Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8B Solutions

Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8B Solutions

EXERCISE – 8B

 

(1) Express in terms of log2 and log 3 :

(i) log 36

Solution :
log 36
log (2×2×3×3)
log (2² ×3²)
log2² + log3²   {∵ log (a×b) = log a + log b}
2 log2 + 3 log3

 

(ii) log 144

Solution :
log (2×2×2×2×3×3)
log (2⁴ ×3²)
log 2⁴ + log 3²
4 log2 + 2 log3

 

(iii) log 4.5

Solution :
log 4.5
log (45/10)
log (9/2)
log 9 – log 2   {log (a/b) = log a – log b}
log 3² – log 2
2 log 3 – log 2

 

(iv) log 26/51 – log 91/119

Solution :
log 26/51 – log 91/119
= log (26/51/91/119)
= log 26/51 × 119/91
= log (2/3)
= log2 – log3

 

(v) log 75/16 – 2log 5/9 + log 32/243

Solution :
log 75/16 – 2 log 5/9 + log 32/243
= log (3×5×5/2⁴) – 2 log5/9 + log 2⁵/3⁵
= log3 + log5² – log2⁴ – 2 {log5 – log3²) + log2⁵ – log 3⁵
= log3 + 2log5 – 4log2 – 2log5 + 4log3 + 5log2 – 5log 3
= 5log3 – 4log2 + 5log2 – 5log3
= – 4log2 + 5log2
= log2

 

(2) Express each of the following in a form free from logarithm:

(i) 2 log x – log y = 1

Solution :
2 log x – log y = 1
log x² – logy = 1
log₁₀ (x²/y) = 1
(10)¹ = x²/y
10 = x²
10y = x²

 

(ii) 2 log x + 3 log y = log a
Solution :
2 log x + 3 log y = log a
=> log x² + log y³ = log a
=> log (x² × y³) = log a
=> x²y³ = a

 

(iii) a log x – b log y = 2 log 3
Solution :
a log x – b log y = 2 log 3
=> log x^a – log y^b = log 3²
=> log x^a/y^b = log 9
=> x^a/y^b = 9
=> x^a = 9y^b

 

(3) Evaluate each of the following without using tables :

(i) log 5 + log 8 – 2 log 2

Solution :
log 5 + log 8 – 2 log 2
=> log (5×8) – 2 log 2
=> log 40 – log 2²
=> log 40 – log 4
=> log (40/4)
=> log 10
=> 1

 

(ii) Log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 – log₁₀ 18

Solution :
log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 – log₁₀ 18
= log₁₀ (8×25) + log 3² – log₁₀ 18
= log₁₀ (8×25×3²) – log₁₀ 18
= log₁₀ (8×25×9) – log₁₀ 18
= log₁₀ (1800) – log₁₀ 18
= log₁₀ (1800/18)
= log₁₀ (100)
= log₁₀ 10²
= 2 log₁₀ 10
= 2 × 1
= 2

 

(iii) log 4 + 1/3 log 125 – 1/5 log 32

Solution :
log4 + 1/3 log 125 – 1/5 log 32
= log4 + log (125)^1/3 – log (32)^1/5
= log4 + log5 – log2
= log (4×5) – log2
= log (4×5/2)
= log (20/2)
= log 10
= 1

 

(4) Prove that :

2 log 15/18 – log 25/162 + log 4/9 = log2

Solution :
2 log 15/18 – log 25/162 + log 4/9 = log2
L.H.S. = 2 log 15/18 – log 25/162 + log 4/9
=> log (15/18)² – log 25/162 + log 4/9
=> log 225/324 – log 25/162 + log 4/9
=> log (225/324 × 4/9 / 25/162)
=> log (25/21 × 162/25)
=> log 2
L.H.S. = R.H.S.

 

(5) Find x, if :

x – log48 + 3 log2 = 1/3 log 125 – log3

Solution :
x – log48 + 3 log2 = 1/3 log125 – log3
x – log48 + log2³ = log (125)^1/3 – log3
x = log48 – log8 + log5 – log3
x = log (48×5/8×3)
x = log10
x = 10  (∵ log₁₀ 10 = 1)

 

(6) Express log₁₀ 2 + 1 in the form of log₁₀ x.
Solution :
log₁₀ 2 + 1
= log₁₀ 2 + log₁₀ 10   (∵ log₁₀ 10 = 1)
= log₁₀ (2 × 10)
= log₁₀ 20

 

(7) Solve for x :

(i) log₁₀ (x – 10) = 1
Solution :
log₁₀ (x – 10) = 1
10¹ = x – 10
– x = – 10 -10
– x = – 20
x = 20

 

(ii) log (x² – 21) = 2

Solution :
log₁₀ (x² – 21) = 2
10² = x² – 21
100 = x² – 21
100 + 21 = x²
121 = x²
Taking square root on both sides,
√121 = √x²
± 11 = x

 

(iii) log(x – 2) + log(x + 2) = log5

Solution :
log(x – 2) + log(x + 2) = log5
log (x – 2) (x + 2) = log5  (∵ log a + log b = log ab)
log ((x² – 4) = log5
x² – 4 = 5
x² = 5 + 4
x² = 9
Taking square root on both sides,
√x² = √9
x = ± 3

 

(iv) log (x + 5) + log (x – 5) = 4 log2 + 2 log3

Solution :
log(x + 5) + log(x – 5) = 4 log2 + 2 log3
= log [(x + 5) (x – 5)] = log2⁴ + log3²
= log [x² – 25] = log16 + log9
= log (x² – 25) = log (16 × 9)
= log (x² – 25) = log (144)
= x² – 25 = 144
= x² = 144 + 25
= x² = 169
Taking square root on both sides,
√x² = √169
x = ± 13

 

(8) Solve for x :

(i) log 81/log 27 = x
Solution :
log  81/log 27 = x
log 3⁴/log 3³ = x
4 log3/3 log3
4/3 = x

 

(ii) log 128/log 32 = x

Solution :
log 128/log 32 = x
log 2⁷/log 2⁵ = x
7 log2/5 log2 = x
7/5 = x

 

(iii) log 64/log 8 = log x

Solution :
log 64/log 8 = log x
log 2⁶/ log 2³ = log x
6 log2/3 log3 = log x
6/3 = log x
2 = log x
log x = 2
log₁₀ x = 2
10² = x

100 = x

 

(iv) log 225/log 15 = log x

Solution :
log 225/log 15 = log x
=> log 15²/log 15 = log x
=> 2 log15/log15 = log x
=> 2 = log x
=> log x = 2
=> log₁₀ x = 2
=> 10² = x
=> 100 = x

 

(9) Given log x = m + n and log y = m – n, express the value of log 10x/y² in terms of m and n.
Solution :
log 10x/y²
log10 + log x – log y²
1 + m + n – 2 log y  (∵ log 10 = 1)
1 + m + n – 2 (m – n)
1 + m + n – 2m – 2n
1 – m + 3n

 

(10) If log₁₀ 2 = a and log₁₀ 3 = b; express each of the following in terms of ‘a’ and ‘b’.
(i) log 12
Solution :
log₁₀ 2 = a and log₁₀ 3 = b
log₁₂ = log (2 × 2 × 3)
= log (2² × 3)
= log2² + log3
= 2 log2 + log3
= 2a + b (∵ Given)

 

(ii) log 2.25

Solution :
log 2.25
= log 225/100
= log (9/4)
= log 9 – log 4
= log 3² – log 2²
= 2 log3 – 2 log2
= 2b – 2a  (∵ Given)

 

(iii) log 2 1/4
Solution :
log 2 1/4
= log 8+1/4
= log (9/4)
= log9 – log4
= log3² – log2²
= 2 log3 – 2 log2
= 2b – 2a (∵ Given)

 

(iv) log 5.4

Solution :
log 5.4
= log 54/10
= log 54 – log 10
= log 54 – 1
= log (3×3×3×2) – 1
= log3³ + log2 – 1
= 3 log3 + log2 – 1
= 3b + a – 1

 

(v) Log 3 1/8

Solution :
log 3 1/8
= log 24 + 1/8
= log 25/8
= log 25 – log 8
= log 5² – log 2³
= 2 log5 – 3 log2
= 2 log (10/2) – 3 log2
= 2 (log10 – log2) – 3 log2
= 2 log10 – 2 log2 – 3 log2
= 2 × 1 – 2a – 3b
= 2 – 2a – 3b (Given)

 

(vi) log60

Solution :
log 60
= log (2×2×3×5)
= log (2×3×10)
= log2 + log3 + log10
= a + b + 1 (∵ Given)

 

(12) If log2 = 0.3010 and log3 = 0.4771 ; find the value of :

(i) log 12

Solution :
log 12 = log (2×2×3)
= log (2² × 3)
= log2² + log3
= 2 log2 + log3
= 2 (0.3010) + 0.4771
= 0.6020 + 0.4771
= 1.0791

 

(ii) log 1.2

Solution :
log 1.2 = log 12/10
= log (2×2×3)/10
= log (2² × 3)/10
= log2² + log3 – log10
= 2 log2 + log3 – log10
= 2 × 0.3010 + 0.4771 – 1
= 1.0791 – 1
= 0.0791

 

(iii) log 3.6

Solution :
log 3.6
= log 36/10
= log (2×2×3×3)/10
= log2² + log3² – log10
= 2 log2 + 2 log3 – 1
= 2 × (0.3010) + 2 × (0.4771) – 1
= 0.6020 + 0.9542 – 1
= 1.5562 – 1
= 0.5562

 

(iv) log 15

Solution :
log 15
= log (15 × 2)/2
= log (30/2)
= log (3×10/2)
= log3 + log10 – log2
= 0.4771 + 1 – 0.3010
= 1.1761

 

(v) log 25

Solution :
log 25 dividing and multiplying by 4.
= log 25 × 4/4
= log 100/4
= log 10²/4
= log 10²/2²
= log 10² – log 2²
= 2 log 10 – 2 log 2
= 2 × 1 – 2 × (0.3010)
= 2 – 0.6020
= 1.398

 

(vi) 2/3 log 8
Solution :
2/3 log 8
= 2/3 log 2³
= 2/3 × 3 log2
= 2 log2
= 2 × (0.3010)
= 0.6020

 

(13) Given 2 log₁₀ x + 1 = log₁₀ 250, find :

(i) x

Solution :
Given that : 2 log₁₀ x + 1 = log₁₀ 250
log₁₀ x² + log₁₀ 10 = log₁₀ 250   {∵ 1 = log₁₀}
log₁₀ (x² × 10) = log₁₀ 250
10x² = 250
x² = 250/10
x² = 25
Taking square root on both sides,
√x² = √25
x = ± 5.

 

(14) Given 3 log x + 1/2 log y = 2, express y in term of x.
Solution :
3 log x + 1/2 log y = 2
log₁₀ x³ + log y^1/2 = 2
log₁₀ x³ + log √y = 2
log₁₀ x² √y= 2
10² = x³ √y
100 = x³ √y
Squaring on both sides,
(100)² = (x³)² (√y)²
10000 = x⁶ y
10000/x⁶ = y
1000x^-6 = y

 

(15) If x = (100)^a, y = (10000)^b and z = (10)^c, find log 10√y/x² z³ in terms of a, b and c.
Solution :
log 10√y/x² z³
= log10 √(10000)^b/((100)^a)² ((10)^c)³
= log10 × √(10^4b)/(10²)^2a (10^3c)
= log10 × √(10)^4b/(10)^4a × (10)^3c
= log10 + log (10^4b)^1/2 – log10^4a – log^3c
= 1 + log10^2b – 4a log10 – 3c log10
= 1 + 2b log10 – 4a × 1 – 3c × 1
= 1 + 2b × 1 – 4a – 3c
= 1 + 2b – 4a – 3c

 

(16) If 3 (log5 – log3) – (log5 – 2log6) = 2 – log x, find x.
Solution :
3 log5 – 3 log3 – log5 + 2 log(2×3) – 2 – log x.
2 log5 – 3 log3 + 2 log2 + 2 log3 – 2 – log x.
log 5² – log 3 + 2 log2 = 2 – log x
log 25 – log 3 + log 2² = 2 – log x
log 25 – log 3 + log 4 = 2 – log x
log (25 × 4/3) = 2 – log x
log 100 – log 3 = 2 – log x
log 10² – log3 = 2 – log x
2 log 10 – log3 = 2 – log x
2 × 1 – log3 = 2 – log x
2 – log 3 = 2 – log x
– log3 = – log x
log3 = log x
log x = log3
x = 3

 

Here is your solution of Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8B

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.