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Exercise 8(A)
1. Express each of the following in logarithmic form:
(i) 53Â = 125
(ii) 3-2Â =Â 1/9
(iii) 10-3Â = 0.001
(iv)Â (81)3/4 = 27
Solution:
We know that,
ab = c ⇒ loga c = b
(i) 53 = 125
log5 125 = 3Â
Â
(ii) 3-2 = 1/9
log3 1/9 = -2Â
Â
(iii) 10-3 = 0.001
log10 0.001 = -3Â
(iv) (81)3/4 = 27
log81 27 = ¾Â
2. Express each of the following in exponential form:
(i)Â log8 0.125 = -1
(ii) log10 0.01 = -2
(iii)Â loga AÂ = x
(iv) log10 1 = 0
Solution:
We know that,
loga c = b ⇒ ab = c
(i)Â log8 0.125 = -1
8-1 = 0.125
(ii) log10 0.01 = -2
8-1 = 0.125
(iii)Â loga AÂ = x
ax = A
(iv) log10 1 = 0
100 = 1
3. Solve for x:Â log10Â x = -2.
Solution:
We have,
log10Â x = -2
10-2 = x [As loga c = b ⇒ ab = c]
x = 10-2
x = 1/102
x = 1/100
Hence, x = 0.01
4. Find the logarithm of:
(i) 100 to the base 10
(ii) 0.1 to the base 10
(iii) 0.001 to the base 10
(iv) 32 to the base 4
(v) 0.125 to the base 2
(vi)Â 1/16Â to the base 4
(vii) 27 to the base 9
(viii)Â 1/81Â to the base 27
Solution;
(i) Let log10 100 = x
So, 10x = 100
10x = 102
Then,
x = 2 [If am = an; then m = n]
Hence, log10 100 = 2
(ii) Let log10 0.1 = x
So, 10x = 0.1
10x = 1/10
10x = 10-1
Then,
x = -1 [If am = an; then m = n]
Hence, log10 0.1 = -1
(iii) Let log10 0.001 = x
So, 10x = 0.001
10x = 1/1000
10x = 1/103
10x = 10-3
Then,
x = -3 [If am = an; then m = n]
Hence, log10 0.001 = -3
(iv) Let log4 32 = x
So, 4x = 32
4x = 2 x 2 x 2 x 2 x 2
(22)x = 25
22x = 25
Then,
2x = 5 [If am = an; then m = n]
x = 5/2
Hence, log4 32 = 5/2
(v) Let log2 0.125 = x
So, 2x = 0.125
2x = 125/1000
2x = 1/8
2x = (½)3
2x = 2-3
Then,
x = -3 [If am = an; then m = n]
Hence, log2 0.125 = -3
(vi) Let log4 1/16 = x
So, 4x = 1/16
4x = (¼)-2
4x = 4-2
Then,
x = -2 [If am = an; then m = n]
Hence, log4 1/16 = -2
(vii) Let log9 27 = x
So, 9x = 27
9x = 3 x 3 x 3
(32)x = 33
32x = 33
Then,
2x = 3 [If am = an; then m = n]
x = 3/2
Hence, log9 27 = 3/2
(viii) Let log27 1/81 = x
So, 27x = 1/81
27x = 1/92
(33)x = 1/(32)2
33x = 1/34
33x = 3-4
Then,
3x = -4 [If am = an; then m = n]
x = -4/3
Hence, log27 1/81 = -4/3
5. State, true or false:
(i) If log10 x = a, then 10x = a
(ii) If xy = z, then y = logz x
(iii)Â log2Â 8 = 3 and log8 2 =Â 1/3
Solution:
(i) We have,
log10Â x = a
So, 10a = x
Thus, the statement 10x = a is false
(ii) We have,
xy = z
So, logx z = y
Thus, the statement y =Â logz x is false
(iii) We have,
log2Â 8 = 3
So, 23 = 8 … (1)
Now consider the equation,
log8 2 = 1/3
81/3 = 2
(23)1/3 = 2 … (2)
Both equations (1) and (2) are correct
Thus, the given statements, log2Â 8 = 3 and log8 2 =Â 1/3 are true
6. Find x, if:
(i) log3Â x = 0
(ii) logx 2 = -1
(iii) log9243 = x
(iv) log5 (x – 7) = 1
(v) log432 = x – 4
(vi) log7 (2x2 – 1) = 2
Solution:
(i) We have, log3Â x = 0
So, 30 = x
1 = x
Hence, x = 1
(ii) we have, logx 2 = -1
So, x-1 = 2
1/x = 2
Hence, x = ½
(iii) We have, log9 243 = x
9x = 243
(32)x = 35
32x = 35
On comparing the exponents, we get
2x = 5
x = 5/2 = 2½
(iv) We have, log5 (x – 7) = 1
So, 51 = x – 7
5 = x – 7
x = 5 + 7
Hence, x = 12
(v) We have, log4 32 = x – 4
So, 4(x – 4) = 32
(22)(x – 4) = 25
2(2x – 8) = 25
On comparing the exponents, we get
2x – 8 = 5
2x = 5 + 8
Hence,
x = 13/2 = 6½
(vi) We have, log7 (2x2 – 1) = 2
So, (2x2 – 1) = 72
2x2 – 1 = 49
2x2 = 49 + 1
2x2 = 50
x2 = 25
Taking square root on both side, we get
x = ±5
Hence, x = 5 (Neglecting the negative value)
Â
7. Evaluate:
(i) log10Â 0.01
(ii) log2 (1 ÷ 8)
(iii) log5Â 1
(iv) log5Â 125
(v) log16Â 8
(vi) log0.5Â 16
Solution:
(i) Let log10Â 0.01 = x
Then, 10x = 0.01
10x = 1/100 = 1/102
So, 10x = 10-2
On comparing the exponents, we get
x = -2
Hence, log10Â 0.01 = -2
(ii) Let log2 (1 ÷ 8) = x
Then, 2x = 1/8
2x = 1/23
So, 2x = 2-3
On comparing the exponents, we get
x = -3
Hence, log10 (1 ÷ 8) = -3
(iii) Let log5Â 1 = x
Then, 5x = 1
5x = 50
On comparing the exponents, we get
x = 0
Hence, log5Â 1 = 0
(iv) Let log5Â 125 = x
Then, 5x = 125
5x = (5 x 5 x 5) = 53
So, 5x = 53
On comparing the exponents, we get
x = 3
Hence, log5Â 125 = 3
(v) Let log16Â 8 = x
Then, 16x = 8
(24)x = (2 x 2 x 2) = 23
So, 24x = 23
On comparing the exponents, we get
4x = 3
x = 3/4
Hence, log16Â 8 = 3/4
(vi) Let log0.5Â 16 = x
Then, 0.5x = 16
(5/10)x = (2 x 2 x 2 x 2)
(1/2)x = 24
So, 2-x = 24
On comparing the exponents, we get
-x = 4
⇒ x = -4
Hence, log0.5Â 16 = -4
Â
8. If loga m = n, express an – 1 in terms in terms of a and m.
Solution:
We have, loga m = n
So,
an = m
Dividing by a on both sides, we get
an/a = m/a
an-1 = m/aÂ
9. Given log2 x = m and log5 y = n
(i) Express 2m-3 in terms of x
(ii) Express 53n+2 in terms of yÂ
Solution:
Given, log2 x = m and log5 y = n
So,
2m = x and 5n = y
(i) Taking, 2m = x
2m/23 = x/23
2m-3 = x/8
(ii) Taking, 5n = y
Cubing on both sides, we have
(5n)3 = y3
53n = y3
Multiplying by 52 on both sides, we have
53n x 52 = y3 x 52
53n+2 = 25y3Â
10. If log2 x = a and log3 y = a, write 72a in terms of x and y.Â
Solution:
Given, log2 x = a and log3 y = a
So,
2a = x and 3a = y
Now, the prime factorization of 72 is
72 = 2 x 2 x 2 x 3 x 3 = 23 x 32
Hence,
(72)a = (23 x 32)a
= 23a x 32a
= (2a)3 x (3a)2
= x3y2 [As 2a = x and 3a = y]
11. Solve for x: log (x – 1) + log (x + 1) = log2 1Â
Solution:
We have,
log (x – 1) + log (x + 1) = log2 1
log (x – 1) + log (x + 1) = 0
log [(x – 1) (x + 1)] = 0
Then,
(x – 1) (x + 1) = 1 [As log 1 = 0]
x2 – 1 = 1
x2 = 1 + 1
x2 = 2
x = ±√2
The value -√2 is not a possible, since log of a negative number is not defined.
Hence, x = √2Â
12. If log (x2 – 21) = 2, show that x = ± 11.
Solution:
Given, log (x2 – 21) = 2
So,
x2 – 21 = 102
x2 – 21 = 100
x2 = 121
Taking square root on both sides, we get
x = ±11
Exercise 8(B)
1. Express in terms of log 2 and log 3:
(i) log 36Â
(ii) log 144Â
(iii) log 4.5
(iv) log 26/51 – log 91/119Â
(v) log 75/16 – 2log 5/9 + log 32/243
Solution:
(i) log 36 = log (2 x 2 x 3 x 3)
= log (22 x 32)
= log 22 + log 32 [Using loga mn = loga m + loga n]
= 2log 2 + 2log 3 [Using loga mn = nloga m]
(ii) log 144 = log (2 x 2 x 2 x 2 x 3 x 3)
= log (24 x 32)
= log 24 + log 32 [Using loga mn = loga m + loga n]
= 4log 2 + 2log 3 [Using loga mn = nloga m]
(iii) log 4.5 = log 45/10
= log (5 x 3 x 3)/ (5 x 2)
= log 32/2
= log 32 – log 2 [Using loga m/n = loga m – loga n]
= 2log 3 – log 2 [Using loga mn = nloga m]
(iv) log 26/51 – log 91/119 = log (26/51)/ (91/119) [Using loga m – loga n = loga m/n]
= log [(26/51) x (119/91)]
= log (2 x 13 x 7 x 117)/ (3 x 17 x 7 x 13)
= log 2/3
= log 2 – log 3 [Using loga m/n = loga m – loga n]
(v) log 75/16 – 2log 5/9 + log 32/243
= log 75/16 – log (5/9)2 + log 32/243 [Using nloga m = loga mn]
= log 75/16 – log 25/81 + log 32/243
= log [(75/16)/ (25/81)] + log 32/243 [Using loga m – loga n = loga m/n]
= log (75 x 81)/ (16 x 25) + log 32/243
= log (3 x 81)/16 + log 32/243
= log 243/16 + log 32/243
= log (243/16) x (32/243) [Using loga m + loga n = loga mn]
= log 32/16
= log 2
2. Express each of the following in a form free from logarithm:
(i) 2 log x – log y = 1
(ii) 2 log x + 3 log y = log a
(iii) a log x – b log y = 2 log 3
Solution:
(i) We have, 2 log x – log y = 1
Then,
log x2 – log y = 1 [Using nloga m = loga mn]
log x2/y = 1 [Using loga m – loga n = loga m/n]
Now, on removing log we have
x2/y = 101
⇒ x2 = 10y
(ii) We have, 2 log x + 3 log y = log a
Then,
log x2 + log y3 = log a [Using nloga m = loga mn]
log x2y3 = log a [Using loga m + loga n = loga mn]
Now, on removing log we have
x2y3 = a
(iii) a log x – b log y = 2 log 3
Then,
log xa – log yb = log 32 [Using nloga m = loga mn]
log xa/yb = log 32 [Using loga m – loga n = loga m/n]
Now, on removing log we have
xa/yb = 32
⇒ x2 = 9yb
3. Evaluate each of the following without using tables:
(i) log 5 + log 8 – 2log 2
(ii) log10 8 + log10 25 + 2log10 3 – log10 18
(iii) log 4 + 1/3 log 125 – 1/5 log 32
Solution:
(i) We have, log 5 + log 8 – 2log 2
= log 5 + log 8 – log 22 [Using nloga m = loga mn]
= log 5 + log 8 – log 4
= log (5 x 8) – log 4 [Using loga m + loga n = loga mn]
= log 40 – log 4
= log 40/4 [Using loga m – loga n = loga m/n]
= log 10
= 1Â
(ii) We have, log10 8 + log10 25 + 2log10 3 – log10 18
= log10 8 + log10 25 + log10 32 – log10 18 [Using nloga m = loga mn]
= log10 8 + log10 25 + log10 9 – log10 18
= log10 (8 x 25 x 9) – log10 18 [Using loga l + loga m + loga n = loga lmn]
= log10 1800 – log10 18
= log10 1800/18 [Using loga m – loga n = loga m/n]
= log10 100
= log10 102
= 2log10 10 [Using loga mn = nloga m]
= 2 x 1
= 2
Â
(iii) We have, log 4 + 1/3log 125 – 1/5log 32
= log 4 + log (125)1/3 – log (32)1/5 [Using nloga m = loga mn]
= log 4 + log (53)1/3 – log (25)1/5
= log 4 + log 5 – log 2
= log (4 x 5) – log 2 [Using loga m + loga n = loga mn]
= log 20 – log 2
= log 20/2 [Using loga m – loga n = loga m/n]
= log 10
= 1
4. Prove that:
2log 15/18 – log 25/162 + log 4/9 = log 2
Solution:
Taking L.H.S.,
= 2log 15/18 – log 25/162 + log 4/9
= log (15/18)2 – log 25/162 + log 4/9 [Using nloga m = loga mn]
= log 225/324 – log 25/162 + log 4/9
= log [(225/324)/(25/162)] + log 4/9 [Using loga m – loga n = loga m/n]
= log (225 x 162)/(324 x 25) + log 4/9
= log (9 x 1)/(2 x 1) + log 4/9
= log 9/2 + log 4/9 [Using loga m + loga n = loga mn]
= log (9/2 x 4/9)
= log 2
= R.H.S.
Â
5. Find x, if:
x – log 48 + 3 log 2 = 1/3 log 125 – log 3.
Solution:
We have,
x – log 48 + 3 log 2 = 1/3 log 125 – log 3
Solving for x, we have
x = log 48 – 3 log 2 + 1/3 log 125 – log 3
= log 48 – log 23 + log 1251/3 – log 3 [Using nloga m = loga mn]
= log 48 – log 8 + log (53)1/3 – log 3
= (log 48 – log 8) + (log 5 – log 3)
= log 48/8 + log 5/3 [Using loga m – loga n = loga m/n]
= log (48/8 x 5/3) [Using loga m + loga n = loga mn]
= log (2 x 5)
= log 10
= 1
Hence, x = 1
Â
6. Express log10 2 + 1 in the form of log10 x.
Solution:
Given, log10 2 + 1
= log10 2 + log10 10 [As, log10 10 = 1]
= log10 (2 x 10) [Using loga m + loga n = loga mn]
= log10 20
7. Solve for x:
(i) log10 (x – 10) = 1
(ii) log (x2 – 21) = 2
(iii) log (x – 2) + log (x + 2) = log 5
(iv) log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3
Solution:
(i) We have, log10 (x – 10) = 1
Then,
x – 10 = 101
x = 10 + 10
Hence, x = 20
(ii) We have, log (x2 – 21) = 2
Then,
x2 – 21 = 102
x2 – 21 = 100
x2 = 100 + 21
x2 = 121
Taking square root on both sides,
Hence, x = ±11
(iii) We have, log (x – 2) + log (x + 2) = log 5
Then,
log (x – 2)(x + 2) = log 5 [Using loga m + loga n = loga mn]
log (x2 – 22) = log 5 [As (x – a)(x + a) = x2 – a2]
log (x2 – 4) = log 5
Removing log on both sides, we get
x2 – 4 = 5
x2 = 5 + 4
x2 = 9
Taking square root on both sides,
x = ±3
(iv) We have, log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3
Then,
log (x + 5) + log (x – 5) = log 24 + log 32 [Using nloga m = loga mn]
log (x + 5)(x – 5) = log 16 + log 9 [Using loga m + loga n = loga mn]
log (x2 – 52) = log (16 x 9) [As (x – a)(x + a) = x2 – a2]
log (x2 – 25) = log 144
Removing log on both sides, we have
x2 – 25 = 144
x2 = 144 + 25
x2 = 169
Taking square root on both sides, we get
x = ±13
8. Solve for x:
(i)Â log 81/log 27 = x
(ii)Â log 128/log 32 = x
(iii)Â log 64/log 8 = log x
(iv)Â log 225/log 15 = log x
Solution:
(i)Â We have, log 81/log 27 = x
x = log 81/log 27
= log (3 x 3 x 3 x 3)/ log (3 x 3 x 3)
= log 34/log 33
= (4log 3)/(3log 3) [Using loga mn = nloga m]
= 4/3
Hence, x = 4/3
(ii)Â We have, log 128/log 32 = x
x = log 128/log 32
= log (2 x 2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2 x 2 x 2)
= log 27/log 25
= (7log 2)/(5log 2) [Using loga mn = nloga m]
= 7/5
Hence, x = 7/5
(iii)Â log 64/log 8 = log x
log x = log 64/log 8
= log (2 x 2 x 2 x 2 x 2 x 2)/ log (2 x 2 x 2)
= log 26/log 23
= (6log 2)/(3log 2) [Using loga mn = nloga m]
= 6/3
= 2
So, log x = 2
Hence, x = 102 = 100
(iv)Â We have, log 225/log 15 = log x
log x = log 225/log 15
= log (15 x 15)/ log 15
= log 152/log 15
= (2log 15)/log 15 [Using loga mn = nloga m]
= 2
So, log x = 2
Hence, x = 102 = 100
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:
Given, log x = m + n and log y = m – n
Now consider log 10x/y2,
log 10x/y2 = log 10x – log y2 [Using loga m/n = loga m – loga n]
= log 10x – 2log y
= log 10 + log x – 2 log y
= 1 + (m + n) – 2 (m – n)
= 1 + m + n – 2m + 2n
= 1 + 3n – m
Â
10. State, true or false:
(i) log 1 × log 1000 = 0
(ii) log x/log y = log x – log y
(iii) If log 25/log 5 = log x, then x = 2
(iv) log x × log y = log x + log y
Solution:
(i) We have, log 1 × log 1000 = 0
Now,
log 1 = 0 and
log 1000 = log 103 = 3log 10 = 3 [Using loga mn = nloga m]
So,
log 1 × log 1000 = 0 x 3 = 0
Thus, the statement log 1 × log 1000 = 0 is true
(ii) We have, log x/log y = log x – log y
We know that,
log x/y = log x – log y
So,
log x/log y ≠log x – log y
Thus, the statement log x/log y = log x – log y is false
Â
(iii) We have, log 25/log 5 = log x
log (5 x 5)/log 5 = log x
log 52/ log 5 = log x
2log 5/log 5 = log x [Using loga mn = nloga m]
2 = log x
So, x = 102
x = 100
Thus, the statement x = 2 is false
Â
(iv) We know, log x + log y = log xy
So,
log x + log y ≠log x × log y
Thus, the statement log x + log y = log x × log y is false
Â
11. If log10 2 = a and log10 3 = b; express each of the following in terms of ‘a’ and ‘b’:
(i) log 12
(ii) log 2.25
(iii) logÂ
Â
(iv) log 5.4
(v) log 60
(iv) logÂ
Â
Solution:
Given that log10 2 = a and log10 3 = b … (1)
(i) log 12 = log (2 x 2 x 3)
= log (2 x 2) + log 3 [Using loga mn = loga m + loga n]
= log 22 + log 3
= 2log 2 + log 3 [Using loga mn = nloga m]
= 2a + b [From 1]
(ii) log 2.25 = log 225/100
= log (25 x 9)/(25 x 4)
= log 9/4
= log (3/2)2
= 2log 3/2 [Using loga mn = nloga m]
= 2(log 3 – log 2) [Using loga m/n = loga m – loga n]
= 2(b – a) [From 1]
= 2b – 2a
(iii) log = log 9/4
= log (3/2)2
= 2log 3/2 [Using loga mn = nloga m]
= 2(log 3 – log 2) [Using loga m/n = loga m – loga n]
= 2(b – a) [From 1]
= 2b – 2a
(iv) log 5.4 = log 54/10
= log (2 x 3 x 3 x 3)/10
= log (2 x 33) – log 10 [Using loga m/n = loga m – loga n]
= log 2 + log 33 – 1 [Using loga mn = loga m + loga n and log 10 = 1]
= log 2 + 3log 3 – 1 [Using loga mn = nloga m]
= a + 3b – 1 [From 1]
Â
(v) log 60 = log (10 x 3 x 2)
= log 10 + log 3 + log 2 [Using loga lmn = loga l + loga m + log10 n]
= 1 + b + a [From 1]
(vi) log = log 25/8
= log 52/23
= log 52 – log 23 [Using loga m/n = loga m – loga n]
= 2log 5 – 3log 2 [Using loga mn = nloga m]
= 2log 10/2 – 3log 2
= 2(log 10 – log 2) – 3log 2 [Using loga m/n = loga m – loga n]
= 2log 10 – 2log 2 – 3log 2
= 2(1) – 2a – 3a [From 1]
= 2 – 5a
Â
12. If log 2 = 0.3010 and log 3 = 0.4771; find the value of:
(i) log 12
(ii) log 1.2
(iii) log 3.6
(iv) log 15
(v) log 25
(vi)Â 2/3 log 8
Solution:
Given, log 2 = 0.3010 and log 3 = 0.4771
(i) log 12 = log (4 x 3)
= log 4 + log 3 [Using loga mn = loga m + loga n]
= log 22 + log 3
= 2log 2 + log 3 [Using loga mn = nloga m]
= 2 x 0.3010 + 0.4771
= 1.0791
(ii) log 1.2 = log 12/10
= log 12 – log 10 [Using loga m/n = loga m – loga n]
= log (4 x 3) – log 10
= log 4 + log 3 – log 10 [Using loga mn = loga m + loga n]
= log 22 + log 3 – log 10
= 2log 2 + log 3 – log 10 [Using loga mn = nloga m]
= 2 x 0.3010 + 0.4771 – 1 [As log 10 = 1]
= 0.6020 + 0.4771 – 1
= 1.0791 – 1
= 0.0791
(iii) log 3.6 = log 36/10
= log 36 – log 10 [Using loga m/n = loga m – loga n]
= log (2 x 2 x 3 x 3) – 1 [As log 10 = 1]
= log (22 x 32) – 1
= log 22 + log 32 – 1 [Using loga mn = loga m + loga n]
= 2log 2 + 2log 3 – 1 [Using loga mn = nloga m]
= 2 x 0.3010 + 2 x 0.4771 – 1
= 0.6020 + 0.9542 – 1
= 1.5562 – 1
= 0.5562
(iv) log 15 = log (15/10 x 10)
= log 15/10 + log 10 [Using loga mn = loga m + loga n]
= log 3/2 + 1 [As log 10 = 1]
= log 3 – log 2 + 1 [Using loga m/n = loga m – loga n]
= 0.4771 – 0.3010 + 1
= 1.1761
(v) log 25 = log (25/4 x 4)
= log 100/4
= log 100 – log 4 [Using loga m/n = loga m – loga n]
= log 102 – log 22
= 2log 10 – 2log 2 [Using loga mn = nloga m]
= (2 x 1) – (2 x 0.3010)
= 2 – 0.6020
= 1.398
13. Given 2 log10Â x + 1 = log10Â 250, find:
(i) x
(ii) log10Â 2x
Solution:
(i) Given equation, 2log10Â x + 1 = log10Â 250
log10Â x2 + log10 10 = log10Â 250 [Using nloga m = loga mn and log10 10 = 1]
log10 10x2 = log10 250 [Using loga m + loga n = loga mn]
Removing log on both sides, we have
10x2 = 250
x2 = 25
x = ±5
As x cannot be a negative value, x = -5 is not possible
Hence, x = 5
(ii) Now, from (i) we have x = 5
So,
log10 2x = log10 2(5)
= log10 10
= 1
Â
14. Given 3log x + ½ log y = 2, express y in term of x.
Solution:
We have, 3log x + ½ log y = 2
log x3 + log y1/2 = 2 [Using loga m + loga n = loga mn]
log x3y1/2 = 2
Removing logarithm, we get
x3y1/2 = 102
y1/2 = 100/x3
On squaring on both sides, we get
y = 10000/x6
y = 10000x-6
Â
15. If x = (100)a, y = (10000)b and z = (10)c, find log 10√y/x2z3 in terms of a, b and c.
Solution:
We have,
x = (100)a, y = (10000)b and z = (10)c
So,
log x = a log 100, log y = b log 10000 and log z = c log 10
⇒ log x = a log 102, log y = b log 104 and log z = c log 10
⇒ log x = 2a log 10, log y = 4b log 10 and log z = c log 10
⇒ log x = 2a, log y = 4b and log z = c … (i)
Now,
log 10√y/x2z3 = log 10√y – log x2z3 [Using loga m/n = loga m – loga n]
= (log 10 + log √y) – (log x2 + log z3) [Using loga mn = loga m + loga n]
= 1 + log y1/2 – log x2 – log z3
= 1 + ½ log y – 2 log x – 3 log z [Using loga mn = nloga m]
= 1 + ½(4b) – 2(2a) – 3c … [Using (i)]
= 1 + 2b – 4a – 3c
Â
16. If 3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x, find x.
Solution:
We have,
3(log 5 – log 3) – (log 5 – 2 log 6) = 2 – log x
3log 5 – 3log 3 – log 5 + 2log 6 = 2 – log x
3log 5 – 3log 3 – log 5 + 2log (3 x 2) = 2 – log x
2log 5 – 3log 3 + 2(log 3 + log 2) = 2 – log x [Using loga mn = loga m + loga n]
2log 5 – log 3 + 2log 3 + 2log 2 = 2 – log x
2log 5 – log 3 + 2log 2 = 2 – log x
2log 5 – log 3 + 2log 2 + log x = 2
log 52 – log 3 + log 22 + log x = 2 [Using nloga m = loga mn]
log 25 – log 3 + log 4 + log x = 2
log (25 × 4 × x)/3 = 2 [Using loga m + loga n = loga mn & loga m – loga n = loga m/n]
log 100x/3 = 2
On removing logarithm,
100x/3 = 102
100x/3 = 100
Dividing by 100 on both sides, we have
x/3 = 1
Hence, x = 3
Exercise 8(C)
1. If log10Â 8 = 0.90; find the value of:
(i) log10Â 4
(ii) log √32
(iii) log 0.125
Solution:
Given, log10Â 8 = 0.90
log10 (2 x 2 x 2) = 0.90
log10 23 = 0.90
3 log10 2 = 0.90
log10 2 = 0.90/3
log10 2 = 0.30 … (1)
Â
(i) log 4 = log10 (2 x 2)
= log10 22
= 2 log10 2
= 2 x 0.60 … [From (1)]
= 1.20
(ii) log √32 = log10 321/2
= ½ log10 (2 x 2 x 2 x 2 x 2)
= ½ log10 25
= ½ x 5 log10 2
= ½ x 5 x 0.30 … [From (1)]
= 0.75
(iii) log 0.125 = log 125/1000
= log10 1/8
= log10 1/23
= log10 2-3
= -3 log10 2
= -3 x 0.30 … [From (1)]
= -0.90
2. If log 27 = 1.431, find the value of:
(i) log 9 (ii) log 300
Solution:
Given, log 27 = 1.431
So, log 33 = 1.431
3log 3 = 1.431
log 3 = 1.431/3
= 0.477 … (1)
(i) log 9 = log 32
= 2log 3
= 2 x 0.477 … [From (1)]
= 0.954
(ii) log 300 = log (3 x 100)
= log 3 + log 100
= log 3 + log 102
= log 3 + 2log 10
= log 3 + 2 [As log 10 = 1]
= 0.477 + 2
= 2.477
Â
3. If log10 a = b, find 103b – 2 in terms of a.
Solution:
Given, log10 a = b
Now,
Let 103b – 2 = x
Applying log on both sides,
log 103b – 2 = log x
(3b – 2)log 10 = log x
3b – 2 = log x
3log10 a – 2 = log x
3log10 a – 2log 10 = log x
log10 a3 – log 102 = log x
log10 a3 – log 100 = log x
log10 a3/100 = log x
On removing logarithm, we get
a3/100 = x
Hence, 103b – 2 = a3/100
4. If log5Â x = y, find 52y+ 3Â in terms of x.
Solution:
Given, log5Â x = y
So, 5y = x
Squaring on both sides, we get
(5y)2 = x2
52y = x2
52y × 53 = x2 × 53
Hence,
52y+3 = 125x2
5. Given: log3Â m = x and log3Â n = y.
(i) Express 32x – 3 in terms of m.
(ii) Write down 31 – 2y + 3x in terms of m and n.
(iii) If 2 log3 A = 5x – 3y; find A in terms of m and n.
Â
Solution:
Given, log3Â m = x and log3Â n = y
So, 3x = m and 3y = n … (1)
(i) Taking the given expression, 32x – 3
32x – 3 = 32x . 3-3
= 32x . 1/33
= (3x)2/33
= m2/33 … [Using (1)]
= m2/27
Hence, 32x – 3 = m2/27
(ii) Taking the given expression, 31 – 2y + 3xÂ
31 – 2y + 3x = 31 . 3-2y . 33xÂ
= 3 . (3y)-2 . (3x)3Â
= 3 . n-2 . m3 … [Using (1)]
= 3m3/n2
Hence, 31 – 2y + 3x = 3m3/n2
(iii) Taking the given equation,
2 log3 A = 5x – 3y
log3 A2 = 5x – 3y
log3 A2 = 5log3 m – 3log3 n … [Using (1)]
log3 A2 = log3 m5 – log3 n3
log3 A2 = log3Â m5/n3
Removing logarithm on both sides, we get
A2 = m5/n3
Hence, by taking square root on both sides
A = √(m5/n3) = m5/2/n3/2
Â
6. Simplify:
(i) log (a)3 – log a
(ii) log (a)3 ÷ log a
Solution:
(i) We have, log (a)3 – log a
= 3log a – log a
= 2log a
(ii) We have, log (a)3 ÷ log a
= 3log a/ log a
= 3
7. If log (a + b) = log a + log b, find a in terms of b.
Solution:
We have, log (a + b) = log a + log b
Then,
log (a + b) = log ab
So, on removing logarithm, we have
a + b = ab
a – ab = -b
a(1 – b) = -b
a = -b/(1 – b)
Hence,
a = b/(b – 1)
8. Prove that:
(i) (log a)2 – (log b)2 = log (a/b). log (ab)
(ii) If a log b + b log a – 1 = 0, then ba. ab = 10
Solution:
(i) Taking L.H.S. we have,
= (log a)2 – (log b)2
= (log a + log b) (log a – log b) [As x2 – y2 = (x + y)(x – y)]
= (log ab). (log a/b)
= R.H.S.
– Hence proved
(ii) We have, a log b + b log a – 1 = 0
So,
log ba + log ab – 1 = 0
log ba + log ab = 1
log baab = 1
On removing logarithm, we get
baab = 10
– Hence proved
Â
Â
9. (i) If log (a + 1) = log (4a – 3) – log 3; find a.
(ii) If 2 log y – log x – 3 = 0, express x in terms of y.
(iii) Prove that: log10 125 = 3(1 – log102).
Solution:
(i) Given, log (a + 1) = log (4a – 3) – log 3
So,
log (a + 1) = log (4a – 3)/3
On removing logarithm on both sides, we have
a + 1 = (4a – 3)/3
3(a + 1) = 4a – 3
3a + 3 = 4a – 3
Hence, a = 6
(ii) Given, 2log y – log x – 3 = 0
So,
log y2 – log x = 3
log y2/x = 3
On removing logarithm, we have
y2/x = 103 = 1000
Hence, x = y2/1000
(iii) Considering the L.H.S., we have
log10Â 125 = log10Â (5 x 5 x 5)
= log10Â 53
= 3log10Â 5
= 3log10Â 10/2
= 3(log10 10 – log10 2)
= 3(1 – log10 2) [Since, log10 10 = 1]
= R.H.S.Â
– Hence proved
Â
10. Given log x = 2m – n, log y = n – 2m and log z = 3m – 2n. Find in terms of m and n, the value of log x2y3/z4.
Solution:
We have, log x = 2m – n, log y = n – 2m and log z = 3m – 2n
Now, considering
log x2y3/z4 = log x2y3 – log z4
= (log x2 + log y3) – log z4
= 2log x + 3log y – 4log z
= 2(2m – n) + 3(n – 2m) – 4(3m – 2n)
= 4m – 2n + 3n – 6m – 12m + 8n
= -14m + 9n
11. Given logx 25 – logx 5 = 2 – logx 1/125; find x.
Solution:
We have, logx 25 – logx 5 = 2 – logx 1/125
logx (5 x 5) – logx 5 = 2 – logx 1/(5 x 5 x 5)
logx 52 – logx 5 = 2 – logx 1/53
2logx 5 – logx 5 = 2 – logx 1/53
logx 5 = 2 – 3logx 1/5
logx 5 = 2 + 3logx (1/5)-1
logx 5 = 2 + 3logx 5
2 = logx 5 – 3logx 5
2 = -2logx 5
-1 = logx 5
Removing logarithm, we get
x-1 = 5
Hence, x = 1/5
Â
Exercise 8(D)
1. If 3/2 log a + 2/3 log b – 1 = 0, find the value of a9.b4
Solution:
Given equation,
3/2 log a + 2/3 log b – 1 = 0
log a3/2 + log b2/3 – 1 = 0
log a3/2 × b2/3 – 1 = 0
log a3/2.b2/3 = 1
Removing logarithm, we have
a3/2.b2/3 = 10
On manipulating,
(a3/2.b2/3)6 = 106
Hence,
a9.b4 = 106
2. If x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5; find the value of a if x + y = 2z.
Solution:
Given, x = 1 + log 2 – log 5, y = 2log3 and z = log a – log 5
Now, considering the given equation x + y = 2z
(1 + log 2 – log 5) + (2log 3) = 2(log a – log 5)
1 + log 2 – log 5 + 2log 3 = 2log a – 2log 5
1 + log 2 – log 5 + 2log 3 + 2log 5 = 2log a
log 10 + log 2 + log 32 + log 5 = log a2
log 10 + log 2 + log 9 + log 5 = log a2
log (10 x 2 x 9 x 5) = log a2
log 900 = log a2
On removing logarithm on both sides, we have
900 = a2
Taking square root, we get
a = ±30
Since, a cannot be a negative value,
Hence, a = 30
3. If x = log 0.6; y = log 1.25 and z = log 3 – 2log 2, find the values of:
(i) x + y – z    (ii) 5x + y – z
Solution:
Given,
x = log 0.6, y = log 1.25 and z = log 3 – 2log 2
So, z = log 3 – log 22
= log 3 – log 4
= log ¾
= log 0.75 … (1)
(i) Considering,
x + y – z = log 0.6 + log 1.25 – log 0.75 … [From (1)]
= log (0.6 x 1.25)/0.75
= log 0.75/0.75
= log 1
= 0 … (2)
Â
(ii) Now, considering
5x+y-z = 50 … [From (2)]
= 1
Â
4. If a2 = log x, b3 = log y and 3a2 – 2b3 = 6 log z, express y in terms of x and z.
Solution:
We have, a2Â = log x and b3Â = log y
Now, considering the equation
3a2 – 2b3 = 6log z
3log x – 2log y = 6log z
log x3 – log y2 = log z6
log x3/y2 = log z6
On removing logarithm on both sides, we get
x3/y2 = z6
So,
y2 = x3/z6
Taking square root on both sides, we get
y = √(x3/z6)
Hence, y = x3/2/z3
Â
5. If log (a – b)/2 = ½ (log a + log b), show that: a2 + b2 = 6ab.
Solution:
We have, log (a – b)/2 = ½ (log a + log b)
log (a – b)/2 = ½ log a + ½ log b
= log a1/2 + log b1/2
= log √a + log √b
= log √(ab)
Now, removing logarithm on both sides, we get
(a – b)/2 = √(ab)
Squaring on both sides, we get
[(a – b)/2]2 = [√(ab)]2
(a – b)2/4 = ab
(a – b)2 = 4ab
a2 + b2 – 2ab = 4ab
a2 + b2 = 4ab + 2ab
a2 + b2 = 6ab
– Hence proved
6. If a2 + b2 = 23ab, show that: log (a + b)/5 = ½ (log a + log b).
Solution:
Given, a2Â + b2Â = 23ab
Adding 2ab on both sides,
a2Â + b2Â + 2ab = 23ab + 2ab
(a + b)2 = 25ab
(a + b)2/25 = ab
[(a + b)/5]2Â = ab
Taking logarithm on both sides, we have
log [(a + b)/5]2Â = log ab
2log (a + b)/5 = log ab
2log (a + b)/5 = log a + log b
Thus,
log (a + b)/5 = ½ (log a + log b)
7. If m = log 20 and n = log 25, find the value of x, so that: 2log (x – 4) = 2m – n.
Solution:
Given, m = log 20 and n = log 25
Now, considering the given expression
2log (x – 4) = 2m – n
2log (x – 4) = 2log 20 – log 25
log (x – 4)2 = log 202 – log 25
log (x – 4)2 = log 400 – log 25
log (x – 4)2 = log 400/25
Removing logarithm on both sides,
(x – 4)2 = 400/25
x2 – 8x + 16 = 16
x2 – 8x = 0
x(x – 8) = 0
So,
x = 0 or x = 8
If x = 0, then log (x – 4) doesn’t exist
Hence, x = 8
8. Solve for x and y; if x > 0 and y > 0; log xy = log x/y + 2log 2 = 2.
Solution:
We have,
log xy = log x/y + 2log 2 = 2
Considering the equation,
log xy = 2
log xy = 2log 10
log xy = log 102
log xy = log 100
On removing logarithm,
xy = 100 … (1)
Now, consider the equation
log x/y + 2log 2 = 2
log x/y + log 22 = 2
log x/y + log 4 = 2
log 4x/y = 2
Removing logarithm, we get
4x/y = 102
4x/y = 100
x/y = 25
(xy)/y2 = 25
100/y2 = 25 … [From (1)]
y2 = 100/25
y2 = 4
y = 2 [Since, y > 0]
From log xy = 2
Substituting the value of y, we get
log 2x = 2
On removing logarithm,
2x = 102
2x = 100
x = 100/2
x = 50
Thus, the values x and y are 50 and 2 respectively
9. Find x, if:
(i) logx 625 = -4Â
(ii) logx (5x – 6) = 2
Solution:
(i) We have, logx 625 = -4
On removing logarithm,
x-4 = 625
(1/x)4 = 54
Taking the fourth root on both sides,
1/x = 5
Hence, x = 1/5
Â
(ii) We have, logx (5x – 6) = 2
On removing logarithm,
x2 = 5x – 6
x2 – 5x + 6 = 0
x2 – 3x – 2x + 6 = 0
x(x – 3) – 2(x – 2) = 0
(x – 2)(x – 3) = 0
Hence,
x = 2 or 3
10. If p = log 20 and q = log 25, find the value of x, if 2log (x + 1) = 2p – q.Â
Solution:
Given, p = log 20 and q = log 25
Considering the equation,
2log (x + 1) = 2p – q
log (x + 1)2 = 2p – q
log (x + 1)2 = 2log 20 – log 25
log (x + 1)2 = log 202 – log 25
log (x + 1)2 = log 400 – log 25
log (x + 1)2 = log 400/25
Removing logarithm on both sides, we have
(x + 1)2 = 400/25 = 16
(x + 1)2 = (4)2
Taking square root on both sides, we have
x + 1 = 4
x = 4 -1
Hence, x = 3Â Â
Â
11. If log2 (x + y) = log3 (x – y) = log 25/log 0.2, find the value of x and y.
Solution:
Considering the relation, log2 (x + y) = log 25/log 0.2
log2 (x + y) = log0.2 25
= log2/10 52
= 2log1/5 5
= 2log5-1 5
= -2log5 5
= -2 x 1
= -2
So, we have
log2 (x + y) = -2
Removing logarithm, we get
x + y = 2-2
x + y = 1/22
x + y = ¼ … (i)
Now, considering the relation log3 (x – y) = log 25/log 0.2
log3 (x – y) = = log0.2 25
= log2/10 52
= 2log1/5 5
= 2log5-1 5
= -2log5 5
= -2 x 1
= -2
So, we have
log3 (x – y) = -2
Removing logarithm, we get
x – y = 3-2
x – y = 1/32
x – y = 1/9 … (ii)
On adding (i) and (ii), we get
x + y = ¼
x – y = 1/9
—————-
2x = ¼ + 1/9
2x = (9 + 4)/36
2x = 13/36
x = 13/(36 x 2)
= 13/72
Now, substituting the value of x in (i), we get
13/72 + y = ¼
y = ¼ – 13/72
= (18 – 13)/72
= 5/72
Hence, the values of x and are is 13/72 and 5/72 respectively
Â
12. Given: log x/log y = 3/2 and log xy = 5; find the values of x and y.Â
Solution:
Given, log x/log y = 3/2 … (i) and log xy = 5 … (ii)
So,
log xy = log x + log y = 5
And, we have log y = (2log x)/3 … [From (i)]
Now,
log x + (2log x)/3 = 5
3log x + 2log x = 5 x 3
5log x = 15
log x = 15/5
log x = 3
Removing logarithm, we get
x = 103 = 1000
Substituting value of x in (ii), we get
log xy = 5
Removing logarithm, we get
xy = 105
(103). y = 105
y = 105/103
y = 102
y = 100
Â
13. Given log10 x = 2a and log10 y = b/2
(i) Write 10a in terms of x
(ii) Write 102b + 1Â in terms of yÂ
(iii) If log10 p = 3a – 2b, express p in terms of x and y.
Solution:
Given, log10 x = 2a and log10 y = b/2
(i) Taking log10 x = 2a
Removing logarithm on both sides,
x = 102a
Taking square root on both sides, we get
x1/2 = 102a/2
Hence, 10a = x1/2
(ii) Taking log10 y = b/2
Removing logarithm on both sides,
y = 10b/2
On manipulating,
y4 = 10b/2 x 4
y4 = 102b
10y4 = 102b x 10
Hence, 102b + 1Â = 10y4
(iii) We have, 10a = x1/2
and y = 10b/2
Considering the equation, log10 p = 3a – 2b
log10 p = 3a – 2b
Removing logarithm, we get
p = 103a – 2b
p = 103a/102b
p = (10a)3/(10b/2)4
p = (x1/2)3/(y)4
Hence, p = x3/2/y4
14. Solve:
log5 (x + 1) – 1 = 1 + log5 (x – 1).
Solution:
Considering the given equation,
log5 (x + 1) – 1 = 1 + log5 (x – 1)
log5 (x + 1) – log5 (x – 1) = 1 + 1
log5 (x + 1)/(x – 1) = 2
Removing logarithm, we have
(x + 1)/(x – 1) = 52
(x + 1)/(x – 1) = 25
(x + 1) = 25(x – 1)
x + 1 = 25x – 25
25x – x = 25 + 1
24x = 26
x = 26/24
Hence, x = 13/12
Â
15. Solve for x, if:
logx 49 – logx 7 + logx 1/343 + 2 = 0Â
Solution:
We have,
logx 49 – logx 7 + logx 1/343 + 2 = 0Â
logx 49/(7 x 343) + 2 = 0
logx 1/49 = -2
logx 1/72 = -2
logx 7-2 = -2
-2logx 7 = -2
So,
logx 7 = 1
Removing logarithm, we get
x = 7
16. If a2 = log x, b3 = log y and a2/2 – b3/3 = log c, find c in terms of x and y. Â
Solution:
Given,
a2 = log x, b3 = log y
Considering the given equation,
a2/2 – b3/3 = log c
(log x)/2 – (log y)/3 = log c
½ log x – 1/3 log y = log c
log x1/2 – log y1/3 = log c
log x1/2/y1/3 = log c
On removing logarithm, we get
x1/2/y1/3 = c
Hence, c = x1/2/y1/3 is the required relationÂ
17. Given: x = log10 12, y = log4 2 x log10 9 and z = log10 0.4, find
(i) x – y – z
(ii) 13x – y – zÂ
Solution:
(i) Considering, x – y – z
= log10 12 – (log4 2 x log10 9) – log10 0.4
= log10 12 – (log4 2 x log10 9) – log10 0.4
= log10 (4 x 3) – (log10 2/ log10 4 x log10 9) – log10 0.4
= log10 4 + log10 3 – (log10 2 x log10 32)/ log10 22 – log10 4/10
= log10 4 + log10 3 – (log10 2 x 2log10 3)/ 2log10 2 – (log10 4 – log10 10)
= log10 4 + log10 3 – log10 3 – log10 4 + log10 10
= log10 4 + log10 3 – log10 3 – log10 4 + 1
= 1
(ii) Now,
13x – y – z = 131 = 13Â
18. Solve for x, logx 15√5 = 2 – logx 3√5Â
Solution:
Considering the given equation,
logx 15√5 = 2 – logx 3√5Â
logx 15√5 + logx 3√5 = 2
logx (15√5 x 3√5) = 2
logx (45 x 5)Â = 2
logx 225Â = 2
Removing logarithm, we get
x2 = 225
Taking square root on both sides,
x = 15Â
19. Evaluate:
(i) logb a x logc b x loga c
(ii) log3 8 ÷ log9 16
(iii) log5 8/(log25 16 x log100 10)Â
Solution:
Using logb a = 1/loga b and logx a/logx b = logb a, we have
(i) logb a x logc b x loga c
= 1
(ii) log3 8 ÷ log9 16
= log3 8/ log9 16
= 3 x ½
= 3/2
(iii) log5 8/(log25 16 x log100 10)
= 3 x ½ x 2
= 3 Â
20. Show that:
loga m ÷ logab m = 1 + loga bÂ
Solution:
Considering the L.H.S.,
loga m ÷ logab m = loga m/logab m
= logm ab/logm a [As logb a = 1/loga b]
= loga ab [As logx a/logx b = logb a]
= loga a + loga b
= 1 + loga bÂ
21. If log√27 x = 2 2/3, find x.
Solution:
We have,
log√27 x = 2 2/3
log√27 x = 8/3
Removing logarithm, we get
x = √278/3
= 271/2 x 8/3
= 274/3
= 33 x 4/3
= 34
Hence, x = 81
22. Evaluate:
1/(loga bc + 1) + 1/(logb ca + 1) + 1/(logc ab + 1)
Solution:
We have,
Selina Solutions for Class 9 Maths Chapter 8- Logarithms
The Chapter 8, Logarithms, contains 4 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.
8.1 Introduction
8.2 Interchanging
8.3 Laws of logarithms with use
8.4 Expansion of expressions with the help of laws of logarithms
8.5 More about logarithms
Selina Solutions for Class 9 Maths Chapter 8- Logarithms
The Chapter 8 of class 9 teaches the students the method of calculating using logarithm.Logarithms are used to make long and complicated calculations easy. Logarithm is related to exponents or indices. I.e., while ab=c is called the exponential form, logac=b is called the logarithmic form. Read and learn the Chapter 8 of Selina textbook to get to know more about Logarithms. Learn the Selina Solutions for Class 9 effectively to attain excellent result in the examination.