nano-	n	10⁻⁹
kilo-	k	10³	pico-	p	10⁻¹²

Converting a prefix is Law 1. 250 megawatts = 250 × 10⁶ W = 2.5 × 10² × 10⁶ W = 2.5 × 10⁸ W. Write the 250 in standard form first, then add the indices.

Walkthrough 16 — how far away is the nearest star? Advanced

Light travels at 3.00 × 10⁸ m/s. A light-year is the distance light travels in one year; take one year to be 3.15 × 10⁷ s. Proxima Centauri, the nearest star to the Sun, is about 4.2 light-years away. Find its distance from the Sun in metres, giving your answer in standard form correct to 3 significant figures.

distance in one year = (3.00 × 10⁸) × (3.15 × 10⁷)

Distance = speed × time, so one light-year is the speed of light multiplied by the number of seconds in a year. Both quantities are already in standard form and both are in SI units, so nothing has to be converted first.
= (3.00 × 3.15) × (10⁸ × 10⁷) = 9.45 × 10¹⁵ m

Numbers with numbers, powers with powers; Law 1 adds the indices, 8 + 7 = 15. A = 9.45 is already between 1 and 10, so one light-year is 9.45 × 10¹⁵ m.
distance to Proxima Centauri = 4.2 × (9.45 × 10¹⁵)

4.2 light-years means 4.2 lots of that distance. The 4.2 is a plain multiplier with no power of ten attached, so it joins the A-number and leaves the 10¹⁵ untouched.
= 39.69 × 10¹⁵

4.2 × 9.45 = 39.69. This is a correct value, but 39.69 is not between 1 and 10, so it is not yet in the form the question asked for.
= 3.969 × 10¹⁶ = 3.97 × 10¹⁶ m (3 s.f.)

39.69 = 3.969 × 10¹, and adding that 1 to the index gives 10¹⁶. Rounding 3.969 to 3 significant figures gives 3.97, and the unit m is part of the answer — the question asked for a distance in metres.

Check yourself

Express (a) 0.000 000 040 7 in standard form, and (b) 6.2 × 10⁴ in ordinary notation.
Answer
(a) The decimal point moves 8 places to the right to sit after the 4, and the number is smaller than 1, so the index is negative: 4.07 × 10⁻⁸.
(b) 6.2 × 10 000 = 62 000.
Calculate (8.4 × 10⁻⁵) ÷ (2.4 × 10³), giving your answer in standard form.
Answer
= $\frac{8.4}{2.4}$ × 10^{−5 − 3}
= 3.5 × 10⁻⁸
= 3.5 × 10⁻⁸. A = 3.5 is already between 1 and 10, so nothing needs adjusting.
Calculate 5.2 × 10⁴ − 7.5 × 10³, giving your answer in standard form.
Answer
Make the powers match: 7.5 × 10³ = 0.75 × 10⁴
= (5.2 − 0.75) × 10⁴
= 4.45 × 10⁴. Subtracting 5.2 − 7.5 without matching the powers first would give a negative answer, which is nonsense here.
Given that P = 4.5 × 10⁶ and Q = 9.0 × 10⁻², find (a) PQ and (b) $\frac{P}{Q}$ , giving each answer in standard form.
Answer
(a) (4.5 × 9.0) × 10^{6 + (−2)} = 40.5 × 10⁴ = 4.05 × 10⁵.
(b) $\frac{4.5}{9.0}$ × 10^{6 − (−2)} = 0.5 × 10⁸ = 5 × 10⁷. Both needed the final tidy-up — one because A was too big, the other because it was too small.
A red blood cell is about 6.2 × 10⁻⁶ m wide. How many of them, laid side by side, would span 1 cm? Give your answer in standard form correct to 3 significant figures.
Answer
1 cm = 1 × 10⁻² m, so the number of cells is
$\frac{1 \times 10^{−2}}{6.2 \times 10^{−6}}$ = $\frac{1}{6.2}$ × 10^{−2 − (−6)}
= 0.161 29… × 10⁴ = 1.612 9… × 10³
≈ 1.61 × 10³ cells (3 s.f.). Converting the centimetre to metres first is essential — the two lengths must be in the same unit.
Explain why 47 × 10⁵ is not in standard form, and write it correctly.
Answer
Standard form requires 1 ≤ A < 10, and here A = 47, which is too big.
47 × 10⁵ = 4.7 × 10¹ × 10⁵
= 4.7 × 10⁶ (Law 1). The value has not changed — only the way it is written.

Simple interest, and what compounding changes

In Book 1 you met simple interest: the interest I a person earns from a bank depends on the money deposited (the principal P), the interest rate R% per annum, and the duration in years T.

I = \frac{P R T}{100}

Compound interest asks a different question: what if the interest earned each year is added to the principal, so that interest is earned on the interest as well?

If the interest paid on a sum of money is compounded annually (or yearly), then

$A = P {(1 + \frac{r}{100})}^{n}$

where A is the total amount, P is the principal, r% is the interest rate per annum and n is the number of years.

Put $1000 in the bank for 3 years at 2% and the two ideas part company:

	Bank A — simple interest 2% per annum	Bank B — 2% per annum compounded yearly
Year 1	interest on $1000 = $20 → $1020	interest on $1000 = $20 → $1020
Year 2	interest on $1000 = $20 → $1040	interest on $1020 = $20.40 → $1040.40
Year 3	interest on $1000 = $20 → $1060	interest on $1040.40 = $20.81 → $1061.21
Interest earned	$60	$61.21

The principal changes every year. In the simple column the interest is worked out on $1000 three times; in the compound column on $1000, then $1020, then $1040.40. The formula does that repeatedly without the table.

Amount, or interest?

A

is everything in the account at the end; the compound interest is what was gained,

I = A - P

. Unless a question says otherwise, leave money to the nearest cent.

Walkthrough 17 — finding compound interest Basic

Find the compound interest on $4000 for 5 years at 4% per annum, compounded annually.

$A = P {(1 + \frac{r}{100})}^{n}$ with P = 4000, r = 4, n = 5

"Compounded annually" means the formula is used in its plain form: r is the rate per year and n counts years. Write the three values down before substituting.
$A = 4000 {(1 + \frac{4}{100})}^{5}$

Substitute. The rate goes in as the number 4, not as 0.04 — the ÷ 100 is already built into the formula.
= $4866.6116…

One calculator line: the bracket is 1.04, raised to the power 5, times 4000. Keep the full display for now — this is a stepping stone, not the answer.
Compound interest $I = A - P$ = $4866.6116… − $4000

The question asked for the interest, and the formula gave the total amount. Subtracting the principal is the step that answers the question actually asked.
= $866.61 (to the nearest cent)

Round once, at the end, to the nearest cent as money questions expect. Rounding A to $4866.61 first happens to give the same answer here, but on a longer question that habit is what shifts a final cent.

Walkthrough 18 — the long way, and why the formula agrees Basic

$800 is invested at 5% per annum, compounded yearly. Find the amount at the end of each of the first three years, and check the third against the compound interest formula.

Year 1: interest = 5% of $800 = $40 → $A_{1}$ = $800 + $40 = $840

The first year is the only one where simple and compound interest agree: the principal really is the original $800, and 5% of 800 is 40.
Year 2: interest = 5% of $840 = $42 → $A_{2}$ = $840 + $42 = $882

Here is the whole difference: the 5% is charged on $840, last year's total, not on the original $800. That is what "compounded" means, and it is why the interest rose from $40 to $42.
Year 3: interest = 5% of $882 = $44.10 → $A_{3}$ = $882 + $44.10 = $926.10

The same move again, now on $882. Each year's interest is bigger than the last because the base it is charged on keeps growing — the growth is not a straight line.
$A = 800 {(1 + \frac{5}{100})}^{3} = 800 \times {1.05}^{3}$

The formula does those same three multiplications in one line: multiplying by 1.05 is exactly "keep the total and add 5% of it", and doing that three times is 1.05³.
= $926.10 — the same answer

The table and the formula must agree, because the formula was built by factorising the table. Use the table when a question asks what happens in a particular year; use the formula when it asks for the total after n years.

Where the formula comes from

Year by year is fine for 3 years and hopeless for 10. Follow the same three years algebraically, taking out a common factor each time, and the pattern appears:

Total amount at the end of the 1^st year, $A_{1} = P_{1} + I_{1} = 1000 + 1000 \times \frac{2}{100} = 1000 (1 + \frac{2}{100})$

Total amount at the end of the 2^nd year, $A_{2} = 1000 (1 + \frac{2}{100}) + 1000 (1 + \frac{2}{100}) \times \frac{2}{100} = 1000 {(1 + \frac{2}{100})}^{2}$

Total amount at the end of the 3^rd year, $A_{3} = 1000 {(1 + \frac{2}{100})}^{3}$

Each line is the previous total, plus 2% of that total — and each time the previous total comes out as a common factor. The index counts the years.

The longer story

For simple interest the interest earned every year is the same, because it is always calculated on the original principal. Compound interest adds each year's interest to the principal, so the next year's interest is charged on a larger sum.

Over the three years above, Bank B pays $1.21 more than Bank A. The gap is small over 3 years at a small rate — which is exactly why the question is worth asking over 10 years, or 20.

The one formula in this chapter that is printed for you. The paper gives "Compound interest — Total amount = $P {(1 + \frac{r}{100})}^{n}$ " as the first entry on its own formulae page, so no marks ride on recalling it. They go on choosing r and n correctly, and on knowing that the formula returns the total amount rather than the interest.

When interest is added more often than once a year

Banks rarely compound only once a year. The formula does not change — but what r and n mean does.

If the interest is compounded monthly, then r% is the interest rate per month and n is the number of months.

The same reading applies to every compounding period: the annual rate is shared evenly across the periods in a year, and n counts periods.

Compounded…	periods in a year	rate `r` per period	`n` for `t` years	6% per annum for 4 years becomes
yearly	1	the annual rate	`t`	`r` = 6, `n` = 4
half-yearly	2	annual rate ÷ 2	2`t`	`r` = 3, `n` = 8
quarterly	4	annual rate ÷ 4	4`t`	`r` = 1.5, `n` = 16
monthly	12	annual rate ÷ 12	12`t`	`r` = 0.5, `n` = 48

Both change, or neither. Divide the annual rate by the periods in a year, and multiply the years by that same number: r is the rate for one period, n the number of periods. Say in your working what one period is.

Walkthrough 19 — compounded half-yearly Intermediate

$9000 is deposited in an account paying 6% compound interest per annum, compounded half-yearly. Find the total amount in the account after 4 years.

One period = 6 months, so r = 6 ÷ 2 = 3

"Half-yearly" means the interest is added twice a year, so each addition is half of the annual 6%. The rate in the formula must be the rate for one compounding period, never the annual headline rate.
n = 4 × 2 = 8 half-year periods

n counts the very periods that r describes. Four years contains eight half-years, so the interest is added eight times. Halving the rate without doubling n would model a bank that pays half as much interest.
$A = 9000 {(1 + \frac{3}{100})}^{8}$

Substitute the period values, not the yearly ones. The principal is untouched by any of this — only r and n are affected by how often the bank compounds.
= $11 400.9307…

1.03 raised to the power 8, times 9000. Keep the full display: the cents are decided by digits not yet on screen.
= $11 400.93 (to the nearest cent)

The question asked for the total amount, so this is the answer as it stands — no subtracting the principal here. (Compounded yearly instead, the same deposit would reach only $11 362.29: compounding more often earns more.)

Walkthrough 20 — compounded monthly Intermediate

Find the compound interest on $2400 for 2 years at 6% per annum, compounded monthly.

r = 6 ÷ 12 = 0.5 (per month)

Twelve compounding periods in a year, so each month carries a twelfth of the annual rate. A rate of 0.5 is perfectly legal in this formula — r does not have to be a whole number.
n = 2 × 12 = 24 months

Two years is twenty-four months, and interest is added at the end of each one. Write the units beside n in your working — "24 months" makes it obvious to you and to the marker that r is a monthly rate.
$A = 2400 {(1 + \frac{0.5}{100})}^{24} = 2400 \times {1.005}^{24}$

0.5 ÷ 100 = 0.005, so the bracket is 1.005 — a monthly growth factor. Notice how close to 1 it is: a small step, taken twenty-four times.
= $2705.1834…

Evaluate in one line. Rounding 1.005²⁴ to a few decimals before multiplying is exactly where money answers drift by cents.
$I = A - P$ = $2705.1834… − $2400 = $305.18

"Compound interest" again means the gain, so subtract the principal and round at the end. It is a shade more than the $296.64 that yearly compounding at the same 6% would have given.

Working backwards, and comparing offers

The formula has four letters. Given any three, a question can ask for the fourth — and in the real-world question at the end of Paper 2 it usually asks you to compare two offers and say which is better.

Walkthrough 21 — finding the interest rate Intermediate

Farid invested $7000 in a fund that pays interest compounded yearly. At the end of 4 years he received a total of $8032.66. Find the interest rate.

$8032.66 = 7000 {(1 + \frac{r}{100})}^{4}$

The same formula, a different unknown. The total received is A, so substitute all three known values and leave r standing — this is now an equation to solve, not a calculation to evaluate.
${(1 + \frac{r}{100})}^{4} = \frac{8032.66}{7000} = 1.147522…$

Divide both sides by the principal to get the bracket on its own. What is left is the growth factor over the whole 4 years: the money ended up about 1.1475 times what it started as.
$1 + \frac{r}{100} = \sqrt[4]{1.147522} = 1.035$

To undo a power of 4, take the 4th root — the root and the index cancel. This is a fractional index doing the work: the 4th root is the power $\frac{1}{4}$ , exactly as in 1.2.
$\frac{r}{100} = 0.035$ , so $r = 3.5$

Subtract the 1 that stands for the original money, leaving only the growth, then multiply by 100 to turn the decimal back into a percentage. Keeping the 1 would read as a rate of 103.5%.
The interest rate is 3.5% per annum.

Answer in the units the question used — a rate, with a percent sign and "per annum". (The root came out as 1.034999… rather than exactly 1.035 only because the $8032.66 in the question was itself rounded to the nearest cent.)

Walkthrough 22 — which offer is better? Advanced

Devi has $12 000 to invest for 5 years. Company A offers 5.2% per annum simple interest. Company B offers 5% per annum compound interest, compounded half-yearly. (i) Which company should she invest in? Explain your answer. (ii) Calculate the difference in interest earned after 5 years.

Company A: $I = \frac{P R T}{100} = \frac{12000 \times 5.2 \times 5}{100} = $ 3120$

A is simple interest, so the compound formula does not apply to it at all — this is the Book 1 formula, and it gives the interest directly, with no subtraction. Deal with the two companies one at a time.
Company B: one period = 6 months, so r = 5 ÷ 2 = 2.5 and n = 5 × 2 = 10

B compounds half-yearly, so both the rate and the count must be converted to half-years before anything is substituted. Note that B's headline rate (5%) is lower than A's 5.2% — so the answer is not obvious, which is the point of the question.
$A = 12000 {(1 + \frac{2.5}{100})}^{10} = $ 15361.01$

1.025 to the power 10, times 12 000. This is B's total amount, so it cannot yet be set against A's $3120 — one is a total, the other an interest.
Company B interest = $15 361.01 − $12 000 = $3361.01

Subtract the principal so that both companies are described by the same quantity: interest earned. Comparing like with like is what this question is really testing.
(i) $3361.01 > $3120, so Devi should invest in Company B.

The explanation being asked for is the comparison itself: B earns more interest over the 5 years. Compounding twice a year beats A's slightly higher rate here, because B's interest goes on to earn interest itself.
(ii) Difference = $3361.01 − $3120 = $241.01

"Difference in interest" is a subtraction of the two interests — not of the two totals, and not a percentage. State it as money, to the nearest cent.

Check yourself — 1.4

Find the compound interest on $3000 for 4 years at 5% per annum, compounded annually.
Answer
A = 3000(1 + 5/100)⁴ = 3000 × 1.05⁴ = $3646.5187…
I = A − P = 3646.5187… − 3000 = $646.52 (to the nearest cent).
Find the total amount when $5000 is invested for 3 years at 4% per annum, compounded quarterly.
Answer
Quarterly: r = 4 ÷ 4 = 1 and n = 3 × 4 = 12.
A = 5000(1 + 1/100)¹² = 5000 × 1.01¹² = $5634.13 (to the nearest cent).
$1200 is invested at 3% per annum compounded yearly. Show that the interest earned in the second year is larger than in the first.
Answer
Year 1: interest = 3% of $1200 = $36, so A₁ = $1236.
Year 2: interest = 3% of $1236 = $37.08.
$37.08 > $36 because the second year's 3% is charged on the larger principal $1236 — the first year's interest is itself now earning interest.
Rahim deposited $2000 in an account paying compound interest yearly. After 3 years he had $2249.73. Find the interest rate.
Answer
(1 + r/100)³ = 2249.73 ÷ 2000 = 1.124865
1 + r/100 = $\sqrt[3]{1.124865}$ = 1.04
r/100 = 0.04, so r = 4 — the rate is 4% per annum.
One bank offers 4% per annum compounded yearly; another offers 4% per annum compounded monthly. Without calculating, say which pays more, and why.
Answer
The monthly one. Both quote the same 4% a year, but the monthly account adds a twelfth of it at the end of every month, so from month 2 onwards the interest is charged on a principal that has already grown — the interest starts earning interest sooner. The more often interest is compounded, the larger the total.

Past-paper questions

G3 N2024 · Paper 1 · Q16 3 marks

Alice invested some money into an account paying compound interest at 3.2% per year.
After 5 years, the money had earned total interest of $2132.16 .

Calculate the amount of money Alice invested in the account. [3]

Worked solution

Let the amount invested be $P. After 5 years the account holds

$P {(1 + \frac{3.2}{100})}^{5} = P \times {1.032}^{5}$

The interest is that total minus the money put in, so

P × 1.032⁵ − P = 2132.16

P(1.032⁵ − 1) = 2132.16, and 1.032⁵ = 1.170 572 95…, so

0.170 572 95… P = 2132.16

$P = \frac{2132.16}{0.170 572 95…} = 12 499.988…$

Alice invested $12 500 (to the nearest cent the division itself gives $12 499.99).

Check: $12 500 × 0.170 572 95… = $2132.1619…, which is the printed $2132.16 to the cent.

Why this works. Everything hinges on the word interest: $2132.16 is what the money earned, so the equation is total − principal = interest. Taking P out as a common factor, P(1.032⁵ − 1), is the credited line, and the bracket is the 5-year interest per dollar invested, about 17.06 cents. Keep 1.032⁵ unrounded — rounding it to 1.17 gives $12 542.

The whole chapter in one table

If you can reproduce this table from memory, you have the skeleton of the topic. Everything else hangs off it.

	What it says	Use it when	The classic slip
Law 1	$a^{m} \times a^{n} = a^{m + n}$	two powers of the same base are multiplied	using it when the bases differ
Law 2	$a^{m} \div a^{n} = a^{m - n}$	two powers of the same base are divided	subtracting the wrong way round
Law 3	${(a^{m})}^{n} = a^{m n}$	a power is raised to another power	adding the indices instead of multiplying
Law 4	$a^{n} b^{n} = {(a b)}^{n}$	a bracketed product is raised to a power	leaving the coefficient unraised
Law 5	${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	a bracketed fraction is raised to a power	raising the numerator only
Compound interest	$A = P {(1 + \frac{r}{100})}^{n}$	interest is added to the principal each period (the one formula here that is printed on the paper)	using the annual rate when it compounds half-yearly or monthly; giving `A` when the interest `A` − `P` was asked for
Zero index	$a^{0} = 1$ , `a` ≠ 0	any base at all carries the index 0	reading 2`x`⁰ as (2`x`)⁰
Negative index	$a^{- n} = \frac{1}{a^{n}}$ , `a` ≠ 0	an index is negative, or "positive index form" is asked for	making the answer negative instead of a reciprocal
Rational index	$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ , `a` ≥ 0	an index is a fraction	swapping the root and the power
Unknown index	if $a^{x} = a^{n}$ then `x` = `n`	the unknown sits in the index	trying to take a root instead of matching the bases
Standard form	$A \times 10^{n}$ , 1 ≤ `A` < 10, `n` an integer	a number is very large or very small	leaving `A` outside the range 1 to 10
× and ÷	multiply or divide the `A`s; add or subtract the indices	the powers of ten may differ freely	forgetting the final re-tidy of `A`
+ and −	make the powers of ten equal, then add or subtract the `A`s	the two terms are added or subtracted	adding the `A`s while the powers still differ

The formulae to know by heart

The five Laws of Indices. Bases a, b real and indices m, n rational; Laws 1–3 then need a > 0, and Laws 4–5 need a > 0 and b > 0.

a^{m} \times a^{n} = a^{m + n} a^{m} \div a^{n} = a^{m - n} {(a^{m})}^{n} = a^{m n}

a^{n} \times b^{n} = {(a \times b)}^{n} a^{n} \div b^{n} = {(\frac{a}{b})}^{n}

The three definitions. The first two need a ≠ 0; the third needs a ≥ 0 with n a positive integer.

a^{0} = 1 a^{- n} = \frac{1}{a^{n}} a^{\frac{1}{n}} = \sqrt[n]{a}

In general $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}$ — bottom is the root, top is the power.

Standard form.

A \times 10^{n}

where 1 ≤ A < 10 and n is an integer.

Reading the index. Large numbers give a positive index, numbers between 0 and 1 give a negative one, and the index counts the places the decimal point has moved.

× and ÷. Deal with the A-numbers and the powers of ten separately, using Laws 1 and 2 on the indices.

+ and −. Rewrite one term so that both carry the same power of ten, then treat that power as a common factor. Either way, finish by checking 1 ≤ A < 10.

The habits that earn the marks

1. Name the law you used — "Law 1 of Indices" beside the line is a free method mark.
2. Clear every bracket before combining anything, and give the outside index to every factor inside, coefficients included.
3. Indices work is exact: leave fractions as fractions and never round an exact answer.
4. "Positive index form" means no negative index survives into the answer.
5. In standard form, check 1 ≤ A < 10 on the last line, every time.
6. Round once, at the end, to 3 s.f. unless the question names another accuracy — and count the figures on A only.
7. Answer in the units the question asked for, and check the magnitude is plausible.

Syllabus at a glance

The three G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus — and the Problems in Real-World Contexts entry that admits compound interest.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
1.8	use of standard form $A \times 10^{n}$ , where `n` is an integer, and $1 \leq A < 10$	1.3 Definition, 1.3 Converting, 1.3 Arithmetic, 1.3 Magnitudes
1.9	positive, negative, zero and fractional indices	1.1 Index notation, 1.2 Zero, 1.2 Negative, 1.2 Rational
1.10	laws of indices	1.1 The five laws, 1.1 Errors, 1.2 Extension, 1.2 Equations
RWC	problems in real-world contexts — "in personal and household finance, including simple and compound interest, taxation, instalments, utilities bills, money exchange, etc."	1.4 Compound interest, 1.4 Other periods, 1.4 Comparing offers

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus. Two notes on scope. Surds are not in this syllabus — the word appears nowhere in K310 or in the by-level document, so the radical sign is used here only to say what a rational index means, never as something to be simplified or rationalised; that is Additional Mathematics. And compound interest, which the textbook develops in 3A §3.5 as an application of indices, is examinable even though it is not one of the numbered N1 outcomes: the syllabus admits it under Problems in Real-World Contexts, which may be "In personal and household finance, including simple and compound interest, taxation, instalments, utilities bills, money exchange, etc." That section appears in the G3 and G2 documents in identical wording, and the formula itself is the first entry on the paper's own formulae list. It is covered in full in 1.4. The G2 Mathematics syllabus lists 1.8, 1.9 and 1.10 in word-for-word identical wording, so — unusually — every part of this chapter is examinable in both bands and nothing here is marked as G3-only.

📐 Indices & Standard Form — Revision Notes

The four questions this chapter answers

Index notation — the recap ⓘ Recap

The five Laws of Indices

Past-paper questions

The four errors this topic is built to catch

Check yourself

The zero index

Negative indices

Past-paper questions

Rational indices

The laws, extended to every rational index

Past-paper questions

Equations with the unknown in the index

Check yourself

Past-paper questions

What standard form is

Converting, both ways

Past-paper questions

Arithmetic in standard form

Past-paper questions

Doing it on the calculator

Past-paper questions

Standard form in the real world

Check yourself

Simple interest, and what compounding changes

Where the formula comes from

When interest is added more often than once a year

Working backwards, and comparing offers

Check yourself — 1.4

Past-paper questions

The whole chapter in one table

The formulae to know by heart

The habits that earn the marks

Syllabus at a glance

Ask CherAI

Your question allowance