Laws of Indices

This lesson covers the fundamental laws of indices (also called exponents or powers) that underpin all work with exponentials and logarithms in the Edexcel 9MA0 A-Level Mathematics specification. You must be fluent with these rules before tackling exponential equations and logarithmic manipulation.

What Are Indices?

An index (plural: indices) tells you how many times a base number is multiplied by itself. For example:

2³ = 2 × 2 × 2 = 8
5² = 5 × 5 = 25

In general, aⁿ means "a multiplied by itself n times," where a is the base and n is the index (or exponent).

The Seven Key Laws

Law 1: Multiplication — aᵐ × aⁿ = aᵐ⁺ⁿ

When you multiply two powers with the same base, you add the indices.

Example: 3⁴ × 3² = 3⁴⁺² = 3⁶ = 729

Why it works: 3⁴ × 3² = (3 × 3 × 3 × 3) × (3 × 3) = 3⁶

Law 2: Division — aᵐ ÷ aⁿ = aᵐ⁻ⁿ

When you divide two powers with the same base, you subtract the indices.

Example: 5⁷ ÷ 5³ = 5⁷⁻³ = 5⁴ = 625

Law 3: Power of a Power — (aᵐ)ⁿ = aᵐⁿ

When you raise a power to another power, you multiply the indices.

Example: (2³)⁴ = 2³ˣ⁴ = 2¹² = 4096

Law 4: Zero Index — a⁰ = 1

Any non-zero number raised to the power zero equals 1.

Why it works: Using Law 2, aⁿ ÷ aⁿ = aⁿ⁻ⁿ = a⁰. But aⁿ ÷ aⁿ = 1. Therefore a⁰ = 1.

Law 5: Negative Index — a⁻ⁿ = 1/aⁿ

A negative index means "one over" the positive power.

Example: 4⁻² = 1/4² = 1/16

Why it works: Using Law 2, a⁰ ÷ aⁿ = a⁰⁻ⁿ = a⁻ⁿ. But a⁰ ÷ aⁿ = 1/aⁿ. Therefore a⁻ⁿ = 1/aⁿ.

Law 6: Fractional Index — a^(1/n) = ⁿ√a

A fractional index with numerator 1 means the nth root.

Examples:

8^(1/3) = ³√8 = 2
25^(1/2) = √25 = 5

Law 7: General Fractional Index — a^(m/n) = (ⁿ√a)ᵐ = ⁿ√(aᵐ)

A fractional index m/n means "take the nth root, then raise to the power m" (or vice versa).

Example: 8^(2/3) = (³√8)² = 2² = 4

Exam Tip: When evaluating expressions like 27^(2/3), always take the root first to keep numbers manageable: ³√27 = 3, then 3² = 9. Do not compute 27² = 729 first.

Combining the Laws

In A-Level questions you will often need to apply several laws in sequence.

Example: Simplify (2x³y²)⁴ ÷ (4x²y)²

Step 1 — Expand brackets:

Numerator: 2⁴ × x¹² × y⁸ = 16x¹²y⁸
Denominator: 4² × x⁴ × y² = 16x⁴y²

Step 2 — Divide:

16x¹²y⁸ ÷ 16x⁴y² = x⁸y⁶

Working with Surds and Indices

Surds can be written using fractional indices:

√x = x^(1/2)
³√x = x^(1/3)
1/√x = x^(-1/2)

This is essential for differentiation and integration later in the course.

Example: Write (4√x³) in index form.

4√x³ = 4x^(3/2)

Common Mistakes to Avoid

Mistake	Correction
(a + b)² = a² + b²	(a + b)² = a² + 2ab + b² — you cannot distribute a power over addition
a⁻² = -a²	a⁻² = 1/a² — a negative index is a reciprocal, not a negative number
a^(1/2) × a^(1/2) = a^(1/4)	a^(1/2) × a^(1/2) = a^(1/2 + 1/2) = a¹ — add the indices
(a²)³ = a⁵	(a²)³ = a⁶ — multiply (not add) the indices

Summary

The seven index laws allow you to simplify any expression involving powers with the same base.
Multiply powers: add indices. Divide powers: subtract indices. Power of a power: multiply indices.
Zero index gives 1. Negative index gives the reciprocal. Fractional index gives roots.
Always take the root before the power when evaluating fractional indices to keep calculations manageable.
Converting surds to fractional indices is a vital skill for calculus.

A-Level Deep Dive: Laws of Indices

Spec mapping

Edexcel 9MA0-01 specification section 2 (Algebra and functions, sub-strand 2.1) covers the laws of indices for all rational exponents (refer to the official specification document for exact wording). This is the foundational statement; the laws themselves are not listed in the formula booklet and must be memorised. The seven canonical laws are: $a^m \cdot a^n = a^{m+n}$ , $a^m / a^n = a^{m-n}$ , $(a^m)^n = a^{mn}$ , $(ab)^n = a^n b^n$ , $a^0 = 1$ (for $a \neq 0$ ), $a^{-n} = 1/a^n$ , and $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$ .

Edexcel 9MA0-01 specification section 6 (Exponentials and logarithms) covers the function $a^x$ and its graph, where $a$ is positive… understand and use the laws of logarithms (refer to the official specification document for exact wording). Every manipulation in section 6 is downstream of section 2: solving $a^{f(x)} = b^{g(x)}$ requires combining indices on a common base; the log laws $\log(xy) = \log x + \log y$ and $\log(x^k) = k \log x$ are nothing more than the index laws $a^m \cdot a^n = a^{m+n}$ and $(a^m)^k = a^{mk}$ read backwards. Logarithms are the inverse operation to exponentiation, so the algebraic structure transfers verbatim.

Synoptic reach across 9MA0: indices are not a single-section topic — they are the connective tissue of the entire Pure paper. Section 7 (Differentiation): $\frac{d}{dx}x^n = nx^{n-1}$ requires confident manipulation of rational $n$ , including negative and fractional. Section 8 (Integration): $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ — every integration of a root or reciprocal goes via a fractional or negative index. Section 4 (Sequences and series — binomial expansion): $(1+x)^n$ for non-integer $n$ produces every term using the index law $(a)^k$ for rational $k$ . Section 6 (Logarithms): the change-of-base formula $\log_a b = \log_c b / \log_c a$ is itself a re-statement of $(a^x)^y = a^{xy}$ . A candidate insecure with index laws cannot pass A-Level Pure; they are the alphabet of the subject.

Worked example with full mark scheme

Question (8 marks):

(a) Solve, giving $x$ as an exact rational, the equation $2^{3x+1} \cdot 4^{x-2} = \dfrac{1}{8}$ . (5)

(b) Hence, or otherwise, solve $\left(\dfrac{1}{9}\right)^{y+1} = 27^{2y-1}$ giving $y$ as an exact rational. (3)

Solution with mark scheme:

(a) Step 1 — convert all terms to a common base.

$4 = 2^2$ , so $4^{x-2} = (2^2)^{x-2} = 2^{2(x-2)} = 2^{2x-4}$ . Also $\dfrac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}$ .

M1 — converting at least one non-base-2 term to a power of 2 using $(a^m)^n = a^{mn}$ . Common error: students leave $4^{x-2}$ untouched and try to "cross-multiply" or take logs prematurely, missing the elegant common-base route.

A1 — both $4^{x-2}$ and $1/8$ correctly rewritten as powers of 2.

Step 2 — combine the LHS using $a^m \cdot a^n = a^{m+n}$ .

$2^{3x+1} \cdot 2^{2x-4} = 2^{(3x+1) + (2x-4)} = 2^{5x-3}$

M1 — correct application of the multiplication law of indices, summing exponents on a common base.

Step 3 — equate indices.

Since the exponential function $x \mapsto 2^x$ is one-to-one, $2^A = 2^B \implies A = B$ . So:

$5x - 3 = -3 \implies 5x = 0 \implies x = 0$

A1 — exact rational answer $x = 0$ .

Total for (a): 5 marks (M2 A2, with 1 method-and-accuracy split as above plus a presentation A1).

(b) Step 1 — convert all terms to base 3.

$\dfrac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}$ , and $27 = 3^3$ . So:

$\left(3^{-2}\right)^{y+1} = \left(3^3\right)^{2y-1}$

M1 — both sides converted to base 3 using $(a^m)^n = a^{mn}$ .

Step 2 — apply the power-of-a-power law to both sides.

$3^{-2(y+1)} = 3^{3(2y-1)} \quad\Longleftrightarrow\quad 3^{-2y-2} = 3^{6y-3}$

Step 3 — equate indices and solve.

$-2y - 2 = 6y - 3 \implies 1 = 8y \implies y = \dfrac{1}{8}$

M1 A1 — correct linear equation in $y$ , solved to give the exact rational $\tfrac{1}{8}$ .

Total for (b): 3 marks (M2 A1).

Grand total: 8 marks. This question rewards the candidate who recognises that every base in sight is a power of a single small prime and acts on that recognition before reaching for logarithms or numerical methods.

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Given that $f(x) = 8^{x+1} \cdot 4^{1-2x}$ for all real $x$ :

(a) Show that $f(x)$ can be written in the form $2^{ax+b}$ where $a$ and $b$ are integers to be found. (3)

(b) Hence solve $f(x) = 32$ , giving $x$ as an exact rational. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — converting $8 = 2^3$ and $4 = 2^2$ to obtain $8^{x+1} = 2^{3(x+1)} = 2^{3x+3}$ and $4^{1-2x} = 2^{2(1-2x)} = 2^{2-4x}$ .
M1 (AO1.1b) — applying $a^m \cdot a^n = a^{m+n}$ to combine the two expressions: $2^{3x+3} \cdot 2^{2-4x} = 2^{(3x+3) + (2-4x)}$ .
A1 (AO1.1b) — simplifying the exponent to obtain $f(x) = 2^{-x+5}$ , i.e. $a = -1$ , $b = 5$ .

(b)

M1 (AO1.1b) — recognising $32 = 2^5$ and using the result from (a) to write the equation as $2^{-x+5} = 2^5$ .
M1 (AO2.1) — equating indices: $-x + 5 = 5$ , justified by the one-to-one nature of $2^x$ .
A1 (AO1.1b) — exact rational $x = 0$ .

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — Edexcel uses index questions almost entirely to test procedural fluency, with the single AO2 mark reserved for explicitly using the injectivity of the exponential to equate indices. Candidates who simply "cancel" without comment often still earn the mark, but a one-line justification is the safest route.

Synoptic links

Connects to:

Section 6 — Logarithms (inverse of indices): the definition $\log_a b = c \Longleftrightarrow a^c = b$ makes logs literally the inverse operation of exponentiation. Every log law restates an index law: $\log_a(xy) = \log_a x + \log_a y$ is $a^m \cdot a^n = a^{m+n}$ read in reverse; $\log_a(x^k) = k \log_a x$ is $(a^m)^k = a^{mk}$ read in reverse.
Section 6 — Exponential growth and decay: models of the form $N(t) = N_0 e^{kt}$ rely on $e^{a+b} = e^a \cdot e^b$ and $e^{ab} = (e^a)^b$ to combine and decompose growth factors. Doubling time, half-life, and continuous compounding all emerge from manipulating these index laws on the natural-exponential base $e$ .
Section 7 — Differentiation (power rule): $\frac{d}{dx}x^n = nx^{n-1}$ extends seamlessly to fractional and negative $n$ — but only after the candidate rewrites $\sqrt[3]{x^2}$ as $x^{2/3}$ and $1/x^4$ as $x^{-4}$ . Without the index laws, the power rule is unreachable for anything other than positive integer powers.
Section 4 — Binomial expansion: $(1+x)^n$ for any rational $n$ produces terms involving $x^k$ for non-negative integer $k$ , but the coefficients themselves involve the index $n$ throughout. Expanding $(1+x)^{-1/2}$ uses fractional-index arithmetic at every step.
Section 2 — Surds: the bridge $\sqrt[n]{a} = a^{1/n}$ allows every surd manipulation to be re-expressed as index manipulation. $\sqrt{x} \cdot \sqrt[3]{x} = x^{1/2} \cdot x^{1/3} = x^{5/6} = \sqrt[6]{x^5}$ — a one-line calculation in index form, several lines if attempted purely in surd form.

Mark-scheme literacy

Index-law questions on 9MA0 are heavily weighted toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	75–85%	Applying the seven index laws, converting between bases, simplifying expressions with rational exponents, equating indices in exponential equations
AO2 (reasoning / interpretation)	10–20%	Justifying the step $a^A = a^B \implies A = B$ via the injectivity of $a^x$ , choosing the most efficient common base, recognising the structure of a "hidden quadratic" in $a^x$
AO3 (problem-solving)	0–5%	Multi-stage modelling using exponentials in unfamiliar contexts; rare in Year 1 but appears in financial-mathematics and population-modelling contexts

Examiner-rewarded phrasing: "writing all terms with a common base of 2"; "applying $a^m \cdot a^n = a^{m+n}$ "; "since $a^x$ is one-to-one, equating indices gives…"; "in the form $2^{ax+b}$ with $a$ , $b$ integers as required". Phrases that lose marks: "cancelling the bases" (the bases are not cancelled, the indices are equated); leaving $\sqrt{x^2}$ instead of $x$ when $x > 0$ ; converting between exponential and index form inconsistently mid-solution.

A pattern worth watching: questions phrased "in the form $a^{f(x)}$ where $a$ is the smallest possible integer" demand a prime base. Writing the answer as $4^{2x+1}$ when the question wants base 2 will lose the final A1 even though the value is identical — read the form constraint precisely.

Grade-band model answers

3-mark question

Question: Simplify $\dfrac{(2x^3 y^2)^4}{4x^2 y}$ , giving your answer in the form $kx^a y^b$ where $k$ , $a$ , $b$ are integers to be found.

Grade C response (~210 words):

Expand the bracket using $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$ :

$(2x^3 y^2)^4 = 2^4 (x^3)^4 (y^2)^4 = 16 x^{12} y^8$ .

So the expression becomes $\dfrac{16 x^{12} y^8}{4 x^2 y}$ .

Simplify the constant: $16 / 4 = 4$ .

For the $x$ terms: $x^{12} / x^2 = x^{12-2} = x^{10}$ .

For the $y$ terms: $y^8 / y^1 = y^{8-1} = y^7$ .

So the simplified expression is $4 x^{10} y^7$ , giving $k = 4$ , $a = 10$ , $b = 7$ .

Examiner commentary: Full marks (3/3). The candidate correctly applies the power-of-a-product and power-of-a-power laws to expand the bracket, then divides constant, $x$ and $y$ separately. Working is brief but every index law is named or visibly used. The answer is presented in the requested form with all three constants identified explicitly. This is the standard Grade C presentation for a procedural question — efficient, correct, no surplus material. Many candidates lose marks here by failing to apply the exponent 4 to every factor inside the bracket (a frequent slip is forgetting to raise the constant 2 to the fourth power, leaving the coefficient as 2 rather than 16).

Grade A response (~250 words):*

Apply the power-of-a-product law $(abc)^n = a^n b^n c^n$ to the numerator:

$(2x^3 y^2)^4 = 2^4 \cdot (x^3)^4 \cdot (y^2)^4$

Then apply the power-of-a-power law $(a^m)^n = a^{mn}$ to each variable factor:

$= 16 \cdot x^{12} \cdot y^8$

Substituting back:

$\dfrac{16 x^{12} y^8}{4 x^2 y} = \dfrac{16}{4} \cdot \dfrac{x^{12}}{x^2} \cdot \dfrac{y^8}{y^1}$

Apply the division law $a^m / a^n = a^{m-n}$ to each variable:

$= 4 \cdot x^{12 - 2} \cdot y^{8 - 1} = 4 x^{10} y^7$

So in the form $kx^a y^b$ : $k = 4$ , $a = 10$ , $b = 7$ , all integers as required.

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate explicitly names every index law as it is applied, separates the simplification into constant, $x$ and $y$ factors before dividing (which avoids algebraic clutter), and closes with "as required" tying back to the question stem. The use of display-math notation and the explicit decomposition $(2x^3y^2)^4 = 2^4 \cdot (x^3)^4 \cdot (y^2)^4$ before evaluating shows examiner-aware presentation discipline. On a longer question this kind of structured working pays dividends because every M-mark is locatable on a single line.

6-mark question

Question: Solve $25^{x+1} = 125^{2x-3}$ , giving $x$ as an exact rational.

Grade B response (~280 words):

Both 25 and 125 are powers of 5: $25 = 5^2$ and $125 = 5^3$ . So:

$25^{x+1} = (5^2)^{x+1} = 5^{2(x+1)} = 5^{2x+2}$ .

$125^{2x-3} = (5^3)^{2x-3} = 5^{3(2x-3)} = 5^{6x-9}$ .

The equation becomes $5^{2x+2} = 5^{6x-9}$ .

Since the bases are equal, the indices must be equal:

$2x + 2 = 6x - 9$ .

Rearranging: $2 + 9 = 6x - 2x$ , so $11 = 4x$ , giving $x = \dfrac{11}{4}$ .

Examiner commentary: Method correct throughout. The candidate identifies the common base, applies the power-of-a-power law on both sides, equates indices and solves the resulting linear equation. The exact rational $\tfrac{11}{4}$ is given correctly. Full marks (6/6) — though the working would be even tighter if the candidate explicitly stated the principle "since $5^x$ is one-to-one, equating bases requires equating indices". This is a classic Grade B presentation: every step is correct but the justification of why bases can be "equated" is implicit rather than stated. On this 6-mark question that is acceptable; on a longer AO2-weighted question the absence of justification can cost the AO2 mark.

Grade A response (~340 words):*

Both bases are powers of 5: $25 = 5^2$ and $125 = 5^3$ . Rewrite each side using the power-of-a-power law $(a^m)^n = a^{mn}$ :

$25^{x+1} = (5^2)^{x+1} = 5^{2(x+1)} = 5^{2x+2}$ $125^{2x-3} = (5^3)^{2x-3} = 5^{3(2x-3)} = 5^{6x-9}$

The equation becomes:

$5^{2x+2} = 5^{6x-9}$

Since the function $f(t) = 5^t$ is strictly increasing (and therefore one-to-one) on $\mathbb{R}$ , equality of values forces equality of inputs:

$2x + 2 = 6x - 9$

Solve linearly:

$2 + 9 = 6x - 2x \quad\Longrightarrow\quad 11 = 4x \quad\Longrightarrow\quad x = \dfrac{11}{4}$

Verification: Substitute $x = \tfrac{11}{4}$ back into the original LHS exponent: $x + 1 = \tfrac{15}{4}$ , so $25^{15/4} = 5^{15/2}$ . RHS exponent: $2x - 3 = \tfrac{22}{4} - 3 = \tfrac{10}{4} = \tfrac{5}{2}$ , so $125^{5/2} = 5^{15/2}$ . Both sides equal, confirming $x = \tfrac{11}{4}$ .

Examiner commentary: Full marks (6/6). The candidate not only solves but justifies the key AO2 step ("since $5^t$ is strictly increasing, equality of values forces equality of inputs") in the language an examiner would recognise from undergraduate analysis. The verification step at the end — substituting back into the original equation — is the kind of self-checking discipline that distinguishes A* candidates: it catches arithmetic slips that propagate through index manipulation undetected. The choice of the prime base 5 (rather than 25 or 125) is also deliberate; the smallest base produces the simplest indices.

9-mark question

Question: $g(x) = \dfrac{16^x \cdot 8^{2-x}}{4^{x+1}}$ for all real $x$ .

(a) Show that $g(x)$ can be written in the form $2^{ax+b}$ where $a$ and $b$ are integers to be found. (4)

(b) Hence solve $g(x) = 64$ , giving $x$ as an exact rational. (3)

Grade A response (~410 words):*

(a) Convert each base to a power of 2: $16 = 2^4$ , $8 = 2^3$ , $4 = 2^2$ . Apply the power-of-a-power law to each factor:

$16^x = (2^4)^x = 2^{4x}$ $8^{2-x} = (2^3)^{2-x} = 2^{3(2-x)} = 2^{6-3x}$ $4^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2}$

The numerator becomes, by the multiplication law $a^m \cdot a^n = a^{m+n}$ :

$2^{4x} \cdot 2^{6-3x} = 2^{(4x) + (6-3x)} = 2^{x+6}$

Then by the division law $a^m / a^n = a^{m-n}$ :

$g(x) = \dfrac{2^{x+6}}{2^{2x+2}} = 2^{(x+6) - (2x+2)} = 2^{-x+4}$

So $g(x) = 2^{-x+4}$ , giving $a = -1$ , $b = 4$ .

(b) Set $g(x) = 64 = 2^6$ :

$2^{-x+4} = 2^6 \quad\Longrightarrow\quad -x + 4 = 6 \quad\Longrightarrow\quad x = -2$

$2^{-x+4} = 2^{-1} \quad\Longrightarrow\quad -x + 4 = -1 \quad\Longrightarrow\quad x = 5$

Examiner commentary: Full marks (9/9). Part (a) is exemplary: each base is converted separately, every index law named at the point of use, the numerator combined before the division — preventing the common slip of trying to combine numerator and denominator in a single line. Parts (b) and (c) reuse the result from (a) directly, honouring the "Hence" command and avoiding wasted work. Both final answers are exact integers and there is no decimal contamination. The structure of the response — "convert, combine numerator, divide, simplify" — is a reusable template that scales to any product/quotient of exponentials with a common-base-able structure. A candidate who internalises this template will rarely lose marks on this question type, regardless of the specific bases chosen by the examiner.

A-Level-depth misconceptions

The errors that distinguish A from A* on index-law questions:

Negative-index sign confusion. $a^{-n} = \dfrac{1}{a^n}$ , but the negative does not attach to the base. $2^{-3} = \tfrac{1}{8}$ , never $-8$ . Candidates write $x^{-2} = -x^2$ under exam pressure; this single slip can cascade through differentiation and integration questions.
Confusing $(a^m)^n$ with $a^m \cdot a^n$ . $(a^m)^n = a^{mn}$ (multiply), but $a^m \cdot a^n = a^{m+n}$ (add). Mid-question, especially when $m$ or $n$ is itself an algebraic expression, the choice of operation slips. Cure: always say aloud "power of a power means multiply; product of powers means add" before writing.
Fractional indices and root order. $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$ — both interpretations agree, but for evaluation it is almost always easier to take the root first. $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ , much cleaner than $\sqrt[3]{64} = 4$ . Candidates who power-first often produce unmanageably large numbers.
Negative bases under fractional indices. $(-8)^{1/3} = -2$ is well-defined for odd root order, but $(-4)^{1/2}$ is not real. Students apply index laws blindly to negative bases without checking the root order, producing answers like " $(-1)^{1/2} = (-1)^{2/4} = ((-1)^2)^{1/4} = 1$ " — a chain of plausible-looking steps that crosses the real–complex boundary illegitimately. The 9MA0 specification keeps to real-valued indices; treat negative bases under non-integer indices with care.
Index laws applied across different bases. $2^x \cdot 3^x = 6^x$ is correct (it follows from $(ab)^n = a^n b^n$ ), but $2^x + 3^x \neq 5^x$ — addition does not combine bases. Equally, $2^x \cdot 3^y$ cannot be simplified further; there is no index law for different bases with different exponents.
$a^0 = 1$ for $a \neq 0$ , but $0^0$ is undefined (or contextually 1). A subtle point: $a^0 = 1$ in the index laws assumes $a \neq 0$ , derived from $a^0 = a^{n-n} = a^n / a^n = 1$ . The case $0^0$ is separately defined (often $1$ for combinatorial purposes, undefined in analysis). On 9MA0 you will not be asked to evaluate $0^0$ , but you should know it is not a free consequence of "anything to the zero is 1".
Mishandling indices in equations vs identities. An identity $a^m \cdot a^n = a^{m+n}$ holds for all valid $a, m, n$ . An equation $2^{f(x)} = 2^{g(x)}$ permits the deduction $f(x) = g(x)$ only because $2^x$ is one-to-one. Some candidates apply the same logic to $\sin(f(x)) = \sin(g(x))$ , producing wrong solutions because $\sin$ is not one-to-one. The injectivity of $a^x$ for $a > 0$ , $a \neq 1$ is the silent but essential premise of every "equate indices" step.

Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Paper 1 index questions. They are all about presentation discipline, not technique.

Premature use of logarithms. Candidates who reach for $\log$ on every exponential equation often miss the cleaner common-base route. $2^{3x+1} \cdot 4^{x-2} = 1/8$ becomes a brutal log calculation if attacked with logs immediately, but a one-line simplification in base 2. Examiners reward the more elegant route with full marks; the log route, even if executed perfectly, can introduce arithmetic slip points and decimal contamination.
Inconsistent base choice. Mid-solution switching between base 2 and base 4 (or base 3 and base 9) produces tangled exponents. The cure is to commit, on the first line, to "I will work in base $p$ " where $p$ is the smallest prime appearing in the question, and convert every term to that base before doing any arithmetic.
Failing to state the injectivity step. When moving from $a^A = a^B$ to $A = B$ , a single sentence ("since $a^x$ is one-to-one for $a > 0$ , $a \neq 1$ ") can be the AO2 mark on a 6-mark question. Candidates who write only "so $A = B$ " usually still earn the mark but are vulnerable on AO2-weighted papers.
Not matching the requested form exactly. A question asking for the answer "in the form $2^{ax+b}$ with $a$ , $b$ integers" is a binding instruction. Writing $4 \cdot 2^{-x+2}$ is not in that form even if the value is identical; the constant must be absorbed into the exponent as $2^2$ .

This pattern is endemic to Paper 1 index questions: candidates know the laws, lose marks on choice of route or presentation.

Going further — university and academic signposting

Index laws and rational-exponent manipulation point directly toward several undergraduate trajectories:

Real analysis (Year 1): the rigorous definition of $x^a$ for irrational exponents $a$ requires the construction of the real numbers via Dedekind cuts or Cauchy sequences, and then the definition $x^a = \lim_{n \to \infty} x^{r_n}$ for any rational sequence $r_n \to a$ . Proving this limit exists and is independent of the choice of sequence is the first non-trivial application of completeness in a university analysis course.
Complex exponentiation (Year 2 complex analysis): $z^w = \exp(w \log z)$ for complex $z, w$ , where $\log z$ is multi-valued. The famous identity $i^i = e^{-\pi/2}$ (a real number!) is a direct consequence of $i = e^{i\pi/2}$ and the complex extension of the index laws — one of the most surprising results undergraduates encounter.
Group theory (Year 2 algebra): $a^n$ for integer $n$ generalises to the iterated group operation in any group: $a \cdot a \cdot \ldots \cdot a$ ( $n$ times) for $n > 0$ , the identity for $n = 0$ , and the inverse $a^{-1} \cdot \ldots \cdot a^{-1}$ for $n < 0$ . The index laws $a^m \cdot a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$ hold in every group, abelian or not — they are the algebraic skeleton beneath the arithmetic.
Numerical analysis (Year 2): computing $x^a$ for irrational $a$ on a computer typically uses $x^a = \exp(a \ln x)$ with the exponential and logarithm computed by Taylor series or argument-reduction tricks. Every time you trust a calculator's value of $2^\pi$ , this conversion is running.

Oxbridge-style interview prompt: "Define $x^a$ for positive real $x$ and arbitrary real $a$ . Now define it for arbitrary complex $a$ . Where do problems arise, and how does the index law $x^{a+b} = x^a \cdot x^b$ continue to hold (or fail)?"

Additional A* practice: hidden quadratic in exponentials

A common A* trap on 9MA0-01 is to give an equation that looks exponential but reduces to a quadratic in a hidden variable. The technique is the substitution $u = a^x$ , which converts hard-looking exponential equations into routine quadratics — exploiting $a^{2x} = (a^x)^2 = u^2$ .

Worked example: Solve $2^{2x} - 5 \cdot 2^x + 4 = 0$ , giving exact answers.

Let $u = 2^x$ . Then by the power-of-a-power law, $2^{2x} = (2^x)^2 = u^2$ . The equation becomes:

$u^2 - 5u + 4 = 0$

Factorise: $(u - 1)(u - 4) = 0$ , so $u = 1$ or $u = 4$ .

Back-substitute:

$2^x = 1 = 2^0$ gives $x = 0$ .
$2^x = 4 = 2^2$ gives $x = 2$ .

So the exact solutions are $x = 0$ and $x = 2$ .

Why A candidates spot this immediately:* the structure "term in $a^{2x}$ + term in $a^x$ + constant" is the signature of a hidden quadratic. Every time this pattern appears, substitute $u = a^x$ first. The same trick converts $e^{2x} - 3e^x + 2 = 0$ (substitute $u = e^x$ ) and $9^x - 4 \cdot 3^x + 3 = 0$ (substitute $u = 3^x$ , noting $9^x = (3^2)^x = (3^x)^2$ ) into routine factorisations. Recognising the pattern across these contexts is exactly the synoptic skill Edexcel rewards.

A subtlety: when back-substituting, check that solutions are valid. $2^x = -3$ has no real solution because $2^x > 0$ for all real $x$ . If the substitution-quadratic produces a negative $u$ root, reject it explicitly with the phrase "since $2^x > 0$ for all real $x$ , we reject $u = -3$ ". This single sentence is often a half-mark or whole mark on its own.

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This content is aligned with the Pearson Edexcel GCE A Level Mathematics (9MA0) specification, Paper 1 — Pure Mathematics, Section 6: Exponentials and logarithms. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Expression or equation<br/>involving indices"] --> B{"What form?"}
    B -->|"Single expression<br/>to simplify"| C["Identify each<br/>index law in play:<br/>aᵐ·aⁿ, aᵐ/aⁿ, (aᵐ)ⁿ"]
    B -->|"Equation aᶠ⁽ˣ⁾ = bᵍ⁽ˣ⁾"| D["Convert all bases<br/>to a common base<br/>(smallest prime)"]
    B -->|"Hidden quadratic:<br/>a²ˣ + a·aˣ + c = 0"| E["Substitute u = aˣ,<br/>so a²ˣ = u²"]
    C --> F["Apply correct law:<br/>add, subtract, or<br/>multiply indices"]
    D --> G["Apply (aᵐ)ⁿ = aᵐⁿ<br/>then combine using<br/>aᵐ·aⁿ = aᵐ⁺ⁿ"]
    E --> H["Solve quadratic in u,<br/>then back-substitute<br/>and check u > 0"]
    F --> I["Simplify to<br/>requested form"]
    G --> J["Equate indices<br/>(injectivity of aˣ)<br/>and solve"]
    H --> K["Final exact answer"]
    I --> K
    J --> K

    style D fill:#27ae60,color:#fff
    style K fill:#3498db,color:#fff

Laws of Indices

Laws of Indices

What Are Indices?

The Seven Key Laws

Law 1: Multiplication — aᵐ × aⁿ = aᵐ⁺ⁿ

Law 2: Division — aᵐ ÷ aⁿ = aᵐ⁻ⁿ

Law 3: Power of a Power — (aᵐ)ⁿ = aᵐⁿ

Law 4: Zero Index — a⁰ = 1

Law 5: Negative Index — a⁻ⁿ = 1/aⁿ

Law 6: Fractional Index — a^(1/n) = ⁿ√a

Law 7: General Fractional Index — a^(m/n) = (ⁿ√a)ᵐ = ⁿ√(aᵐ)

Combining the Laws

Working with Surds and Indices

Common Mistakes to Avoid

Summary

A-Level Deep Dive: Laws of Indices

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns

Going further — university and academic signposting

Additional A* practice: hidden quadratic in exponentials

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Visual summary

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