ML Aggarwal Solutions Class 9 Math 9th Chapter Logarithms Exercise 9.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Ninth Chapter Logarithms Exercise 9.1. APC Solution Class 9 Exercise 9.1.
Exercise 9.1
(1) Convert the following to logarithmic form:
(i) 52 = 25
Solution:
52 = 25
Or, log5 25 = 2
(ii) a5 = 64
Solution:
a5 = 64
Or, loga 64 = 5
(iii) 7x = 100
Solution:
7x = 100
Or, log7 = 100 = x
(iv) 90 = 1
Solution:
90 = 1
Or, log9 1 = 0
(v) 61 = 6
Solution:
61 = 6
Or, log6 6 = 1
(vi) 3-2 = 1/9
Solution:
3-2 = 1/9
Or, log3 1/9 = -2
(vii) 10-2 = 0.01
Solution:
10-2 = 0.01
Or, log10 0.01 = -2
(viii) (81)3/4 = 27
Solution:
(81)3/4 = 27
Or, log81 27 = 3/4
(2) Convert the following into exponential form:
(i) log2 32 = 5
Solution:
Log2 32 = 5
Or, 25 = 32
(ii) log3 81 = 4
Solution:
log3 81 = 4
Or, 34 = 81
(iii) log3 1/3 = -1
Solution:
log3 1/3 = -1
Or, 3-1 = 1/5
(iv) log8 4 = 2/3
Solution:
log8 4 = 2/3
Or, 82/3 = 4
(v) log8 32 = 5/3
Solution:
log8 32 = 5/3
Or, 85/3 = 32
(vi) log10 (0.001) = -3
Solution:
log10 (0.001) = -3
Or, 10-3 = 0.001
(vii) log2 0.25 = -2
Solution:
log2 0.25 = -2
Or, 2-2 = 0.25
(viii) loga (1/a) = -1
Solution:
loga (1/a) = -1
Or, a-1 = 1/a
(3) By Converting to exponential form, find the values of:
(i) log2 16
Solution:
log2 16
Let, log2 16 = x
∴ 2x = 16
Or, 2x = 24
∴ x = 4
(ii) log5 125
Solution:
log5 125
Let, log5 125 = x
∴ 5x = 125
Or, 5x = 53
∴ x = 3
(iii) log4 8
Solution:
log4 8
Let, log4 8 = x
∴ 4x = 8
Or, 4x = 4 3/2
∴ x = 3/2
(iv) log9 27
Solution:
log9 27
Let, log9 27 = x
∴ 9x = 27
Or, 9x = 93/2
∴ x = 3/2
(v) log10 (0.01)
Solution:
log10 (0.01)
∴ 10x = 0.01
Or, 10x = 1/100
Or, 10x = 1/100
Or, 10x = 10-2
∴ x = -2
(vi) log7 1/7
Solution:
log7 1/7
Let, log7 1/7 = x
∴ 7x = 1/7
Or, 7x = 7 – 1
∴ x = -1
(vii) log0.5 256
Solution:
log0.5 256
Let, log0.5 256 = x
∴ (0.5)x = 256
Or, (1/2)x = 256
Or, 2-x = 28
∴ -x = 8
Or, x = -8
(viii) log2 0.25
Solution:
log2 0.25
Let, log2 0.25 = x
∴ 2x = 0.25
Or, 2x = 25/100
Or, 2x = 1/4
Or, 2x = 1/22
Or, 2x = 2-2
∴ x = -2
(4) solve the following equations for x;
(i) log3 x = 2
Solution:
log3 x = 2
∴ 32 = x
Or, x = 9
(ii) logx25 = 2
Solution:
logx25 = 2
∴ x2 = 25
Or, x2 = 52
∴ x = 5
(iii) log10 x = -2
Solution:
log10 x = -2
∴ 10-2 = x
Or, x = 1/102
Or, x = 1/100
Or, x = 0.01
(iv) log4 x = 1/2
Solution:
log4 x = 1/2
∴ 4 1/2 = x
Or, 2 = 1/2 = x
Or, x = 2
(v) logx 11= 1
Solution:
logx 11= 1
∴ x1 = 11
Or, x = 11
(vi) logx 1/4 = -1
Solution:
logx 1/4 = -1
∴ x-1 = 1/4
Or, 1/x = 1/4
Or, x = 4
(vii) log81 x = 3/2
Solution:
log81 x = 3/2
∴ 813/2 = x
Or, 92.3/2 = x
Or, x = 93
Or x = 729
(viii) log9 x = 2.5
Solution:
log9 x = 2.5
∴ 92.5 = x
Or, 9 25/10 = x
Or, 9 5/2 = x
Or, 3 5/2 = x
Or, x = 35
Or, x = 243
(ix) log4 x = – 1.5
Solution:
Or, log4 x = -15/10
Or, log4 x = -3/2
∴ 4 – 3/2 = x
Or, 2 – 3/2 = x
Or, x = 1/2
Or, x = 1/8
Or, x = 0.125
(x) log√5 x = 2
Solution:
log√5 x = 2
∴ (√5)2 = x
Or, x = 5
(xi) logx 0.001 = -3
Solution:
logx 0.001 = -3
∴ x-3 = 0.001
Or, 1/x3 = 1/1000
Or, 1/x3 = 1/103
Or, x3 = 10
Or, x = 10
(xii) log√3 (x + 1) = 2
Solution:
log√3 (x + 1) = 2
∴ (√3)2 = x + 1
Or, x + 1 = 3
Or, x = 2
(xiii) log4 (2x + 3) = 3/2
Solution:
log4 (2x + 3) = 3/2
∴ 4 3/2 = 2x + 3
Or, 2.3/2 = 2x + 3
Or, 2x + 3 = 23
Or, 2x = 8-3
Or, x = 5/2
(xiv) log 3√2 x = 3
Solution:
log 3√2 x = 3
∴ (3√2)3 = x
Or, x = 2 1/3.3
Or, x = 2
(xv) log2 (x2 – 1) = 3
Solution:
log2 (x2 – 1) = 3
∴ 23 = x3 – 1
Or, x2 = 8 + 1
Or, x = √9
Or, x = 3
(xvi) log x = -1
Solution:
log x = -1
Or, log10 x = -1
∴ 10-1 = x
Or, x = 1/10
Or, x = 0.1
(xvii) log (2x – 3) = 1
Solution:
log (2x – 3) = 1
Or, log10 = (2x – 3) = 1
∴ 101 = 2x – 3
Or, 2x = 10 + 3
Or, x = 13/2
(xviii) log x = -2, 0, 1/3
Solution:
Logx = -2, 6, 1/3
∴ either, log x = -2
Or, log10 x = -2
∴ 10-2 = x
Or, x = 1/100
Or, log x = 0
Or, log10 x = 0
∴ 100 = x
Or, x = 1
Or, log x = 1/3
Or, log10 x = 1/3
∴ 10 1/3 = x
Or, x = 3√10
(5) Given log10a = b, express 102b-3 in terms of a.
Solution:
Given, log10a = b
∴ 10b = a
Now, 102b-3
= 102b/103
= 102.10b/103
= 10b/10
= a/10
(6) Given log10 x = a, log10 y = b and log10 z = c,
(i) Write down 102a-3 in terms of x.
(ii) Write down 103b-1 in terms of y
(iii) if log10 P = 2a + b/2 – 3c, express P in terms of x, y and z.
Solution:
Given, log10 x = a, log10 y = b
∴ 10a = x —– (i)
∴ 10b y —— (ii)
Also, log10 z = c
∴ 10c = z —– (iii)
(i) 1024 – 3
= 1024/103 —- from (i)
= x2/1000
(ii) 1036-1
= 1036/101 —– from (ii)
= y3/10
(iii) Given, log10 P = 2a + b/2 – 3C
∴ 10(2a + b/2 – 3c) = P
Or, 1024 × 10b/2/10 3c = P … from (i), (ii), (iii)
Or, x2 × y1/2/Z3 = P
Or, P = x2√y/z3
(7) If log10 x = a, log10 y = b, find the value of xy.
Solution:
Given, log10 x = a, log10 y = b
∴ 10a = x — (i) ∴ 10b = y —- (ii)
∴ xy = 10a.10b — From (i), (ii)
Or, xy = 10a+b
(8) Given log10 a = m and log10 b = n, express a3/b2 in terms of m and n.
Solution:
Given, log10 a = m, log10 b = n
∴ 10m = a —– (i) ∴ 10n = b —- (ii)
Now, a3/b2 = (10m)3/(10)2 = 103m/102n [From (i), (ii)]
(9) Given log10 x = 2a and log10 y = b/2,
(i) Write 10a in terms of x.
(ii) Write 102b+1 in terms of y
(iii) If log10 P = 3a – 2b, express P in terms of x and y.
Solution:
Given, log10 x = 2a
∴ 102a = x
Or, 10a = 100√a —– (i)
Log10 y = b/2
∴ 10 b/2 = y
Or, 10b = y2 —- (ii)
(i) 10a = √x — from — (i) (ii) 102b+1
= 2 (y2)2 . 101
= y4 . 10
= 10y4
(iii) Given, log10 P = 3a – 2b
∴ 103a-2b = P
Or, 103a/102b = P
Or, P = (√x)3/(y2)2
Or, P = (x3x1/2)/y4
Or, P = x3/2/y4
(10) If log2 y = x and log3 z = x, find 72x in terms of y and z.
Solution:
Given, log2 y = x
∴ 2x = y
∴ 72x
= (23.32)x —- from (i)
= 23x.32x
= y3.72
Log3 7 = x
∴ 3x = 7
∴ 72 = 2×2×2×3×3
= 23 × 32 —– (i)
(11) If log2 x = a and log5 y = a, write 1002a-1 in terms of x and y.
Solution:
Given, log2 x = a, log5 y = a
∴ 2a = x —– (i) ∴ 5a = y (ii)
∴ 1002a-1
= (22 × 52)2a-1 —- from (iii)
= 22(2a – 1) × 52(2a-1)
= 24a-2 × 54a-2
= 24a/22 × 54a/52
= x4/4 × y4/25 —- from (i), (ii)
= x4y4/100
Log5 y = a
∴ 5 = y —- (ii)
∴ 100 = 2×2×5×5
= 22 × 52 —- (iii)
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