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 xa – log yb = log 32
=> log xa/yb = log 9
=> xa/yb = 9
=> xa = 9yb

 

(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 y1/2 = 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 × √(104b)/(102)2a (103c)
= log10 × √(10)4b/(10)4a × (10)3c
= log10 + log (104b)1/2 – log104a – log3c
= 1 + log102b – 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

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