What is 2.1: Laws of indices?

For a non-zero base, a^m a^n=a^m+n, a^m/a^n=a^m-n and (a^m)^n=a^mn. Interpret rational powers through roots: a^p/q=[q]a^p=([q]a)^p whenever the real-valued expression is defined.

How many marks are 2.1 questions worth?

2.1 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 2.1 appeared in?

We have not yet indexed a verified past-paper appearance for 2.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.1

Understand and use the laws of indices for all rational exponents.

Draft — not yet indexed

Laws of indices

Worked answers and methods for 2.1 on Edexcel A-level Maths 9MA0.

Explanation

For a non-zero base, $a^m a^n=a^{m+n}$ , $a^m/a^n=a^{m-n}$ and $(a^m)^n=a^{mn}$ . Interpret rational powers through roots: $a^{p/q}=\sqrt[q]{a^p}=(\sqrt[q]{a})^p$ whenever the real-valued expression is defined.
For example, $16^{3/4}=(\sqrt[4]{16})^3=2^3=8$ .
A negative exponent means reciprocal, not a negative value: $a^{-r}=1/a^r$ ; also avoid applying index laws across addition.
The base must be non-zero for division and negative powers.
When an equation contains different-looking powers, rewrite them with a common positive base before equating exponents; this makes the index law being used visible and avoids premature decimal approximations.

Worked example

Given $x>0$ , simplify $\dfrac{x^{5/2}}{x^{-1/2}}$ to a single power of $x$ .

1.For division with a common base, subtract the exponents: $x^{5/2}/x^{-1/2}=x^{5/2-(-1/2)}=x^{6/2}=x^3$ .

Answer: $x^3$

Common mistakes

Don't apply $(a^m)^n=a^{m+n}$ instead of multiplying the exponents.
Don't treat $a^{-r}$ as a negative number instead of the reciprocal $1/a^r$ .

Exam tip

When solving an exponential equation, rewrite both sides with the same base and show the equation formed by equating exponents.

Worked practice

Tier 1 · Easy

Evaluate

81^{3/4}

without using a calculator.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$27$	2
Notes
$81^{3/4}=(\sqrt[4]{81})^3=3^3=27$ .

(2 marks)

Tier 2 · Standard

Given

x>0

, simplify

\dfrac{\left(16x^{12}\right)^{3/4}}{2x^{-2}}

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$4x^{11}$	3
Notes
Since $x>0$ , $\left(16x^{12}\right)^{3/4}=16^{3/4}x^9=8x^9$ . Therefore $\dfrac{8x^9}{2x^{-2}}=4x^{9-(-2)}=4x^{11}$ .

(3 marks)

Tier 3 · Hard

Solve

8^{2x-1}=4^{x+2}

, giving the exact value of

x

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=\dfrac{7}{4}$	4
Notes
Write both sides with base $2$ : $8^{2x-1}=2^{3(2x-1)}$ and $4^{x+2}=2^{2(x+2)}$ . Equal positive bases have equal exponents, so $6x-3=2x+4$ . Hence $4x=7$ and $x=7/4$ .

(4 marks)

Tier 1 · Easy

Given

x>0

, solve

x^{3/2}=125

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x=25$	2
Notes
Since $x>0$ , $x^{3/2}=(\sqrt{x})^3$ . Therefore $\sqrt{x}=5$ , so $x=25$ .

(2 marks)

Tier 2 · Standard

Given

x>0

, simplify

\dfrac{(27x^{-9})^{2/3}}{9x^{-1}}

as a single power of

x

, and then with positive indices.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$x^{-5}=\dfrac1{x^5}$	3
Notes
$(27x^{-9})^{2/3}=27^{2/3}x^{-6}=9x^{-6}$ . Dividing by $9x^{-1}$ gives $x^{-6-(-1)}=x^{-5}=1/x^5$ .

(3 marks)

Tier 3 · Hard

Solve

(27^{2x-1})^{1/2}9^{-x/3}=81^{x/4-1}

, giving the exact value of

x

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x=-\dfrac{15}{8}$	5
Notes
Write every term with base $3$ . The left-hand side is $3^{3(2x-1)/2}3^{-2x/3}=3^{7x/3-3/2}$ , while the right-hand side is $3^{x-4}$ . Equating exponents gives $7x/3-3/2=x-4$ . Multiplying by $6$ gives $14x-9=6x-24$ , so $8x=-15$ and $x=-15/8$ .

(5 marks)

Tier 2 · Standard

For

x>0

, the identity

(x^{p-1/2})^4x^{-3}\equiv x^7

is true for all values of

x

. Find the rational number

p

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$p=3$	3
Notes
The exponent on the left is $4(p-1/2)-3=4p-5$ . Equality for all positive $x$ requires $4p-5=7$ , so $4p=12$ and $p=3$ .

(3 marks)

Tier 3 · Hard

Given

x>0

and

y>0

, simplify

\left(\dfrac{(x^{-1/2}y^{2/3})^6}{x^{-5}y}\right)^{3/5}

as a product of powers with positive indices.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x^{6/5}y^{9/5}$	4
Notes
Inside the outer power, $(x^{-1/2}y^{2/3})^6=x^{-3}y^4$ . Dividing by $x^{-5}y$ gives $x^{-3-(-5)}y^{4-1}=x^2y^3$ . Raising this to the power $3/5$ gives $x^{6/5}y^{9/5}$ .

(4 marks)

Tier 3 · Hard

Given

x>0

and

y>0

, the identity

\left((x^py^q)^{-2}x^{1/2}y^3\right)^{1/3}\equiv x^{-3/2}y^{5/3}

holds for all such

x

and

y

. Find the rational numbers

p

and

q

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$p=\dfrac52$ and $q=-1$	5
Notes
The exponent of $x$ on the left is $(-2p+1/2)/3$ , so $(-2p+1/2)/3=-3/2$ . Hence $-2p+1/2=-9/2$ , giving $p=5/2$ . The exponent of $y$ is $(-2q+3)/3$ , so $(-2q+3)/3=5/3$ . Hence $-2q+3=5$ , giving $q=-1$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Given

x>0

, solve

\dfrac{(x^{3/2})^{-2}(x^{-1/3})^6}{x^{1/4}}=32

, giving the exact value of

x

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$x=2^{-20/21}$	4
Notes
Using the index laws, the exponent on the left is $-3-2-1/4=-21/4$ . Thus $x^{-21/4}=32=2^5$ . Raising both sides to the power $-4/21$ gives $x=(2^5)^{-4/21}=2^{-20/21}$ , which is positive as required.

(4 marks)

Verified exam appearances

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Other points in 2 Algebra and functions

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