Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8C Solutions
EXERCISE – 8C
(1) If log10 8 = 0.90, find the value of :
(i) log10 4
Solution :
log10 8 = 0.90
log10 23 = 0.90
3 log10 2 = 0.90
log10 2 = 0.90/3
log10 2 = 0.30
(i) log10 4 = log10 22
= 2 log10 2
= 2 (0.30)
log10 4 = 0.60
(ii) log √32 = log (32)1/2
= log (25)1/2
= log (2)5/2
= 5/2 log 2
= 5/2 × 0.30
= 5 × 0.15
= 0.75
(iii) log 0.125 = log 125/1000
= log 1/8
= log (1/23)
= log (2-3)
= -3 log2
= – 3 (0.30)
= – 0.10
(2) If log27 = 1.431, find the value of :
(i) log9
Solution :
Given that :
log27 = 1.431
log33 = 1.431
3 log3 = 1.431
log3 = 1.431/3
log3 = 0.477 …..(i)
(i) log 9 = log 32
= 2 log3
= 2 × (0.477) (from i)
= 0.954
(ii) log 300 = log (3 × 100)
= log3 + log100
= 0.477 + log102
= 0.477 + 2 log10
= 0.477 + 2 × 1
= 0.477 + 2
= 2.477
(3) If log10 a = b, find 103b-2 in terms of a
Solution :
Given that,
log10 a = b
(10)b = a
Cubing on both side,
((10)b)3 = a3
(10)3b = a3
dividing both sides by 102
(10)3b/102 = a3/102
(10)3b-2 = a3/100
(4) If log5 x = y, find 52y+3 in terms of x
Solution :
Given that,
log5 x = y
5y = x
Squaring on both sides,
(5y)2 = x2
52y = x2
Multiplying both sides by 53
52y × 53 = x2 ×53
52y + 3 = x2 ×125
52y + 3 = 125x2
(5) Given : log3 m = x and log3 n = y.
(i) Express 32x-3 in terms of m
Solution :
Given that, log3 m = x and log3 n = y
=> 3x = m
3y = n
Now,
(i) 32x-3 in terms of m
32x-3 = 32x-33
= (3x)2/33
= m2/27
(ii) Write down 31-2y+3x in terms of m and n
Solution :
31-2y+3x
= 31 × 33x/32y
= 3 × (3x)3/(3y)2
= 3×m3/n2 {∵ 3x = m 3y = n}
(iii) If 2 log3 A = 5x – 3y ; find A in terms of m and n
Solution :
2 log3 A = 5x – 3y
log3 A2 = 5x – 3y
35x-3y = A2
35x/33y = A2
(3x)5/(3y)3 = A2
m5/n3 = A2
Taking square root on both sides,
√m5/n3 = √A2
√m5/n3 = A
(6) Simplify :
(i.) log (a)3 – log a
Solution :
Log (a)3 – log a
=> log a3 = log a
=> log (a3/a)
=> log a2
=> 2 log a.
(ii) Log (a)3 ÷ log a
Solution :
log (a)3 ÷ log a
=> log a3/log a
=> 3 log a/log a
=> 3
(7) If log (a + b) = log a + log b, find a in terms of b.
Solution :
log (a + b) = log a + log b
log (a + b) = log (ab)
a + b = ab
a – ab = – b
a (1 – b) = – b
a = – b/(1 – b)
a = – b/ – (b – 1)
a = b/b – 1
(8.) Prove that :
(i.) (log a)2 – (log b)2 =log (a/b), log (ab)
=> L.H.S. = (log a)2 – (log b)2
(log a + log b) (log a – log b)
log (ab) log (a/b)
L.H.S = R.H.S
Hence proved.
(ii.) If a log b + b log a – 1 = 0, then ba – ab = 10.
Solution :
=> a log b + b log a – 1 = 0
log ba + log ab = 1
log10 (ba ab) = 1
(10)1 = ba ab
10 = ba ab
Hence proved.
(9) (i.) If log (a + 1) = log (4a – 3) – log3; find a.
Solution :
Given that,
log (a + 1) = log (4a – 3) – log3
log (a + 1) = log (4a – 3)/3
(a + 1) = 4a – 3/3
(a + 1) = 4a – 3/3
3 (a + 1) = 4a – 3
3a + 3 = 4a – 3
– 4a + 3a = – 3 – 3
– a = – 6
a = 6
(ii) If 2 log y – log x – 3 = 0, express x in terms of y.
Solution :
Given that,
2 log y – log x – 3 = 0
log y2 – log x = 3
log10 y2/x = 3
(10)3 = y2/x
x = y2/103
x = y2/1000
(iii) Prove that log10 125 = 3 (1 – log10 2)
Solution :
log10 125 = 3 (1 – log10 2)
R.H.S = 3 (1 – log10 2)
= 3 [log10 10 – log10 2]
= 3 [log10 (10/2)]
= 3 log10 5
= log10 53
= log10 125
(10) Given log x = 2m – n, logy = n – 2m and log z = 3m – 2n. Find in terms of m and n, the value of log x2 y3/z4.\
Solution :
log x2 y3/z4
=> log x2 + log y3 – log z4
=> 2 log x + 3 log y – 4 log z
=> 2 (2m – n) + 3 (n – 2m) – 4(3m – 2n) (∵ Given)
=> 4m – 2n + 3n – 6m – 12m + 8n
=> 9n – 14m
(11) Given logx 25 – logx 5 = 2 – logx 1/125; find x
Solution :
Given that,
logx 25 – logx 5 = 2 – logx 1/125
logx (25/5) = 2 – logx (1/53)
logx 5 = 2 – logx 5-3
logx 5 = 2 + 3 logx 5
logx 5 – 3 logx 5 = 2
– 2 logx 5 = 2
logx 5 = 2/-2
logx 5 = -1
x-1 = 5
1/x = 5
x = 1/5
Here is your solution of Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8C
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