IGCSEMaths: ExtendedCIETopic QuestionsAlgebra & SequencesAlgebraic Roots & Indices

Syllabus Edition

First teaching 2023

First exams 2025

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Algebraic Roots & Indices (CIE IGCSE Maths: Extended)

Topic Questions

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1a
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Simplify t to the power of 21 divided by t to the power of 7.

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1b
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Simplify left parenthesis u to the power of 5 right parenthesis to the power of 5.

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2
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Simplify left parenthesis x to the power of 8 right parenthesis cubed.

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3
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t to the power of x cross times t squared equals t to the power of 10
 
Find the value of x.
 

x = ................................................ 

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4
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Simplify fully (3e)0

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5
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Simplify  w squared over w cubed.

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1
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25 equals 125 to the power of k

Find the value of k.
 

k equals................................................ 

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2
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Simplify  open parentheses 27 x to the power of 9 close parentheses to the power of 2 over 3 end exponent.

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3
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Find the value of  left parenthesis 64 x to the power of 4 right parenthesis to the power of 0.5 end exponent space cross times space 4 x to the power of negative 2 end exponent.

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4
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2 cubed equals 4 to the power of p

Find the value of p.
 

p= ............................................ 

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5
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Find the value of n when 5 to the power of n equals 1 over 125.
 

n equals.................................................... 

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6
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Simplify open parentheses 64 over m cubed close parentheses to the power of negative 1 third end exponent.

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7
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2 to the power of p equals 1 over 8 to the power of 4

Find the value of p.


p = ............................................ 

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8
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Given that  open parentheses cube root of 1 over x end root close parentheses to the power of 4 equals x to the power of m

find the value of m

m = .................................................

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9
Sme Calculator1 mark

Simplify fully  open parentheses fraction numerator 9 t to the power of 4 w to the power of 9 over denominator 18 t to the power of 6 w to the power of 10 end fraction close parentheses to the power of negative 2 end exponent

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1
Sme Calculator2 marks

Simplify.

open parentheses 3 x squared y to the power of 4 close parentheses cubed

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2
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Simplify.

2 x to the power of 2 space end exponent cross times space 5 x to the power of 5

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3
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Simplify open parentheses 4 x y squared close parentheses cubed.

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4
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Simplify 8 t to the power of 8 divided by 4 t to the power of 4.

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5
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Simplify open parentheses 5 x to the power of 4 close parentheses cubed.

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6
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Simplify space left parenthesis 27 x to the power of 6 right parenthesis to the power of 1 third end exponent.

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7
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Simplify 2 x cubed cross times 3 x squared.

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8
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3 to the power of negative 2 end exponent cross times 3 to the power of x equals 81

Find the value of x.

 

x = ................................................... 

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9
Sme Calculator3 marks

Simplify.

i)
a cubed cross times a to the power of 6
 
[1]
ii)
left parenthesis 5 x y squared right parenthesis cubed 
 
[2]

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10
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Simplify open parentheses 4 over x close parentheses to the power of negative 2 end exponent space.

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11
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Simplify a cubed b to the power of 7 divided by a to the power of 6 b squared.

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12
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Simplify 5 a cubed c squared cross times 2 a squared c to the power of 7.

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13
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Simplify  left parenthesis 3 w cubed right parenthesis cubed.

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1
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2 to the power of 12 divided by 2 to the power of k over 2 end exponent equals 32

Find the value of k.


k equals ................................................. 

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2
Sme Calculator3 marks

Simplify open parentheses fraction numerator 27 x to the power of 12 over denominator 64 y cubed end fraction close parentheses to the power of negative 1 third end exponent.

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3
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      3 to the power of negative q end exponent cross times 1 over 27 equals 81

Find the value of q.

q equals.............................................. 

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4
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Simplify fully  open parentheses fraction numerator 64 x to the power of 6 over denominator 25 y squared end fraction close parentheses to the power of negative 1 half end exponent

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5
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Given that  4 to the power of k plus 3 space end exponent equals space 16 cross times 2 to the power of k

find the value of k.
Show your working clearly.

k = .................................................. 

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