Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8A Solutions
EXERCISE – 8A
(1) Express each of the following in logarithmic form.
(i) 53 = 125
=> We know that,
the formula of logarithm.
xy = z => logx z = y
∴ x = 5, y = 3, z = 125
∴ 53 = 125 => log5 125 = 3
(ii) 10-3 = 0.001
=> We know that,
the formula of logarithm,
xy = z => logx z = y
∴ x = 10, y = 3, z = 0.001
∴ 10-3 = 0.001 => log10 0.001 = -3
(iii) 3-2 = 1/9
=> We know that,
the formula of logarithm,
xy = z => logx z= y
∴ x = 3, y = -2, z = 1/9
∴ 3-2 = 1/9 => log3 1/9 = -2
(iv) (81)3/4 = 27
=> We know that,
the formula of logarithm,
xy = z => logx z = y
∴ x = 81, y = 3/4, z = 27
∴ (81)3/4 = 27 => log81 27 = 3/4
(2) Express each of the following in exponents form :
(i) Log8 0.125 = -1
=> We know that,
Logx z = y => xy = z
∴ loy8 0.125 = -1 => 8-1 = 0.125
(ii) log10 0.01 = -2
=> We know that,
Logx z = y => xy = z
∴ log10 0.01 = -2 => 10-2 = 0.01
(iii) Loga A = x
=> We know that,
Logx z = y => xy = z
∴ loga A = x => ax = A
(iv) log10 1 = 0
=> we know that,
logx z = y => xy = z
∴ log10 1 = 0 => 100 = 1
(3) Solve for x : log10 x = -2
Solution :
Log10 x = -2
=> (10)-2 = x
=> 1/102 = x
=> 1/100 = x
=> 0.01 = x
(4) Find the logarithm of :
(i) 100 to the base 10
Solution :
log10 100 = x
=> (10)x = 100
=> (10)x = (10)2
=> x = 2 (∵ base same)
(ii) 0.1 to the base 10
Solution :
log10 0.1 = x
=> (10)x = 0.1
=> 10x = 1/10
=> 10x = 10-1
=> x = -1
(iii) 0.001 to the base 10
Solution :
log10 0.001 = x
=> 10x = 0.001
=> 10x = 1/1000
=> 10x = .1/103
=> 10x = 10-3
=> x = -3
(iv) 32 to the base 4
Solution :
log4 32 = x
=> 4x = 32
=> 22x = 25
=> 2x = 5
=> x = 5/2
(v) 0.124 to the base 2
Solution :
log2 0.125 = x
=> 2x = 0.125
=> 2x = 125/1000
=> 2x = 1/8
=> 2x = 1/23
=> 2x = 2-3
=> x = -3
(vi) 1/16 to the base 4
Solution :
log4 1/16 = x
=> 4x = 1/16
=> 4x = 1/42
=> 22x = 1/24
=> 22x = 2-4
=> 2x = -4
=> x = -4/2
=> x = -2
(vii) 27 to the base 9
solution :
log9 27 = x
=> 9x = 27
=> 32x = 33
=> 2x = 3
=> x = 3/2
=> x = 3/2
(viii) 1/81 to the base 27
Solution :
log27 1/81 = x
=> 27x = 1/81
=> 33x = 1/34
=> 33x = 3-4
=> 3x = -4
=> x = -4/3
(6) Find x, if
(i) log3 x = 0
Solution :
log3 x = 0
=> 30 = x
=> 1 = x ∵ a0 = 1
(ii) logx 2 = -1
Solution :
logx 2 = -1
=> (x)-1 = 2
=> 1/x = 2
=> x = 1/2
(iii) log9 243 = x
Solution :
log9 243 = x
=> 9x = 243
=> (3)2x = (3)5
=> 2x = 5
=> x = 5/2
(iv) log5 (x-7) = 1
Solution :
log5 (x-7) = 1
=> (5)1 = x – 7
=> 5 = x – 7
=> 7 + 5 = x
=> 12 = x
(v) log4 32 = x – 4
Solution :
log4 32 = x – 4
=> 4x-4 = 32
=> 4x-4 = 32
=> (22)x-4 = 25
=> 22x-8 = 25
=> 2x – 8 = 5
=> 2x = 5 + 8
=> 2x = 13
=> x = 13/2
(vi) Log7 (2x2 – 1) = 2
Solution :
log7 (2x2 – 1) = 2
=> 72 = 2x2 -1
=> 49 = 2x2 – 1
=> 49 + 1 = 2x2
=> 50 = 2x2
=> 50/2 x2
=> 25 = x2
Taking square root on both sides,
=> √25 = √x2
=> 5 = x
(7) Evaluate :
(i) log10 0.01
Solution :
log10 =0.01 = x
=> 10x = 0.01
=> 10x = 1/100
=> 10x = 1/102
=> 10x = 10-2
=> x = -2
(ii) log2 (1 ÷ 8)
Solution :
log2 (1 ÷ 8) = x
=> log2 (1/8) x
=> 2x = 1/8
=> 2x = 1/23
=> 2x = 2-3
=> x = -3
(iii) log5 1
Solution :
log5 1 = x
=> 5x = 1
=> x = 0 ∵ a0 = 1
(iv) log5 125
Solution :
log5 125 = x
=> 5x = 125
=> 5x = 53
=> x = 3
(v) log16 8 = x
Solution :
log16 8 = x
=> 16x = 8
=> 24x = 23
=> 4x = 3
=> x = 3/4
(vi) log0.5 16 = x
Solution :
log0.5 16 = x
=> (0.5)x = 16
=> (5/10)x = 16
=> (1/2)x = 24
=> (2-1)x = 24
=> 2-x = 24
=> -x = 4
=> x = – 4
(8) If loga m = n, express an-1 in terms in terms of a and m.
Solution :
Loga m = n
an = m
Dividing on both sides by a,
an/a1 = m/a
an-1 = m/a
(9) Given log2 x = m and log5 y = n
(i) Express 2m-3 in terms of x.
(ii) Express 53n+2 in terms of y.
Solution :
(i) log2 x = m
2m = x
2m/23 = x/23 (∵ Dividing both sides by 23)
2m-3 = x/8
(ii) log5 y = n
5n = y
cube on both sides,
53n = y3
multiplying both sides by 52
53n × 52 = y3 × 52
53n + 2 = 25y3
(10) If log2 x = a and log3 y = a, write 72a in terms of x and y.
Solution :
log2 x = a
2a = x
Log3 y = a
3a = y
Now, 72a = (2×2×2×3×3)a
= 23a × 32a
= (2a)3 × (3a)2
72a = x3 y2
(11) Solve for x : log (x-1) + log (x+1) = log2 1
Solution :
log (x-1) + log (x+1) = log2 1
by using product rule,
log [(x-1) (x+1)] = 0 ∵ log1 = 0
log [(x2 – 1)] = log1
x2 – 1 = 1
x2 = 1 + 1
x2 = 2
x = √2
(12) If log (x2 – 21) = 2, show that x = ± 11
Solution :
log10 (x2 – 21) = 2
102 = x2 – 21
100 = x2 – 21
100 + 21 = x2
121 = x2
√121 = x2
Taking square root on both sides,
± 11 = x
Here is your solution of Selina Concise Class 9 Maths Chapter 8 Logarithms Exercise 8A
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