ICSE Class 9 Maths Selina Solutions Chapter 8: Here are the Selina answers to the problems found in the ICSE Class 9 Maths Selina Solutions Chapter 8 Logarithms. Students get in-depth knowledge on the subject of logarithms in this chapter. By completing all of the questions in the Selina textbook, students can easily receive a perfect score on their exams.
The ICSE Class 9 Maths Selina Solutions Chapter 8 is quite simple to comprehend. All the exercise questions in the book are addressed in these solutions, which follow the ICSE or CISCE syllabus. The ICSE Class 9 Maths Selina Solutions Chapter 8 are available here in PDF format, which can be accessed online or downloaded. Additionally, students can download these Selina solutions for free and use them offline for practice.ICSE Class 9 Maths Selina Solutions Chapter 8 PDF
1. Express each of the following in logarithmic form:
(i) 5 3 = 125
(ii) 3 -2 = 1/9
(iii) 10 -3 = 0.001
(iv) (81) 3/4 = 27
Solution:
We know that, a b = c ⇒ log a c = b (i) 5 3 = 125 log 5 125 = 3 (ii) 3 -2 = 1/9 log 3 1/9 = -2 (iii) 10 -3 = 0.001 log 10 0.001 = -3 (iv) (81) 3/4 = 27 log 81 27 = ¾2. Express each of the following in exponential form:
(i) log 8 0.125 = -1
(ii) log 10 0.01 = -2
(iii) log a A = x
(iv) log 10 1 = 0
Solution:
We know that, log a c = b ⇒ a b = c (i) log 8 0.125 = -1 8 -1 = 0.125 (ii) log 10 0.01 = -2 8 -1 = 0.125 (iii) log a A = x a x = A (iv) log 10 1 = 0 10 0 = 13. Solve for x: log 10 x = -2.
Solution:
We have, log 10 x = -2 10 -2 = x [As log a c = b ⇒ a b = c] x = 10 -2 x = 1/10 2 x = 1/100 Hence, x = 0.014. 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) 1/16 to the base 4
(vii) 27 to the base 9
(viii) 1/81 to the base 27
Solution;
(i) Let log 10 100 = x So, 10 x = 100 10 x = 10 2 Then, x = 2 [If a m = a n ; then m = n] Hence, log 10 100 = 2 (ii) Let log 10 0.1 = x So, 10 x = 0.1 10 x = 1/10 10 x = 10 -1 Then, x = -1 [If a m = a n ; then m = n] Hence, log 10 0.1 = -1 (iii) Let log 10 0.001 = x So, 10 x = 0.001 10 x = 1/1000 10 x = 1/10 3 10 x = 10 -3 Then, x = -3 [If a m = a n ; then m = n] Hence, log 10 0.001 = -3 (iv) Let log 4 32 = x So, 4 x = 32 4 x = 2 x 2 x 2 x 2 x 2 (2 2 ) x = 2 5 2 2x = 2 5 Then, 2x = 5 [If a m = a n ; then m = n] x = 5/2 Hence, log 4 32 = 5/2 (v) Let log 2 0.125 = x So, 2 x = 0.125 2 x = 125/1000 2 x = 1/8 2 x = (½) 3 2 x = 2 -3 Then, x = -3 [If a m = a n ; then m = n] Hence, log 2 0.125 = -3 (vi) Let log 4 1/16 = x So, 4 x = 1/16 4 x = (¼) -2 4 x = 4 -2 Then, x = -2 [If a m = a n ; then m = n] Hence, log 4 1/16 = -2 (vii) Let log 9 27 = x So, 9 x = 27 9 x = 3 x 3 x 3 (3 2 ) x = 3 3 3 2x = 3 3 Then, 2x = 3 [If a m = a n ; then m = n] x = 3/2 Hence, log 9 27 = 3/2 (viii) Let log 27 1/81 = x So, 27 x = 1/81 27 x = 1/9 2 (3 3 ) x = 1/(3 2 ) 2 3 3x = 1/3 4 3 3x = 3 -4 Then, 3x = -4 [If a m = a n ; then m = n] x = -4/3 Hence, log 27 1/81 = -4/35. State, true or false:
(i) If log 10 x = a, then 10 x = a
(ii) If x y = z, then y = log z x
(iii) log 2 8 = 3 and log 8 2 = 1/3
Solution:
(i) We have, log 10 x = a So, 10 a = x Thus, the statement 10 x = a is false (ii) We have, x y = z So, log x z = y Thus, the statement y = log z x is false (iii) We have, log 2 8 = 3 So, 2 3 = 8 … (1) Now consider the equation, log 8 2 = 1/3 8 1/3 = 2 (2 3 ) 1/3 = 2 … (2) Both equations (1) and (2) are correct Thus, the given statements, log 2 8 = 3 and log 8 2 = 1/3 are true6. Find x, if:
(i) log 3 x = 0
(ii) log x 2 = -1
(iii) log 9 243 = x
(iv) log 5 (x – 7) = 1
(v) log 4 32 = x – 4
(vi) log 7 (2x 2 – 1) = 2
Solution:
(i) We have, log 3 x = 0 So, 3 0 = x 1 = x Hence, x = 1 (ii) we have, log x 2 = -1 So, x -1 = 2 1/x = 2 Hence, x = ½ (iii) We have, log 9 243 = x 9 x = 243 (3 2 ) x = 3 5 3 2x = 3 5 On comparing the exponents, we get 2x = 5 x = 5/2 = 2½ (iv) We have, log 5 (x – 7) = 1 So, 5 1 = x – 7 5 = x – 7 x = 5 + 7 Hence, x = 12 (v) We have, log 4 32 = x – 4 So, 4 (x – 4) = 32 (2 2 ) (x – 4) = 2 5 2 (2x – 8) = 2 5 On comparing the exponents, we get 2x – 8 = 5 2x = 5 + 8 Hence, x = 13/2 = 6½ (vi) We have, log 7 (2x 2 – 1) = 2 So, (2x 2 – 1) = 7 2 2x 2 – 1 = 49 2x 2 = 49 + 1 2x 2 = 50 x 2 = 25 Taking square root on both side, we get x = ±5 Hence, x = 5 (Neglecting the negative value)7. Evaluate:
(i) log 10 0.01
(ii) log 2 (1 ÷ 8)
(iii) log 5 1
(iv) log 5 125
(v) log 16 8
(vi) log 0.5 16
Solution:
(i) Let log 10 0.01 = x Then, 10 x = 0.01 10 x = 1/100 = 1/10 2 So, 10 x = 10 -2 On comparing the exponents, we get x = -2 Hence, log 10 0.01 = -2 (ii) Let log 2 (1 ÷ 8) = x Then, 2 x = 1/8 2 x = 1/2 3 So, 2 x = 2 -3 On comparing the exponents, we get x = -3 Hence, log 10 (1 ÷ 8) = -3 (iii) Let log 5 1 = x Then, 5 x = 1 5 x = 5 0 On comparing the exponents, we get x = 0 Hence, log 5 1 = 0 (iv) Let log 5 125 = x Then, 5 x = 125 5 x = (5 x 5 x 5) = 5 3 So, 5 x = 5 3 On comparing the exponents, we get x = 3 Hence, log 5 125 = 3 (v) Let log 16 8 = x Then, 16 x = 8 (2 4 ) x = (2 x 2 x 2) = 2 3 So, 2 4x = 2 3 On comparing the exponents, we get 4x = 3 x = 3/4 Hence, log 16 8 = 3/4 (vi) Let log 0.5 16 = x Then, 0.5 x = 16 (5/10) x = (2 x 2 x 2 x 2) (1/2) x = 2 4 So, 2 -x = 2 4 On comparing the exponents, we get -x = 4 ⇒ x = -4 Hence, log 0.5 16 = -48. If log a m = n, express a n – 1 in terms in terms of a and m.
Solution:
We have, log a m = n So, a n = m Dividing by a on both sides, we get a n /a = m/a a n-1 = m/a9. Given log 2 x = m and log 5 y = n
(i) Express 2 m-3 in terms of x
(ii) Express 5 3n+2 in terms of y
Solution:
Given, log 2 x = m and log 5 y = n So, 2 m = x and 5 n = y (i) Taking, 2 m = x 2 m /2 3 = x/2 3 2 m-3 = x/8 (ii) Taking, 5 n = y Cubing on both sides, we have (5 n ) 3 = y 3 5 3n = y 3 Multiplying by 5 2 on both sides, we have 5 3n x 5 2 = y 3 x 5 2 5 3n+2 = 25y 310. If log 2 x = a and log 3 y = a, write 72 a in terms of x and y.
Solution:
Given, log 2 x = a and log 3 y = a So, 2 a = x and 3 a = y Now, the prime factorization of 72 is 72 = 2 x 2 x 2 x 3 x 3 = 2 3 x 3 2 Hence, (72) a = (2 3 x 3 2 ) a = 2 3a x 3 2a = (2 a ) 3 x (3 a ) 2 = x 3 y 2 [As 2 a = x and 3 a = y]11. Solve for x: log (x – 1) + log (x + 1) = log 2 1
Solution:
We have, log (x – 1) + log (x + 1) = log 2 1 log (x – 1) + log (x + 1) = 0 log [(x – 1) (x + 1)] = 0 Then, (x – 1) (x + 1) = 1 [As log 1 = 0] x 2 – 1 = 1 x 2 = 1 + 1 x 2 = 2 x = ±√2 The value -√2 is not a possible, since log of a negative number is not defined. Hence, x = √212. If log (x 2 – 21) = 2, show that x = ± 11.
Solution:
Given, log (x 2 – 21) = 2 So, x 2 – 21 = 10 2 x 2 – 21 = 100 x 2 = 121 Taking square root on both sides, we get x = ±111. Express in terms of log 2 and log 3:
(i) log 36
(ii) log 144
(iii) log 4.5
(iv) log 26/51 – log 91/119
(v) log 75/16 – 2log 5/9 + log 32/243
Solution:
(i) log 36 = log (2 x 2 x 3 x 3) = log (2 2 x 3 2 ) = log 2 2 + log 3 2 [Using log a mn = log a m + log a n] = 2log 2 + 2log 3 [Using log a m n = nlog a m] (ii) log 144 = log (2 x 2 x 2 x 2 x 3 x 3) = log (2 4 x 3 2 ) = log 2 4 + log 3 2 [Using log a mn = log a m + log a n] = 4log 2 + 2log 3 [Using log a m n = nlog a m] (iii) log 4.5 = log 45/10 = log (5 x 3 x 3)/ (5 x 2) = log 3 2 /2 = log 3 2 – log 2 [Using log a m/n = log a m – log a n] = 2log 3 – log 2 [Using log a m n = nlog a m] (iv) log 26/51 – log 91/119 = log (26/51)/ (91/119) [Using log a m – log a n = log a 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 log a m/n = log a m – log a n] (v) log 75/16 – 2log 5/9 + log 32/243 = log 75/16 – log (5/9) 2 + log 32/243 [Using nlog a m = log a m n ] = log 75/16 – log 25/81 + log 32/243 = log [(75/16)/ (25/81)] + log 32/243 [Using log a m – log a n = log a 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 log a m + log a n = log a mn] = log 32/16 = log 22. 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 x 2 – log y = 1 [Using nlog a m = log a m n ] log x 2 /y = 1 [Using log a m – log a n = log a m/n] Now, on removing log we have x 2 /y = 10 1 ⇒ x 2 = 10y (ii) We have, 2 log x + 3 log y = log a Then, log x 2 + log y 3 = log a [Using nlog a m = log a m n ] log x 2 y 3 = log a [Using log a m + log a n = log a mn] Now, on removing log we have x 2 y 3 = a (iii) a log x – b log y = 2 log 3 Then, log x a – log y b = log 3 2 [Using nlog a m = log a m n ] log x a /y b = log 3 2 [Using log a m – log a n = log a m/n] Now, on removing log we have x a /y b = 3 2 ⇒ x 2 = 9y b3. Evaluate each of the following without using tables:
(i) log 5 + log 8 – 2log 2
(ii) log 10 8 + log 10 25 + 2log 10 3 – log 10 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 2 2 [Using nlog a m = log a m n ] = log 5 + log 8 – log 4 = log (5 x 8) – log 4 [Using log a m + log a n = log a mn] = log 40 – log 4 = log 40/4 [Using log a m – log a n = log a m/n] = log 10 = 1 (ii) We have, log 10 8 + log 10 25 + 2log 10 3 – log 10 18 = log 10 8 + log 10 25 + log 10 3 2 – log 10 18 [Using nlog a m = log a m n ] = log 10 8 + log 10 25 + log 10 9 – log 10 18 = log 10 (8 x 25 x 9) – log 10 18 [Using log a l + log a m + log a n = log a lmn] = log 10 1800 – log 10 18 = log 10 1800/18 [Using log a m – log a n = log a m/n] = log 10 100 = log 10 10 2 = 2log 10 10 [Using log a m n = nlog a m] = 2 x 1 = 2 (iii) We have, log 4 + 1/3log 125 – 1/5log 32 = log 4 + log (125) 1/3 – log (32) 1/5 [Using nlog a m = log a m n ] = log 4 + log (5 3 ) 1/3 – log (2 5 ) 1/5 = log 4 + log 5 – log 2 = log (4 x 5) – log 2 [Using log a m + log a n = log a mn] = log 20 – log 2 = log 20/2 [Using log a m – log a n = log a m/n] = log 10 = 14. 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 nlog a m = log a m n ] = log 225/324 – log 25/162 + log 4/9 = log [(225/324)/(25/162)] + log 4/9 [Using log a m – log a n = log a 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 log a m + log a n = log a mn] = log (9/2 x 4/9) = log 2 = R.H.S.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 2 3 + log 125 1/3 – log 3 [Using nlog a m = log a m n ] = log 48 – log 8 + log (5 3 ) 1/3 – log 3 = (log 48 – log 8) + (log 5 – log 3) = log 48/8 + log 5/3 [Using log a m – log a n = log a m/n] = log (48/8 x 5/3) [Using log a m + log a n = log a mn] = log (2 x 5) = log 10 = 1 Hence, x = 16. Express log 10 2 + 1 in the form of log 10 x.
Solution:
Given, log 10 2 + 1 = log 10 2 + log 10 10 [As, log 10 10 = 1] = log 10 (2 x 10) [Using log a m + log a n = log a mn] = log 10 207. Solve for x:
(i) log 10 (x – 10) = 1
(ii) log (x 2 – 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, log 10 (x – 10) = 1 Then, x – 10 = 10 1 x = 10 + 10 Hence, x = 20 (ii) We have, log (x 2 – 21) = 2 Then, x 2 – 21 = 10 2 x 2 – 21 = 100 x 2 = 100 + 21 x 2 = 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 log a m + log a n = log a mn] log (x 2 – 2 2 ) = log 5 [As (x – a)(x + a) = x 2 – a 2 ] log (x 2 – 4) = log 5 Removing log on both sides, we get x 2 – 4 = 5 x 2 = 5 + 4 x 2 = 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 2 4 + log 3 2 [Using nlog a m = log a m n ] log (x + 5)(x – 5) = log 16 + log 9 [Using log a m + log a n = log a mn] log (x 2 – 5 2 ) = log (16 x 9) [As (x – a)(x + a) = x 2 – a 2 ] log (x 2 – 25) = log 144 Removing log on both sides, we have x 2 – 25 = 144 x 2 = 144 + 25 x 2 = 169 Taking square root on both sides, we get x = ±138. Solve for x:
(i) log 81/log 27 = x
(ii) log 128/log 32 = x
(iii) log 64/log 8 = log x
(iv) log 225/log 15 = log x
Solution:
(i) 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 3 4 /log 3 3 = (4log 3)/(3log 3) [Using log a m n = nlog a m] = 4/3 Hence, x = 4/3 (ii) 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 2 7 /log 2 5 = (7log 2)/(5log 2) [Using log a m n = nlog a m] = 7/5 Hence, x = 7/5 (iii) 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 2 6 /log 2 3 = (6log 2)/(3log 2) [Using log a m n = nlog a m] = 6/3 = 2 So, log x = 2 Hence, x = 10 2 = 100 (iv) We have, log 225/log 15 = log x log x = log 225/log 15 = log (15 x 15)/ log 15 = log 15 2 /log 15 = (2log 15)/log 15 [Using log a m n = nlog a m] = 2 So, log x = 2 Hence, x = 10 2 = 1009. Given log x = m + n and log y = m – n, express the value of log 10x/y 2 in terms of m and n.
Solution:
Given, log x = m + n and log y = m – n Now consider log 10x/y 2 , log 10x/y 2 = log 10x – log y 2 [Using log a m/n = log a m – log a 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 – m10. 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 10 3 = 3log 10 = 3 [Using log a m n = nlog a 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 (iii) We have, log 25/log 5 = log x log (5 x 5)/log 5 = log x log 5 2 / log 5 = log x 2log 5/log 5 = log x [Using log a m n = nlog a m] 2 = log x So, x = 10 2 x = 100 Thus, the statement x = 2 is false (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 false11. If log 10 2 = a and log 10 3 = b; express each of the following in terms of ‘a’ and ‘b’:
(i) log 12
(ii) log 2.25
(iii) log
(iv) log 5.4
(v) log 60
(iv) log
Solution:
Given that log 10 2 = a and log 10 3 = b … (1) (i) log 12 = log (2 x 2 x 3) = log (2 x 2) + log 3 [Using log a mn = log a m + log a n] = log 2 2 + log 3 = 2log 2 + log 3 [Using log a m n = nlog a 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 log a m n = nlog a m] = 2(log 3 – log 2) [Using log a m/n = log a m – log a n] = 2(b – a) [From 1] = 2b – 2a
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) 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 log a mn = log a m + log a n] = log 2 2 + log 3 = 2log 2 + log 3 [Using log a m n = nlog a m] = 2 x 0.3010 + 0.4771 = 1.0791 (ii) log 1.2 = log 12/10 = log 12 – log 10 [Using log a m/n = log a m – log a n] = log (4 x 3) – log 10 = log 4 + log 3 – log 10 [Using log a mn = log a m + log a n] = log 2 2 + log 3 – log 10 = 2log 2 + log 3 – log 10 [Using log a m n = nlog a 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 log a m/n = log a m – log a n] = log (2 x 2 x 3 x 3) – 1 [As log 10 = 1] = log (2 2 x 3 2 ) – 1 = log 2 2 + log 3 2 – 1 [Using log a mn = log a m + log a n] = 2log 2 + 2log 3 – 1 [Using log a m n = nlog a 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 log a mn = log a m + log a n] = log 3/2 + 1 [As log 10 = 1] = log 3 – log 2 + 1 [Using log a m/n = log a m – log a n] = 0.4771 – 0.3010 + 1 = 1.1761 (v) log 25 = log (25/4 x 4) = log 100/4 = log 100 – log 4 [Using log a m/n = log a m – log a n] = log 10 2 – log 2 2 = 2log 10 – 2log 2 [Using log a m n = nlog a m] = (2 x 1) – (2 x 0.3010) = 2 – 0.6020 = 1.39813. Given 2 log 10 x + 1 = log 10 250, find:
(i) x
(ii) log 10 2x
Solution:
(i) Given equation, 2log 10 x + 1 = log 10 250 log 10 x 2 + log 10 10 = log 10 250 [Using nlog a m = log a m n and log 10 10 = 1] log 10 10x 2 = log 10 250 [Using log a m + log a n = log a mn] Removing log on both sides, we have 10x 2 = 250 x 2 = 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, log 10 2x = log 10 2(5) = log 10 10 = 114. Given 3log x + ½ log y = 2, express y in term of x.
Solution:
We have, 3log x + ½ log y = 2 log x 3 + log y 1/2 = 2 [Using log a m + log a n = log a mn] log x 3 y 1/2 = 2 Removing logarithm, we get x 3 y 1/2 = 10 2 y 1/2 = 100/x 3 On squaring on both sides, we get y = 10000/x 6 y = 10000x -615. If x = (100) a , y = (10000) b and z = (10) c , find log 10√y/x 2 z 3 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 10 2 , log y = b log 10 4 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/x 2 z 3 = log 10√y – log x 2 z 3 [Using log a m/n = log a m – log a n] = (log 10 + log √y) – (log x 2 + log z 3 ) [Using log a mn = log a m + log a n] = 1 + log y 1/2 – log x 2 – log z 3 = 1 + ½ log y – 2 log x – 3 log z [Using log a m n = nlog a m] = 1 + ½(4b) – 2(2a) – 3c … [Using (i)] = 1 + 2b – 4a – 3c16. 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 log a mn = log a m + log a 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 5 2 – log 3 + log 2 2 + log x = 2 [Using nlog a m = log a m n ] log 25 – log 3 + log 4 + log x = 2 log (25 × 4 × x)/3 = 2 [Using log a m + log a n = log a mn & log a m – log a n = log a m/n] log 100x/3 = 2 On removing logarithm, 100x/3 = 10 2 100x/3 = 100 Dividing by 100 on both sides, we have x/3 = 1 Hence, x = 31. If log 10 8 = 0.90; find the value of:
(i) log 10 4
(ii) log √32
(iii) log 0.125
Solution:
Given, log 10 8 = 0.90 log 10 (2 x 2 x 2) = 0.90 log 10 2 3 = 0.90 3 log 10 2 = 0.90 log 10 2 = 0.90/3 log 10 2 = 0.30 … (1) (i) log 4 = log 10 (2 x 2) = log 10 2 2 = 2 log 10 2 = 2 x 0.60 … [From (1)] = 1.20 (ii) log √32 = log 10 32 1/2 = ½ log 10 (2 x 2 x 2 x 2 x 2) = ½ log 10 2 5 = ½ x 5 log 10 2 = ½ x 5 x 0.30 … [From (1)] = 0.75 (iii) log 0.125 = log 125/1000 = log 10 1/8 = log 10 1/2 3 = log 10 2 -3 = -3 log 10 2 = -3 x 0.30 … [From (1)] = -0.902. If log 27 = 1.431, find the value of:
(i) log 9 (ii) log 300
Solution:
Given, log 27 = 1.431 So, log 3 3 = 1.431 3log 3 = 1.431 log 3 = 1.431/3 = 0.477 … (1) (i) log 9 = log 3 2 = 2log 3 = 2 x 0.477 … [From (1)] = 0.954 (ii) log 300 = log (3 x 100) = log 3 + log 100 = log 3 + log 10 2 = log 3 + 2log 10 = log 3 + 2 [As log 10 = 1] = 0.477 + 2 = 2.4773. If log 10 a = b, find 10 3b – 2 in terms of a.
Solution:
Given, log 10 a = b Now, Let 10 3b – 2 = x Applying log on both sides, log 10 3b – 2 = log x (3b – 2)log 10 = log x 3b – 2 = log x 3log 10 a – 2 = log x 3log 10 a – 2log 10 = log x log 10 a 3 – log 10 2 = log x log 10 a 3 – log 100 = log x log 10 a 3 /100 = log x On removing logarithm, we get a 3 /100 = x Hence, 10 3b – 2 = a 3 /1004. If log 5 x = y, find 5 2y+ 3 in terms of x.
Solution:
Given, log 5 x = y So, 5 y = x Squaring on both sides, we get (5 y ) 2 = x 2 5 2y = x 2 5 2y × 5 3 = x 2 × 5 3 Hence, 5 2y+3 = 125x 25. Given: log 3 m = x and log 3 n = y.
(i) Express 3 2x – 3 in terms of m.
(ii) Write down 3 1 – 2y + 3x in terms of m and n.
(iii) If 2 log 3 A = 5x – 3y; find A in terms of m and n.
Solution:
Given, log 3 m = x and log 3 n = y So, 3 x = m and 3 y = n … (1) (i) Taking the given expression, 3 2x – 3 3 2x – 3 = 3 2x . 3 -3 = 3 2x . 1/3 3 = (3 x ) 2 /3 3 = m 2 /3 3 … [Using (1)] = m 2 /27 Hence, 3 2x – 3 = m 2 /27 (ii) Taking the given expression, 3 1 – 2y + 3x 3 1 – 2y + 3x = 3 1 . 3 -2y . 3 3x = 3 . (3 y ) -2 . (3 x ) 3 = 3 . n -2 . m 3 … [Using (1)] = 3m 3 /n 2 Hence, 3 1 – 2y + 3x = 3m 3 /n 2 (iii) Taking the given equation, 2 log 3 A = 5x – 3y log 3 A 2 = 5x – 3y log 3 A 2 = 5log 3 m – 3log 3 n … [Using (1)] log 3 A 2 = log 3 m 5 – log 3 n 3 log 3 A 2 = log 3 m 5 /n 3 Removing logarithm on both sides, we get A 2 = m 5 /n 3 Hence, by taking square root on both sides A = √(m 5 /n 3 ) = m 5/2 /n 3/26. 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 = 37. 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 b a . a b = 10
Solution:
(i) Taking L.H.S. we have, = (log a) 2 – (log b) 2 = (log a + log b) (log a – log b) [As x 2 – y 2 = (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 b a + log a b – 1 = 0 log b a + log a b = 1 log b a a b = 1 On removing logarithm, we get b a a b = 10 – Hence proved9. (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: log 10 125 = 3(1 – log 10 2).
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 y 2 – log x = 3 log y 2 /x = 3 On removing logarithm, we have y 2 /x = 10 3 = 1000 Hence, x = y 2 /1000 (iii) Considering the L.H.S., we have log 10 125 = log 10 (5 x 5 x 5) = log 10 5 3 = 3log 10 5 = 3log 10 10/2 = 3(log 10 10 – log 10 2) = 3(1 – log 10 2) [Since, log 10 10 = 1] = R.H.S. – Hence proved10. 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 x 2 y 3 /z 4 .
Solution:
We have, log x = 2m – n, log y = n – 2m and log z = 3m – 2n Now, considering log x 2 y 3 /z 4 = log x 2 y 3 – log z 4 = (log x 2 + log y 3 ) – log z 4 = 2log x + 3log y – 4log z = 2(2m – n) + 3(n – 2m) – 4(3m – 2n) = 4m – 2n + 3n – 6m – 12m + 8n = -14m + 9n11. Given log x 25 – log x 5 = 2 – log x 1/125; find x.
Solution:
We have, log x 25 – log x 5 = 2 – log x 1/125 log x (5 x 5) – log x 5 = 2 – log x 1/(5 x 5 x 5) log x 5 2 – log x 5 = 2 – log x 1/5 3 2log x 5 – log x 5 = 2 – log x 1/5 3 log x 5 = 2 – 3log x 1/5 log x 5 = 2 + 3log x (1/5) -1 log x 5 = 2 + 3log x 5 2 = log x 5 – 3log x 5 2 = -2log x 5 -1 = log x 5 Removing logarithm, we get x -1 = 5 Hence, x = 1/51. If 3/2 log a + 2/3 log b – 1 = 0, find the value of a 9 .b 4
Solution:
Given equation, 3/2 log a + 2/3 log b – 1 = 0 log a 3/2 + log b 2/3 – 1 = 0 log a 3/2 × b 2/3 – 1 = 0 log a 3/2 .b 2/3 = 1 Removing logarithm, we have a 3/2 .b 2/3 = 10 On manipulating, (a 3/2 .b 2/3 ) 6 = 10 6 Hence, a 9 .b 4 = 10 62. 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 3 2 + log 5 = log a 2 log 10 + log 2 + log 9 + log 5 = log a 2 log (10 x 2 x 9 x 5) = log a 2 log 900 = log a 2 On removing logarithm on both sides, we have 900 = a 2 Taking square root, we get a = ±30 Since, a cannot be a negative value, Hence, a = 303. If x = log 0.6; y = log 1.25 and z = log 3 – 2log 2, find the values of:
(i) x + y – z (ii) 5 x + y – z
Solution:
Given, x = log 0.6, y = log 1.25 and z = log 3 – 2log 2 So, z = log 3 – log 2 2 = 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) (ii) Now, considering 5 x+y-z = 5 0 … [From (2)] = 14. If a 2 = log x, b 3 = log y and 3a 2 – 2b 3 = 6 log z, express y in terms of x and z.
Solution:
We have, a 2 = log x and b 3 = log y Now, considering the equation 3a 2 – 2b 3 = 6log z 3log x – 2log y = 6log z log x 3 – log y 2 = log z 6 log x 3 /y 2 = log z 6 On removing logarithm on both sides, we get x 3 /y 2 = z 6 So, y 2 = x 3 /z 6 Taking square root on both sides, we get y = √(x 3 /z 6 ) Hence, y = x 3/2 /z 35. If log (a – b)/2 = ½ (log a + log b), show that: a 2 + b 2 = 6ab.
Solution:
We have, log (a – b)/2 = ½ (log a + log b) log (a – b)/2 = ½ log a + ½ log b = log a 1/2 + log b 1/2 = 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 a 2 + b 2 – 2ab = 4ab a 2 + b 2 = 4ab + 2ab a 2 + b 2 = 6ab – Hence proved6. If a 2 + b 2 = 23ab, show that: log (a + b)/5 = ½ (log a + log b).
Solution:
Given, a 2 + b 2 = 23ab Adding 2ab on both sides, a 2 + b 2 + 2ab = 23ab + 2ab (a + b) 2 = 25ab (a + b) 2 /25 = ab [(a + b)/5] 2 = ab Taking logarithm on both sides, we have log [(a + b)/5] 2 = 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 20 2 – 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 x 2 – 8x + 16 = 16 x 2 – 8x = 0 x(x – 8) = 0 So, x = 0 or x = 8 If x = 0, then log (x – 4) doesn’t exist Hence, x = 88. 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 10 2 log xy = log 100 On removing logarithm, xy = 100 … (1) Now, consider the equation log x/y + 2log 2 = 2 log x/y + log 2 2 = 2 log x/y + log 4 = 2 log 4x/y = 2 Removing logarithm, we get 4x/y = 10 2 4x/y = 100 x/y = 25 (xy)/y 2 = 25 100/y 2 = 25 … [From (1)] y 2 = 100/25 y 2 = 4 y = 2 [Since, y > 0] From log xy = 2 Substituting the value of y, we get log 2x = 2 On removing logarithm, 2x = 10 2 2x = 100 x = 100/2 x = 50 Thus, the values x and y are 50 and 2 respectively9. Find x, if:
(i) log x 625 = -4
(ii) log x (5x – 6) = 2
Solution:
(i) We have, log x 625 = -4 On removing logarithm, x -4 = 625 (1/x) 4 = 5 4 Taking the fourth root on both sides, 1/x = 5 Hence, x = 1/5 (ii) We have, log x (5x – 6) = 2 On removing logarithm, x 2 = 5x – 6 x 2 – 5x + 6 = 0 x 2 – 3x – 2x + 6 = 0 x(x – 3) – 2(x – 2) = 0 (x – 2)(x – 3) = 0 Hence, x = 2 or 310. If p = log 20 and q = log 25, find the value of x, if 2log (x + 1) = 2p – q.
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 20 2 – 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 = 311. If log 2 (x + y) = log 3 (x – y) = log 25/log 0.2, find the value of x and y.
Solution:
Considering the relation, log 2 (x + y) = log 25/log 0.2 log 2 (x + y) = log 0.2 25 = log 2/10 5 2 = 2log 1/5 5 = 2log 5 -1 5 = -2log 5 5 = -2 x 1 = -2 So, we have log 2 (x + y) = -2 Removing logarithm, we get x + y = 2 -2 x + y = 1/2 2 x + y = ¼ … (i) Now, considering the relation log 3 (x – y) = log 25/log 0.2 log 3 (x – y) = = log 0.2 25 = log 2/10 5 2 = 2log 1/5 5 = 2log 5 -1 5 = -2log 5 5 = -2 x 1 = -2 So, we have log 3 (x – y) = -2 Removing logarithm, we get x – y = 3 -2 x – y = 1/3 2 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 respectively12. Given: log x/log y = 3/2 and log xy = 5; find the values of x and y.
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 = 10 3 = 1000 Substituting value of x in (ii), we get log xy = 5 Removing logarithm, we get xy = 10 5 (10 3 ). y = 10 5 y = 10 5 /10 3 y = 10 2 y = 10013. Given log 10 x = 2a and log 10 y = b/2
(i) Write 10 a in terms of x
(ii) Write 10 2b + 1 in terms of y
(iii) If log 10 p = 3a – 2b, express p in terms of x and y.
Solution:
Given, log 10 x = 2a and log 10 y = b/2 (i) Taking log 10 x = 2a Removing logarithm on both sides, x = 10 2a Taking square root on both sides, we get x 1/2 = 10 2a/2 Hence, 10 a = x 1/2 (ii) Taking log 10 y = b/2 Removing logarithm on both sides, y = 10 b/2 On manipulating, y 4 = 10 b/2 x 4 y 4 = 10 2b 10y 4 = 10 2b x 10 Hence, 10 2b + 1 = 10y 4 (iii) We have, 10 a = x 1/2 and y = 10 b/2 Considering the equation, log 10 p = 3a – 2b log 10 p = 3a – 2b Removing logarithm, we get p = 10 3a – 2b p = 10 3a /10 2b p = (10 a ) 3 /(10 b/2 ) 4 p = (x 1/2 ) 3 /(y) 4 Hence, p = x 3/2 /y 414. Solve:
log 5 (x + 1) – 1 = 1 + log 5 (x – 1).
Solution:
Considering the given equation, log 5 (x + 1) – 1 = 1 + log 5 (x – 1) log 5 (x + 1) – log 5 (x – 1) = 1 + 1 log 5 (x + 1)/(x – 1) = 2 Removing logarithm, we have (x + 1)/(x – 1) = 5 2 (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 = 13/1215. Solve for x, if:
log x 49 – log x 7 + log x 1/343 + 2 = 0
Solution:
We have, log x 49 – log x 7 + log x 1/343 + 2 = 0 log x 49/(7 x 343) + 2 = 0 log x 1/49 = -2 log x 1/7 2 = -2 log x 7 -2 = -2 -2log x 7 = -2 So, log x 7 = 1 Removing logarithm, we get x = 716. If a 2 = log x, b 3 = log y and a 2 /2 – b 3 /3 = log c, find c in terms of x and y.
Solution:
Given, a 2 = log x, b 3 = log y Considering the given equation, a 2 /2 – b 3 /3 = log c (log x)/2 – (log y)/3 = log c ½ log x – 1/3 log y = log c log x 1/2 – log y 1/3 = log c log x 1/2 /y 1/3 = log c On removing logarithm, we get x 1/2 /y 1/3 = c Hence, c = x 1/2 /y 1/3 is the required relation17. Given: x = log 10 12, y = log 4 2 x log 10 9 and z = log 10 0.4, find
(i) x – y – z
(ii) 13 x – y – z
Solution:
(i) Considering, x – y – z = log 10 12 – (log 4 2 x log 10 9) – log 10 0.4 = log 10 12 – (log 4 2 x log 10 9) – log 10 0.4 = log 10 (4 x 3) – (log 10 2/ log 10 4 x log 10 9) – log 10 0.4 = log 10 4 + log 10 3 – (log 10 2 x log 10 3 2 )/ log 10 2 2 – log 10 4/10 = log 10 4 + log 10 3 – (log 10 2 x 2log 10 3)/ 2log 10 2 – (log 10 4 – log 10 10) = log 10 4 + log 10 3 – log 10 3 – log 10 4 + log 10 10 = log 10 4 + log 10 3 – log 10 3 – log 10 4 + 1 = 1 (ii) Now, 13 x – y – z = 13 1 = 1318. Solve for x, log x 15√5 = 2 – log x 3√5
Solution:
Considering the given equation, log x 15√5 = 2 – log x 3√5 log x 15√5 + log x 3√5 = 2 log x (15√5 x 3√5) = 2 log x (45 x 5) = 2 log x 225 = 2 Removing logarithm, we get x 2 = 225 Taking square root on both sides, x = 1519. Evaluate:
(i) log b a x log c b x log a c
(ii) log 3 8 ÷ log 9 16
(iii) log 5 8/(log 25 16 x log 100 10)
Solution:
Using log b a = 1/log a b and log x a/log x b = log b a, we have (i) log b a x log c b x log a c20. Show that:
log a m ÷ log ab m = 1 + log a b
Solution:
Considering the L.H.S., log a m ÷ log ab m = log a m/log ab m = log m ab/log m a [As log b a = 1/log a b] = log a ab [As log x a/log x b = log b a] = log a a + log a b = 1 + log a b21. 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 = √27 8/3 = 27 1/2 x 8/3 = 27 4/3 = 3 3 x 4/3 = 3 4 Hence, x = 8122. Evaluate:
1/(log a bc + 1) + 1/(log b ca + 1) + 1/(log c ab + 1)
Solution:
We have,1. Fundamental Understanding of Logarithms:
Concept Clarity: The ICSE Class 9 Maths Selina Solutions Chapter 8 introduces the fundamental concepts of logarithms, including the definition, properties, and rules. Understanding these basics is crucial as logarithms are foundational in many advanced mathematical concepts.
Building Blocks: ICSE Class 9 Maths Selina Solutions Chapter 8 helps students build a solid foundation for more complex topics in higher classes, including calculus and algebra.
2. Practical Applications:
Real-World Applications: Logarithms have real-world applications in fields like science (e.g., pH scale in chemistry), engineering, and computer science. Learning this chapter helps students understand how logarithms are used to solve practical problems.
Problem-Solving Skills: By solving a variety of problems, students improve their analytical and problem-solving skills, which are applicable in various scientific and technological fields.