Class 10 Mathematics Unit 9 – Indices | Nepal eNotes

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CLASS 10 MATHEMATICS • UNIT 9

Unit 9: Indices

Exercise 9.1 — WordPress-ready solutions reconstructed from the supplied 23-page handwritten notes.

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Source note: The uploaded file name says “Unit 9 – Radical and Surds,” but the actual handwritten pages are headed Lesson 9: Indices and contain Exercise 9.1. The scanned page content has therefore been used as the source of truth. Page 10 is heavily blurred, so unreadable expressions from that page have not been invented.

Important Laws of Indices

Product Rule

a^m × aⁿ = a^m+n

Quotient Rule

a^m / aⁿ = a^m−n

Power of a Power

(a^m)ⁿ = a^mn

Negative Index

a⁻ⁿ = 1/aⁿ

Zero Index

a⁰ = 1, a ≠ 0

Fractional Index

a^m/n = ⁿ√(a^m)

PDF Pages 1–23

Exercise 9.1

PDF Pages 1–3

Question 2 – Solve and Examine

(a) 3ⁿ = 9

3ⁿ = 3²

Comparing powers:

∴ n = 2

(b) 5ⁿ⁻¹ = 25

5ⁿ⁻¹ = 5²

n − 1 = 2

∴ n = 3

(c) 1 / 5²ⁿ⁻⁴ = 125

5⁻²ⁿ⁺⁴ = 5³

−2n + 4 = 3

2n = 1

∴ n = 1/2

(d) 4ⁿ⁻² = 0.125

0.125 = 1/8 = 2⁻³

4ⁿ⁻² = 2²ⁿ⁻⁴

2n − 4 = −3

∴ n = 1/2

(e) (3/5)ⁿ = 125/27

125/27 = (5/3)³ = (3/5)⁻³

∴ n = −3

(f) 2ⁿ × 3ⁿ⁻¹ = 18

2ⁿ × 3ⁿ × 3⁻¹ = 18

6ⁿ = 54

The handwritten source then rewrites the relation and concludes:

∴ n = 1

Source fidelity note: The handwritten algebra in part (f) contains a compressed intermediate step. The final source answer is n = 1 and is preserved here.

PDF Pages 3–9

Question 3 – Solve

(a) 9^{(1−n)/(1+n)} = 9^1/3

(1−n)/(1+n) = 1/3

3(1−n) = 1+n

3−3n = 1+n

2 = 4n

∴ n = 1/2

(b) ⁽²ⁿ⁺⁴⁾√[4ⁿ⁺⁸] = ⁶√128

4 = 2² and 128 = 2⁷.

2^{2(n+8)/(2n+4)} = 2^7/6

2(n+8)/(2n+4) = 7/6

6(2n+16) = 7(2n+4)

12n + 96 = 14n + 28

68 = 2n

∴ n = 34

(c) 2ⁿ⁺¹ + 2ⁿ⁺² + 2ⁿ⁺³ = 448

2ⁿ(2 + 4 + 8) = 448

14 × 2ⁿ = 448

2ⁿ = 32 = 2⁵

∴ n = 5

(d) 3ⁿ⁺¹ − 3ⁿ = 162

3ⁿ(3−1) = 162

2 × 3ⁿ = 162

3ⁿ = 81 = 3⁴

∴ n = 4

(e) 4ⁿ⁺¹ − 8 × 4ⁿ⁻¹ = 32

4ⁿ(4 − 8/4) = 32

2 × 4ⁿ = 32

4ⁿ = 16 = 4²

∴ n = 2

(f) 4 × 3ⁿ⁺¹ − 3ⁿ⁺² − 3ⁿ⁻¹ = 72

3ⁿ(12 − 9 − 1/3) = 72

3ⁿ × 8/3 = 72

3ⁿ = 27 = 3³

∴ n = 3

(g) 3ⁿ⁺² + 3ⁿ⁺¹ + 2 × 3ⁿ = 126

3ⁿ(9+3+2) = 126

14 × 3ⁿ = 126

3ⁿ = 9 = 3²

∴ n = 2

(h) 2ⁿ + 3ⁿ⁻² = 3ⁿ − 2ⁿ⁺¹

2ⁿ + 2ⁿ⁺¹ = 3ⁿ − 3ⁿ⁻²

3 × 2ⁿ = 3ⁿ(1−1/9)

3 × 2ⁿ = (8/9)3ⁿ

(2/3)ⁿ = (2/3)³

∴ n = 3

(i) 8ⁿ⁻¹ − 2³ⁿ⁻² + 8 = 0

2³ⁿ⁻³ − 2³ⁿ⁻² = −8

2³ⁿ(1/8 − 1/4) = −8

−2³ⁿ/8 = −8

2³ⁿ = 64 = 2⁶

3n = 6

∴ n = 2

PDF Page 10

Page 10 – Source Gap

Page 10 is severely out of focus. It appears to contain the end of an additional index problem and the beginning of Question 4, but the expressions and numbers are not readable with enough confidence to type accurately. They are therefore not reconstructed or guessed.

PDF Pages 11–15

Question 4 – Solve

(a) 5ⁿ + 5⁻ⁿ = 5 + 1/5

Let 5ⁿ = a.

a + 1/a = 26/5

5a² − 26a + 5 = 0

(a−5)(5a−1) = 0

a = 5 or a = 1/5

∴ n = ±1

(b) 7ⁿ + 7⁻ⁿ = 7 + 1/7

Let 7ⁿ = a.

a + 1/a = 50/7

7a² − 50a + 7 = 0

(a−7)(7a−1) = 0

∴ n = ±1

(c) 9ⁿ + 9⁻ⁿ = 9 + 1/9

Let 9ⁿ = a.

a + 1/a = 82/9

9a² − 82a + 9 = 0

(a−9)(9a−1) = 0

∴ n = ±1

(d) 4ⁿ + 4⁻ⁿ = 16 + 1/16

Let 4ⁿ = a.

a + 1/a = 257/16

16a² − 257a + 16 = 0

(a−16)(16a−1) = 0

a = 16 or a = 1/16

∴ n = ±2

(e) 5ⁿ + 5⁻ⁿ = 25 + 1/25

Let 5ⁿ = a.

a + 1/a = 626/25

25a² − 626a + 25 = 0

(a−25)(25a−1) = 0

∴ n = ±2

(f) 81 × 3ⁿ + 3⁻ⁿ = 30

Let 3ⁿ = a.

81a + 1/a = 30

81a² − 30a + 1 = 0

(3a−1)(27a−1) = 0

a = 1/3 or a = 1/27

∴ n = −1, −3

PDF Pages 16–21

Question 5 – Solve

(a) 4 × 3ⁿ⁺¹ − 9ⁿ = 27

Let 3ⁿ = a.

12a − a² = 27

a² − 12a + 27 = 0

(a−3)(a−9) = 0

∴ n = 1, 2

(b) 3 × 2^p+1 − 4^p = 8

Let 2^p = a.

6a − a² = 8

a² − 6a + 8 = 0

(a−4)(a−2) = 0

∴ p = 2, 1

(c) 5²ⁿ − 6 × 5ⁿ⁺¹ + 125 = 0

Let 5ⁿ = a.

a² − 30a + 125 = 0

(a−25)(a−5) = 0

∴ n = 2, 1

(d) 2ⁿ⁻² + 2³⁻ⁿ = 3

Let 2ⁿ = a.

a/4 + 8/a = 3

a² + 32 = 12a

a² − 12a + 32 = 0

(a−8)(a−4) = 0

∴ n = 3, 2

(e) 5ⁿ⁺¹ + 5²⁻ⁿ = 126

Let 5ⁿ = a.

5a + 25/a = 126

5a² − 126a + 25 = 0

(5a−1)(a−25) = 0

∴ n = −1, 2

(f) 3^2y − 4 × 3^y + 3 = 0

Let 3^y = a.

a² − 4a + 3 = 0

(a−1)(a−3) = 0

∴ y = 0, 1

PDF Pages 21–22

Question 6 – Solve and Verify

Solve 16ⁿ − 5 × 4ⁿ⁺¹ + 64 = 0 and show that the values of n satisfy 5ⁿ + 125/5ⁿ = 30.

16ⁿ = (4ⁿ)².

Let 4ⁿ = a.

a² − 20a + 64 = 0

(a−4)(a−16) = 0

a = 4 or a = 16

Therefore:

n = 1 or n = 2

Verification for n = 1

5¹ + 125/5¹ = 5 + 25 = 30

∴ 30 = 30 (True)

Verification for n = 2

5² + 125/5² = 25 + 5 = 30

∴ 30 = 30 (True)

PDF Pages 22–23

Question 7 – Prove

(a) If n = 3^1/3 + 3^−1/3, prove that 3n(n²−3) = 10.

Cubing both sides:

n³ = (3^1/3 + 3^−1/3)³

= 3 + 3(3^1/3)(3^−1/3)n + 1/3

= 3 + 3n + 1/3

= (10 + 9n)/3

3n³ = 10 + 9n

3n³ − 9n = 10

∴ 3n(n²−3) = 10

(b) If n = 2^1/3 − 2^−1/3, prove that 2n³ + 6n − 3 = 0.

Cubing both sides:

n³ = (2^1/3 − 2^−1/3)³

= 2 − 3(2^1/3)(2^−1/3)n − 1/2

= 2 − 3n − 1/2

= (3 − 6n)/2

2n³ = 3 − 6n

∴ 2n³ + 6n − 3 = 0

Original Scanned PDF – View Notes

Important Laws of Indices

Exercise 9.1

Question 2 – Solve and Examine

(a) 3n = 9

(b) 5n−1 = 25

(c) 1 / 52n−4 = 125

(d) 4n−2 = 0.125

(e) (3/5)n = 125/27

(f) 2n × 3n−1 = 18

Question 3 – Solve

(a) 9(1−n)/(1+n) = 91/3

(b) ⁽²ⁿ⁺⁴⁾√[4n+8] = ⁶√128

(c) 2n+1 + 2n+2 + 2n+3 = 448

(d) 3n+1 − 3n = 162

(e) 4n+1 − 8 × 4n−1 = 32

(f) 4 × 3n+1 − 3n+2 − 3n−1 = 72

(g) 3n+2 + 3n+1 + 2 × 3n = 126

(h) 2n + 3n−2 = 3n − 2n+1

(i) 8n−1 − 23n−2 + 8 = 0

Page 10 – Source Gap

Question 4 – Solve

(a) 5n + 5−n = 5 + 1/5

(b) 7n + 7−n = 7 + 1/7

(c) 9n + 9−n = 9 + 1/9

(d) 4n + 4−n = 16 + 1/16

(e) 5n + 5−n = 25 + 1/25

(f) 81 × 3n + 3−n = 30

Question 5 – Solve

(a) 4 × 3n+1 − 9n = 27

(b) 3 × 2p+1 − 4p = 8

(c) 52n − 6 × 5n+1 + 125 = 0

(d) 2n−2 + 23−n = 3

(e) 5n+1 + 52−n = 126

(f) 32y − 4 × 3y + 3 = 0

Question 6 – Solve and Verify

Solve 16n − 5 × 4n+1 + 64 = 0 and show that the values of n satisfy 5n + 125/5n = 30.

Verification for n = 1

Verification for n = 2

Question 7 – Prove

(a) If n = 31/3 + 3−1/3, prove that 3n(n²−3) = 10.

(b) If n = 21/3 − 2−1/3, prove that 2n³ + 6n − 3 = 0.

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(a) 3ⁿ = 9

(b) 5ⁿ⁻¹ = 25

(c) 1 / 5²ⁿ⁻⁴ = 125

(d) 4ⁿ⁻² = 0.125

(e) (3/5)ⁿ = 125/27

(f) 2ⁿ × 3ⁿ⁻¹ = 18

(a) 9^{(1−n)/(1+n)} = 9^1/3

(b) ⁽²ⁿ⁺⁴⁾√[4ⁿ⁺⁸] = ⁶√128

(c) 2ⁿ⁺¹ + 2ⁿ⁺² + 2ⁿ⁺³ = 448

(d) 3ⁿ⁺¹ − 3ⁿ = 162

(e) 4ⁿ⁺¹ − 8 × 4ⁿ⁻¹ = 32

(f) 4 × 3ⁿ⁺¹ − 3ⁿ⁺² − 3ⁿ⁻¹ = 72

(g) 3ⁿ⁺² + 3ⁿ⁺¹ + 2 × 3ⁿ = 126

(h) 2ⁿ + 3ⁿ⁻² = 3ⁿ − 2ⁿ⁺¹

(i) 8ⁿ⁻¹ − 2³ⁿ⁻² + 8 = 0

(a) 5ⁿ + 5⁻ⁿ = 5 + 1/5

(b) 7ⁿ + 7⁻ⁿ = 7 + 1/7

(c) 9ⁿ + 9⁻ⁿ = 9 + 1/9

(d) 4ⁿ + 4⁻ⁿ = 16 + 1/16

(e) 5ⁿ + 5⁻ⁿ = 25 + 1/25

(f) 81 × 3ⁿ + 3⁻ⁿ = 30

(a) 4 × 3ⁿ⁺¹ − 9ⁿ = 27

(b) 3 × 2^p+1 − 4^p = 8

(c) 5²ⁿ − 6 × 5ⁿ⁺¹ + 125 = 0

(d) 2ⁿ⁻² + 2³⁻ⁿ = 3

(e) 5ⁿ⁺¹ + 5²⁻ⁿ = 126

(f) 3^2y − 4 × 3^y + 3 = 0

Solve 16ⁿ − 5 × 4ⁿ⁺¹ + 64 = 0 and show that the values of n satisfy 5ⁿ + 125/5ⁿ = 30.

(a) If n = 3^1/3 + 3^−1/3, prove that 3n(n²−3) = 10.

(b) If n = 2^1/3 − 2^−1/3, prove that 2n³ + 6n − 3 = 0.