Indices I — Paper 2 · IGCSE Mathematics 0580

16 Simplify Examiner's Use 0.75  p 4  (a)   ,  16  Answer(a) [2] (b) 32q-3 ÷ 23q-2. Answer(b) [2]

4 marks

Mark scheme: p 16 (a) or 0.125p3 1, 1 Independent marks for letter and no. 8 9 (b) q–1 1, 1 Independent marks for letter and no. 8 1 9 Allow 1 q–1 or 8 8q

This question in 0580/21 May/June 2010

5 Write 28 × 82 × 4-2 in the form 2n. Answer [2]

2 marks

Mark scheme: 5 210 2 M1 26 or 2–4 seen

This question in 0580/22 May/June 2010

6 3x × 94 = 3n. Find n in terms of x. Answer n = [2]

2 marks

Mark scheme: 6 x + 8 2 M1 38 seen

This question in 0580/23 May/June 2010

14 Find the value of n in the following equations. (a) 2n = 1024 Answer(a) n = [1] (b) 42n – 3 = 16 Answer(b) n = [2]

3 marks

Mark scheme: 14 (a) 10(.0) 1 1 (b) 2 , 2.5(0) 2 M1 2n – 3 = 2 2 1

This question in 0580/21 Oct/Nov 2010

16 Simplify 1  16 16  2 (a)  x  ,  81  Answer(a) [2] 16 y 10 × 4 y −4 (b) 7 . 32 y Answer(b) [2]

4 marks

Mark scheme: 4 4 16 (a) x8 2 B1 B1 x8 9 9 (b) 2y–1 2 B1 2 B1 y–1

This question in 0580/22 Oct/Nov 2010

13 a × 107 + b × 106 = c × 106 Find c in terms of a and b. Give your answer in its simplest form. Answer c = [2]

2 marks

Mark scheme: 13 10a + b or a × 101 + b (× 100) 2 M1 [a × 107 + b × 106] ÷ 106 IGCSE – October/November 2010 0580 23 70

This question in 0580/23 Oct/Nov 2010

4 (a) Find m when 4m × 42 = 412 . Answer(a) m = [1] p 5 (b) Find p when 6 ÷ 6 = 6 . Answer(b) p = [1]

2 marks

Mark scheme: 4 (a) 10 1 (b) 5.5 oe 1

This question in 0580/21 May/June 2011

3 Use your calculator to find the value of (a) 30 × 2.52, Answer(a) [1] (b) 2.5 – 2. Answer(b) [1]

2 marks

Mark scheme: 3 (a) 6.25 cao 1 (b) 0.16 cao 1

This question in 0580/22 Oct/Nov 2011

11 Find the values of m and n. For Examiner's (a) 2m = 0.125 Use Answer(a) m = [2] (b) 24n × 22n = 512 Answer(b) n = [2]

4 marks

Mark scheme: 11 (a) –3 2 M1 1/23 or 2–3 (b) 1.5 2 M1 26n or 6n = 9

This question in 0580/22 Oct/Nov 2011

14 Simplify the following. For Examiner's 2 Use (a) (4pq )3 Answer(a) [2] 1 − (b) (16 x 8 ) 4 Answer(b) [2]

4 marks

Mark scheme: 14 (a) 64p3q6 2 B1 64puqv or kp3q6 1 1 1 (b) 0.5x–2 or 2 oe 2 B1 u oe or 2 oe 2 x 2 x kx

This question in 0580/23 May/June 2012

 3  83  3  18 q14 (a)   ×   = p  8   8  Find the value of p and the value of q. Answer(a) p = q = [2] O3 O4 O4 (b) 5 + 5 = k × 5 Find the value of k. Answer(b) k = [2]

4 marks

Mark scheme: 14 3 1 2 9 1 3 1 (a) p = q = B2 p = and q = or B1 p = q ≠ 8 2 64 4 8 2 5 −+3 5 −4 2 M1 for a correct statement for k e.g. − 4 or for (b) k = 6 5 the factorisation 5–4 (5 + 1) = k × 5–4 or 1 k (5 + )1 = 625 625

This question in 0580/22 Oct/Nov 2012

4 11 Simplify (256w256) . Answer [2]

2 marks

Mark scheme: 11 4w64 2 B1 for 4wn or kw64

This question in 0580/23 Oct/Nov 2012

16 (a) (224) 2 = p4 Find the value of p. Answer(a) p = … [2] q2 + q 2 (b) Simplify 1 1 . q 4 # q 4 Answer(b) … [3] __________________________________________________________________________________________

5 marks

Mark scheme: 16 (a) 8 2 B1 for 212 or 4096 3 3 (b) 2 q 2 3 B2 for kq 2 as the answer or 1 B1 for 2q2 and B1 for q 2 oe nfww

This question in 0580/21 May/June 2014

17 (a) Simplify (3125t 125) 5. Answer(a) … [2] p 1 (b) Find the value of p when 3 = . 9 Answer(b) p = … [1] (c) Find the value of w when x72 ÷ xw = x8. Answer(c) w = … [1] __________________________________________________________________________________________

4 marks

Mark scheme: 17 (a) 5t25 2 B1 for 5tk or mt25 (m ≠ 0) (b) –2 1 (c) 64 1 M1 f 1458 3456

This question in 0580/22 May/June 2014

11 (a) Simplify x8 ÷ x2. Answer(a) … [1] x6 13 (b) Simplify . 27 ` j Answer(b) … [2] __________________________________________________________________________________________

3 marks

Mark scheme: 11 (a) x6 1 2 x (b) 2 B1 for answer kx2 or x3k or 13 3 5

This question in 0580/21 Oct/Nov 2014

21 (a) Simplify (i) x0, Answer(a)(i) … [1] (ii) m4 × m3, Answer(a)(ii) … [1] 1 (iii) ( 8p6) 3 . Answer(a)(iii) … [2] (b) 243 x = 3 2 Find the value of x. Answer(b) x = … [2] __________________________________________________________________________________________ Question 22 is printed on the next page.

6 marks

Mark scheme: 21 (a) (i) 1 1 (ii) 7 1 m (iii) 2p 2 2 SC1 for 2 p k or kp 2 k ≠ 0 2 1 2 5 or 243 5 seen (b) or 0.4 2 B1 for 53 or 3 5 x or 243 5

This question in 0580/22 Feb/March 2015

14 Simplify. (a) x 3 y 4 # x 5 y 3 … [2] (b) ^3p 2 m 5h3 … [2]

4 marks

Mark scheme: 14 (a) x 8 y 7 final answer 2 B1 for answer x 8 y k or kx y 7 (k ≠ 0) (b) 27p 6 m15 final answer 2 B1 for 2 correct of 27, p 6 , m15 in a product as answer 2.8 2 + 3.6 2 − 5.32

This question in 0580/22 Feb/March 2016

2 Simplify. n2 × n5 … [1]

1 marks

Mark scheme: 2 n7 final answer 1

This question in 0580/23 Oct/Nov 2016

15 Work out. (a) t 24 ' t 4 … [1] (b) x5 2 ^ h … [1] 8 43 (c) 81m ^ h … [2]

4 marks

Mark scheme: 15 (a) t 20 final answer 1 (b) x10 final answer 1 (c) 27m6 final answer 2 B1 for 27 m k or km 6 as final answer

This question in 0580/22 Feb/March 2017

r 16 (a) 2 = 16 Find the value of r. r = … [1] (b) 3 t = 5 3 Find the value of t. t = … [1]

2 marks

Mark scheme: 6(a) − 4 1 6(b) 1 1 or 0.2 5

This question in 0580/21 May/June 2017

20 Simplify. (a) 6w0 … [1] (b) 5x3 − 3x3 … [1] (c) 3y6 # 5y-2 … [2]

4 marks

Mark scheme: 20(a) 6 1 20(b) 2x 3 final answer 1 20(c) 15y 4 final answer 2 B1 for 15 ky or ky 4 as final answer (k ≠ 0)

This question in 0580/23 May/June 2017

2 Work out. 2 -4 # 2 5 … [1]

1 marks

Mark scheme: 2 2 1

This question in 0580/21 Oct/Nov 2017

13 Simplify. (a) m5 2 ^ h … [1] (b) 4x 3 y # 5 x 2 y … [2]

3 marks

Mark scheme: 13(a) m10 final answer 1 13(b) 20x5y2 final answer 2 B1 for 2 out of 3 elements correct in final answer or correct answer spoiled

This question in 0580/21 Oct/Nov 2017

-q 19 3 # = 81 27 Find the value of q. q = … [2]

2 marks

Mark scheme: 9 −7 2 B1 for 3−3 or 34 or 73 or 3–7 seen or SC1 for final answer 7

This question in 0580/21 May/June 2018

17 (a) t x # t 2 = t 10 Find the value of x. x = … [1] (b) Simplify. 4 -2 (i) c x m … [1] (ii) a 3 b 7 ' a 6 b 2 … [2]

4 marks

Mark scheme: 17(a) 8 1 17(b)(i) x 2 1 final answer 16 17(b)(ii) −3 5 b 5 2 a b or final answer −3 k k 5 3 a B1 for a b or a b

This question in 0580/21 Oct/Nov 2018

23 f (x) = 7 + 3x g (x) = x4 h (x) = 3x (a) h (3x) = k x Find the value of k. k = … [2] (b) Find the value of x when f (x) = g (2) . x = … [2] (c) Find f -1 (x) . f -1 (x) = … [2]

6 marks

Mark scheme: 23(a) 27 2 M1 for 33x seen 23(b) 3 2 M1 for 7 + 3x = 24 23(c) x − 7 2 M1 for x = 7 + 3y oe final answer y 7 3 or y – 7 = 3x or –3x = 7 – y or = + x 3 3

This question in 0580/21 Oct/Nov 2018

w 216 (a) Simplify 3 . w … [1] (b) Simplify 3w 3 3 . ^ h … [2]

3 marks

Mark scheme: 16(a) 1 −1 1 or w w 16(b) 27w9 final answer 2 B1 for kw9 or 27 w k

This question in 0580/23 Oct/Nov 2018

6 Simplify. (a) t21 ÷ t7 … [1] (b) (u5)5 … [1]

2 marks

Mark scheme: 6(a) t14 final answer 1 6(b) u 25 final answer 1

This question in 0580/21 May/June 2019

3 18 (a) Simplify 81y 16 4 . ` j … [2] (b) 2 3 = 4 p Find the value of p. p = … [1]

3 marks

Mark scheme: 18(a) 27y 12 final answer 2 B1 for ky12 or 27 ky in final answer 18(b) 3 1 oe 2

This question in 0580/21 May/June 2019

12 Simplify. (a) 5m 2 # 2m 3 … [2] 3 (b) x 8 ` j … [1]

3 marks

Mark scheme: 12(a) 10m 5 final answer 2 B1 for 10 m k or km 5 as final answer 12(b) x 24 final answer 1

This question in 0580/22 May/June 2019

10 Simplify. 4 e x83 o- 3 … [2]

2 marks

Mark scheme: 1 4 210 16 − 4 3 or 16x−4 3  x12  –  8   x  4 x M1 for   or  3  or   or  2   x   4096  better 16 k or B1 for or 16xk or or kx–4 final kx x 4 answer

This question in 0580/21 Oct/Nov 2019

21 (a) 3 -2 # 3 x = 81 Find the value of x. x = … [2] 1 (b) x - 3 = 32 x -2 Find the value of x. x = … [3]

5 marks

Mark scheme: 21(a) 6 2 B1 for 34 or 3x–2 or M1 for 3x = 81 × 32 or better 21(b) 8 3 5 M2 for x 3 = 32 or better 1 32 or M1 for = or better 1 x 2 x 3 1 − 2 or x 3 −−= 32 or better

This question in 0580/22 Oct/Nov 2019

13 Simplify 8t 8 ' 4t 4 . … [2]

2 marks

Mark scheme: 13 2t 4 2 B1 for 2tn or kt4 (n,k ≠ 0)

This question in 0580/22 May/June 2020

6 Simplify. (a) p 2 # p 4 … [1] (b) m 15 ' m 5 … [1] (c) ( k3 ) 5 … [1]

3 marks

Mark scheme: 6(a) p 6 1 6(b) m10 1 6(c) k 15 1

This question in 0580/23 May/June 2020

11 Simplify. 2x 2 # 5x 5 … [2]

2 marks

Mark scheme: 11 10x7 final answer 2 B1 for kx7 or 10xk final answer or for correct answer then spoilt

This question in 0580/22 Oct/Nov 2020

18 (a) Simplify. 3 4xy 2 ` j … [2] (b) 25 = 125k Find the value of k. k = … [1]

3 marks

Mark scheme: 18(a) 64x3y6 final answer 2 B1 for kx3y6 or 64xky6 or 64x3yk final answer or correct answer then spoilt 18(b) 2 1 3

This question in 0580/23 Oct/Nov 2020

11 (a) Simplify fully. ( 4ab 5 ) 4 … [2] 3 = 6 (b) 2p 1 Find the value of p. p = … [1] (c) 812 ' 3 t = 9 Find the value of t. t = … [2]

5 marks

Mark scheme: 11(a) 256a 4 b 20 final answer 2 B1 for two correct elements in final answer 11(b) 27 1 11(c) 6 2 M1 for 3k ÷ 3t = 32 or 38 ÷ 3t = 3 k oe or better or 3t = 729 oe

This question in 0580/21 May/June 2021

22 (a) Simplify. 2 x 3 8 x 3 … [1] (b) 16 = 64k Find the value of k. k = … [1] (c) Solve. 4 - 3 x 3 x 1 3 # = 3 e 9 o x = … [3]

5 marks

Mark scheme: 22(a) –2 1 1 x or final answer x 2 22(b) 2 1 3 22(c) 1 nfww 3 M1 for 3–2(4 – 3x) oe or better 3 x 1 or 9 2 × 9 − ( 4 −x3 ) = 9 2 oe or better M1 for 3x + (their– 2)× (4 – 3x)= 1 oe or better 3x 1 or their − ( 4 − 3 x ) = their oe or better 2 2

This question in 0580/21 Oct/Nov 2021

13 (a) Write 243 # 27 2 n as a single power of 3 in terms of n. … [2] (b) k = 2 # 32 # p 3 , where p is a prime number greater than 3. Write 6k 2as a product of prime factors in terms of p. … [2]

4 marks

Mark scheme: 13(a) 36n + 5 final answer 2 B1 for 35 or (33)2n or better or answer 6n + 5 13(b) 23 × 35 × p6 final answer 2 B1 for two parts correct or 2 × 3 × 2 × 32 × p3 × 2 × 32 × p3 or 1944p6 or k2 = 22 × 34 × p6

This question in 0580/22 Oct/Nov 2021

13 (a) Simplify h 2 # h 5 . … [1] 7 -3 (b) Simplify . b x l … [1] (c) a 8 ' a p = a 2 Find the value of p. p = … [1]

3 marks

Mark scheme: 13(a) h 7 final answer 1 13(b) x 3 1 final answer 343 13(c) 6 1

This question in 0580/22 Feb/March 2022

x 115 4 = 64 Find the value of x. x = … [1]

1 marks

Mark scheme: 15 –3 1

This question in 0580/21 May/June 2022

11 Find the value of p when 6 p # 6 4 = 6 28 . p = … [1]

1 marks

Mark scheme: 11 24 1

This question in 0580/22 May/June 2022

6 Simplify. (a) y 3 ' y 5 … [1] (b) 7x 0 … [1]

2 marks

Mark scheme: 6(a) –2 1 1 y or final answer y 2 6(b) 7 1

This question in 0580/23 May/June 2022

1 25 m - 4 = 27m -1 Find the value of m. m = … [3]

3 marks

Mark scheme: 25 81 3 3 M2 for m 4  27 or better 1 27 or M1 for or better 1 m m 4 1  1 or m 4 = 27 1 If 0 scored SC1 for answer 81

This question in 0580/23 May/June 2022

10 Simplify 18 x 18 ' 9 x 9 . … [2]

2 marks

Mark scheme: 10 2x 9 final answer 2 B1 for kx 9 or 2 kx as final answer or 2x 9 spoiled

This question in 0580/21 Oct/Nov 2022

5 Simplify d 8 ' d 2 . … [1]

1 marks

Mark scheme: 5 d 6 1

This question in 0580/23 May/June 2023

10 Simplify. (a) n 5 # n … [1] (b) 8x 6 ' 2x 2 … [2] 2 (c) ( 243y 20 ) 5

5 marks

Mark scheme: 10(a) n6 final answer 1 10(b) 4x4 final answer 2 B1 for kx4 or 4xk as final answer or correct answer seen and then spoiled 10(c) 9y8 final answer 2 B1 for ky8 or 9yk final answer or correct answer seen and spoiled

This question in 0580/21 Oct/Nov 2023

13 (a) 3 3 p # 3 2p = 729 Find the value of p. p = … [2] (b) Simplify. 1 ( 32x 10 ) 5 … [2]

4 marks

Mark scheme: 13(a) 1.2 oe 2 B1 for 32 p + 3 p or 3 6 soi 13(b) 2x 2 final answer 2 B1 for kx 2 or 2 kx as final answer or correct answer spoiled

This question in 0580/22 Oct/Nov 2023

21 (a) 5 3 = 3h Write down the value of h. h = … [1] 3 (b) Simplify 4x 3 . ` j … [2]

3 marks

Mark scheme: 21(a) 1 1 oe 5 21(b) 64x9 2 B1 for 64 kx or kx 9 as final answer or correct answer spoiled

This question in 0580/22 Feb/March 2024

12 Simplify. 32g 16 (a) 16g 8 … [2] 3 (b) `625k 8j4 … [2]

4 marks

Mark scheme: 12(a) 2g 8 final answer 2 B1 for final answer kg 8 or 2 g k or correct answer seen then spoilt 12(b) 125k 6 final answer 2 B1 for final answer ck 6 or 125k c or correct answer seen then spoilt

This question in 0580/23 May/June 2024

4 V = 4 mp 2 (a) Find V when m = 10 and p =- 3 . V = … [2] (b) Find the positive value of p when V = 3200 and m = 2 . p = … [2]

4 marks

Mark scheme: 4(a) 360 2 M1 for 4 × 10 × (–3)2 oe If 0 scored, SC1 for answer –360 4(b) 20 2 M1 for [p2 = ] 3200 ÷ (2 × 4) oe

This question in 0580/22 Feb/March 2025

x 2 223 (a) Simplify e o . 4 … [2] x 1 x x + 3 (b) 16 # b l = 4 2 Find the value of x. x = … [4]

6 marks

Mark scheme: 23(a) 3 2 3x kx [ x] final answer B1 for or 8 k 8 23(b) 6 nfww 4 B3 for 4x – x = 2(x + 3) oe or for correctly combining to a single base on each side with no brackets in the powers e.g. 23x = 22x + 6 oe or better OR M1 for (24)x or (2–1)x or 22(x + 3) or better seen M1 for correctly forming a linear equation in x from their powers of their consistent base

This question in 0580/22 Feb/March 2025

6 Find the value of (a) 5 -5 # 5 5 … [1] 3 (b) 125 2. … [2]

3 marks

Mark scheme: 6(a) 1 cao 1 6(b) 25 cao 2 2 2 3 3 125 5 3 M1 for or 3 1252 or ( ) ( ) or B1 for 3 125 = 5

This question in 0580/21 Oct/Nov 2025

13 M = 2 7 # 3 3 # 5 2 (a) Write 14M as a product of its prime factors. Give your answer in index form. … [2] (b) R is an integer. M is a cube number. R Find the smallest possible value of R. R = … [2]

4 marks

Mark scheme: 13(a) 28 × 33 × 52 × 7[1] cao 2 M1 for [14 =] 2 × 7 soi 13(b) 50 2 M1 for 23 × 33 [× 50] or 26 × 33 [× 50] oe or or 24 × 52 or 2 × 52 oe –86 400 or SC1 for answers –50, [–]400, [–]1350, [–]3200, [–]10 800, 86 400 oe

This question in 0580/22 Oct/Nov 2025

14 Find the value of (a) 3 2 ' 3 -2 … [2] - 23 (b) 16 . … [2]

4 marks

Mark scheme: 14(a) 81 2 B1 for 34 or 32 × 9 or 9 × 9 14(b) 1 2 1 1 [  ]64 M1 for 4 3 or 64–1 or 4096

This question in 0580/22 Oct/Nov 2025

24 27 nx = ( 9 x) 2 Find the value of n. n = … [2]

2 marks

Mark scheme: 24 4 2 B1 for 33n[x] or 34[x] [n =] 3

This question in 0580/22 Oct/Nov 2025

9 n 15 ' n x = n 5 Find the value of x. x = … [1]

1 marks

Mark scheme: 9 10 1

This question in 0580/23 Oct/Nov 2025

12 (a) Write down the value of 730. … [1] 7 (b) Find the value of -2 . 3 … [2] (c) Write 27 # 812 in the form 3n. … [2]

5 marks

Mark scheme: 12(a) 1 1 12(b) 63 2 1 M1 for 3–2 = soi 9 12(c) 311 2 M1 for 33 or (34)2 or better

This question in 0580/23 Oct/Nov 2025