ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths APC Understanding Solutions. Solutions of Exercise-9.2. This post is the Solutions of ML Aggarwal Chapter 9- Logarithms for ICSE Maths Class-9. APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-9 Logarithms for ICSE Board Class-9. Visit official website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 9th |
| Chapter-9 | Logarithms |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-9.2 Questions |
| Edition | 2021-2022 |
Exe-9.2 Solutions of ML Aggarwal for ICSE Class-9 Ch-9, Logarithms
Note:- Before viewing Solutions of Chapter -9 Logarithms Class-9 of ML Aggarwal Solutions . Read the Chapter Carefully. Then solve all example given in Exercise-9.1, Exercise-9.2, MCQs, Chapter Test.
Logarithms Exe-9.2
ML Aggarwal Class 9 ICSE Maths Solutions
Page 175
Question 1. Simplify the following:
(i) log a3 – log a2
(ii) log a3 ÷ log a2
(iii) log 4/log 2
(iv) (log 8 log 9)/log 27
(v) log 27/log √3
(vi) (log 9 – log 3)/log 27
Answer:
(i) log a3 – log a2
By using Quotient law,
log a3 – log a2 = log (a3/a2)
= log a
(ii) log a3 ÷ log a2
By using power law,
log a3 ÷ log a2 = 3log a ÷ 2 log a
= 3log a / 2log a
= 3/2
(iii) log 4/log 2
Let us simplify the expression,
log 4/log 2 = log(2×2)/log 2
By using power law,
= 2 log 2/log 2
= 2
(iv) (log 8 log 9)/log 27
Let us simplify the expression,
(log 8 log 9)/log 27 = (log 23. log 32)/log 33
By using power law,
= [(3 log 2).(2 log 3)]/(3 log 3)
= [(log 2).2] / 1
= 2 log 2
= log 22
= log 4
(v) log 27/log √3
Let us simplify the expression,
log 27/log √3 = log(3×3×3)/log(3)1/2
= log 33/log 31/2
By using power law
= 3log 3/((1/2)log 3)
= (3×2)/1 (log 3/log 3)
= (6) (1)
= 6
(vi) (log 9 – log 3)/log 27
Let us simplify the expression,
(log 9 – log 3)/log 27 = [log(3×3) – log 3] / log(3×3×3)
= [log 32 – log 3] / log 33
By using power law
= [2 log 3 – log 3]/3log 3
= log 3/3log 3
= 1/3
Question 2. Evaluate the following:
(i) log (10 ÷ ∛10)
(ii) 2 + ½ log(10-3)
(iii) 2 log 5 + log 8 – ½ log 4
(iv) 2 log 103 + 3 log 10-2 – 1/3 log 5-3 + ½ log 4
(v) 2 log 2 + log 5 – ½ log 36 – log 1/30
(vi) 2 log 5 + log 3 + 3 log 2 – ½ log 36 – 2 log 10
(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80
(viii) 2 log10 5 + log10 8 – ½ log10 4
Answer:
(i) log (10 ÷ ∛10)
Let us simplify the expression,
log (10 ÷ ∛10) = log (10 ÷ 101/3)
= log (101 – 1/3)
= log (102/3)
= 2/3 log 10
= 2/3 (1)
= 2/3
(ii) 2 + ½ log(10-3)
Let us simplify the expression,
2 + ½ log(10-3) = 2 + ½ × (-3) log 10
= 2 – 3/2 log 10
= 2 – 3/2 (1)
= 2 – 3/2
= (4-3)/2
= ½
(iii) 2 log 5 + log 8 – ½ log 4
Let us simplify the expression,
2 log 5 + log 8 – ½ log 4 = log 52 + log 8 – ½ log 22
= log 25 + log 8 – ½ 2log 2
= log 25 + log 8 – log 2
= log (25×8)/2
= log (25×4)
= log 100
= log 102
= 2 log 10
= 2 (1)
= 2
(iv) 2 log 103 + 3 log 10-2 – 1/3 log 5-3 + ½ log 4
Let us simplify the expression,
2 log 103 + 3 log 10-2 – 1/3 log 5-3 + ½ log 4 = 2×3 log 10 + 3(-2)log 10 – 1/3 (-3) log 5 + ½ log 22
= 6 log 10 – 6 log 10 + log 5 + ½ 2 log 2
= 6 log 10 – 6 log 10 + log 5 + log 2
= 0 + log 5 + log 2
= log (5×2)
= log 10
= 1
(v) 2 log 2 + log 5 – ½ log 36 – log 1/30
Let us simplify the expression,
2 log 2 + log 5 – ½ log 36 – log 1/30 = log 22 + log 5 – ½ log 62 – log (1/30)
= log 4 + log 5 – log 6 – log 1/30
= log 4 + log 5 – log 6 – (log 1 – log 30)
= log 4 + log 5 – log 6 – log 1 + log 30
= log 4 + log 5 + log 30 – (log 6 + log 1)
= log (4×5×30) – log (6×1)
= log (4×5×30)/(6×1)
= log (4×5×5)
= log 100
= log 102
= 2 log 10
= 2 (1)
= 2
(vi) 2 log 5 + log 3 + 3 log 2 – ½ log 36 – 2 log 10
Let us simplify the expression,
2log 5 + log 3 + 3log 2 – ½ log 36 – 2log 10 = log 52 + log 3 + log 23 – ½ log 62 – log 102
= log 25 + log 3 + log 8 – log 6 – log 100
= log (25×3×8) – log(6×100)
= log (25×3×8)/ (6×100)
= log (1×3×8) / (6×4)
= log 24/24
= log 1
= 0
(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80
Let us simplify the expression,
log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80 = log 2 + 16(log 16 – log 15) + 12(log 25 – log 24) + 7(log 81 – log 80)
= log 2 + 16 (log 24 – log (3×5)) + 12 (log 52 – log (3×2×2×2)) + 7 (log (3×3×3×3) – log (24 × 5))
= log 2 + 16(4log 2 – (log 3 + log 5)) + 12 (2log 5 – log (3×23)) + 7 (log 34 – (log 24 + log 5))
= log 2 + 16 (4log 2 – log 3 – log 5) + 12 (2log 5 – (log 3 + 3log 2)) + 7 (4log 3 – 4log 2 – log 5)
= log 2 + 64log 2 – 16log 3 – 16log 5 + 24log 5 – 12log 3 – 36log 2 + 28log 3 – 28log 2 – 7log 5
= (log 2 + 64log 2 – 36log 2 – 28log 2) + (-16log 3 – 12log 3 + 28log 3) + (-16log 5 + 24log 5 – 7log 5)
= (65log 2 – 64log 2) + (-28log 3 + 28log 3) + (-23log 5 + 24log 5)
= log 2 + 0 + log 5
= log (2×5)
= log 10
= 1
(viii) 2 log10 5 + log10 8 – ½ log10 4
Let us simplify the expression,
2 log10 5 + log10 8 – ½ log10 4 = log10 52 + log10 8 – log10 41/2
= log10 25 + log10 8 – log10 (2)2×1/2
= log10 25 + log10 8 – log10 2
= log10 [(25×8)/2]
= log10 (25×4)
= log10 100
= log10 102
= 2 log10 10
= 2 (1)
= 2
Logarithms Exe-9.2
ML Aggarwal Class 9 ICSE Maths Solutions
Page 176
Question 3. Express each of the following as a single logarithm:
(i) 2 log 3 – ½ log 16 + log 12
(ii) 2 log10 5 – log10 2 + 3 log10 4 + 1
(iii) ½ log 36 + 2 log 8 – log 1.5
(iv) ½ log 25 – 2 log 3 + 1
(v) ½ log 9 + 2 log 3 – log 6 + log 2 – 2
Answer:
(i) 2 log 3 – ½ log 16 + log 12
Let us simplify the expression into single logarithm,
2 log 3 – ½ log 16 + log 12 = 2 log 3 – ½ log 42 + log 12
= 2 log 3 – log 4 + log 12
= log 32 – log 4 + log 12
= log 9 – log 4 + log 12
= log (9×12)/4
= log (9×3)
= log 27
(ii) 2 log10 5 – log10 2 + 3 log10 4 + 1
Let us simplify the expression into single logarithm,
2 log10 5 – log10 2 + 3 log10 4 + 1 = log10 52 – log10 2 + log10 43 + log10 10
= log10 25 – log10 2 + log10 64 + log10 10
= log10 (25×64×10) – log10 2
= log10 (16000) – log10 2
= log10 (16000/2)
= log10 8000
(iii) ½ log 36 + 2 log 8 – log 1.5
Let us simplify the expression into single logarithm,
½ log 36 + 2 log 8 – log 1.5 = log 361/2 + log 82 – log 1.5
= log (6)2×1/2 + log 64 – log 1.5
= log 6 + log 64 – log (15/10)
= log 6 + log 64 – (log 15 – log 10)
= log (6×64) – log 15 + log 10
= log (6×64×10) – log 15
= log [(6×64×10)/15]
= log (4×64)
= log 256
(iv) ½ log 25 – 2 log 3 + 1
Let us simplify the expression into single logarithm,
½ log 25 – 2 log 3 + 1 = log 251/2 – log 32 + log 10
= log (5)2×1/2 – log 9 + log 10
= log 5 – log 9 + log 10
= log (5×10) – log 9
= log ((5×10)/9)
= log 50/9
(v) ½ log 9 + 2 log 3 – log 6 + log 2 – 2
Let us simplify the expression into single logarithm,
½ log 9 + 2 log 3 – log 6 + log 2 – 2 = log 91/2 + log 32 – log 6 + log 2 – log 100
= log 32×1/2 + log 9 – log 6 + log 2 – log 100
= log 3 + log 9 – log 6 + log 2 – log 100
= log [(3×9×2)/(6×100)]
= log 9/100
Question 4. Prove the following:
(i) log10 4 ÷ log10 2 = log3 9
(ii) log10 25 + log10 4 = log5 25
Answer:
(i) log10 4 ÷ log10 2 = log3 9
Let us consider LHS, log10 4 ÷ log10 2
log10 4 ÷ log10 2 = log10 22 ÷ log10 2
= 2 log10 2 ÷ log10 2
= 2 log10 2/ log10 2
= 2 (1)
= 2
Now let us consider RHS,
log3 9 = log3 32
= 2 log3 3
= 2(1)
= 2
∴ LHS = RHS
Hence proved.
(ii) log10 25 + log10 4 = log5 25
Let us consider LHS, log10 25 + log10 4
log10 25 + log10 4 = log10 (25×4)
= log10 100
= log10 102
= 2 log10 10
= 2(1)
= 2
Now, let us consider RHS,
log5 25 = log5 52
= 2 log5 5
= 2 (1)
= 2
∴ LHS = RHS
Hence proved.
Question 5. If x = (100)a, y = (10000)b and z = (10)c, express log [(10√y)/x2z3] in terms of a, b, c.
Answer:
Given:
x = (100)a = (102)a = 102a
y = (10000)b = (104)b = 104b
z = (10)c
It is given that, log [(10√y)/x2z3]
log [(10√y)/x2z3] = (log 10 + log √y) – (log x2 + log z3)
= (1 + log(y)1/2) – (log x2 + log z3) [we know that, log 10 = 1]
= (1 + ½ log y) – (2 log x + 3 log z)
Now substitute the values of x, y, z, we get
= (1 + ½ log 104b) – (2 log 102a + 3 log 10c)
= (1 + ½ 4b log 10) – (2×2a log 10 + 3×c log 10)
= (1 + ½ 4b) – (2×2a + 3c) [Since, log 10 = 1]
= (1 + 2b) – (4a + 3c)
= 1 + 2b – 4a – 3c
Question 6. If a = log10x, find the following in terms of a :
(i) x
(ii) log10 5√x2
(iii) log10 5x
Answer:
Given:
a = log10x
(i) x
10a = x
∴ x = 10a
(ii) log10 5√x2
log10 5√x2 = log10 (x2)1/5
= log10 (x)2/5
= 2/5 log10 x
= 2/5 (a)
= 2a/5
(iii) log10 5x
x = (10)a
= log10 5x
= log10 5(10)a
= log10 5 + log10 10
= log10 5 + a(1)
= a + log10 5
Question 7. If a =log 2/3, b = log 3/5 and c = 2 log √(5/2). Find the value of
(i) a + b + c
(ii) 5a+b+c
Answer:
Given:
a =log 2/3
b = log 3/5
c = 2 log √(5/2)
(i) a + b + c
Let us substitute the given values, we get
a + b + c = log 2/3 + log 3/5 + 2 log √(5/2)
= (log 2 – log 3) + (log 3 – log 5) + 2 log (5/2)1/2
= log 2 – log 3 + log 3 – log 5 + 2 × ½ (log 5 – log 2)
= log 2 – log 3 + log 3 – log 5 + log 5 – log 2
= 0
(ii) 5a+b+c
5a+b+c = 50
= 1
Question 8. If x = log 3/5, y = log 5/4 and z = 2 log √3/2, find the value of
(i) x + y – z
(ii) 3x+y-z
Answer:
Given:
x = log 3/5 = log 3 – log 5
y = log 5/4 = log 5 – log 4
z = 2 log √3/2 = log (√3/2)2 = log ¾ = log 3 – log 4
(i) x + y – z
Let us substitute the given values, we get
x + y – z = log 3 – log 5 + log 5 – log 4 – (log 3 – log 4)
= log 3 – log 5 + log 5 – log 4 – log 3 + log 4
= 0
(ii) 3x+y-z
3x+y-z = 30
= 1
Question 9. If x = log10 12, y = log4 2 × log10 9 and z = log10 0.4, find the values of
(i) x – y – z
(ii) 7x-y-z
Answer:
Given:
x = log10 12
y = log4 2 × log10 9
z = log10 0.4
(i) x – y – z
Let us substitute the given values, we get
x – y – z = log10 12 – log4 2 × log10 9 – log10 0.4
= log10 (3×4) – log4 41/2 × log10 32 – log10 4/10
= log10 3 + log10 4 – ½ log4 4 × 2 log10 3 – (log10 4 – log10 10)
= log10 3 + log10 4 – ½ × 1 × 2 log10 3 – log10 4 + 1
= log10 3 + log10 4 – log10 3 – log10 4 + 1
= 1
(ii) 7x-y-z
7x-y-z = 71
= 7
Question 10. If log V + log 3 = log π + log 4 + 3 log r, find V in terms of other quantities.
Answer:
Given:
log V + log 3 = log π + log 4 + 3 log r
Let us simplify the given expression to find V,
log (V × 3) = log (π × 4 × r3)
log 3V = log 4πr3
3V = 4πr3
V = 4πr3/3
Question 11. Given 3 (log 5 – log 3) – (log 5 – 2 log 6) = 2 – log n, find n.
Answer:
Given:
3 (log 5 – log 3) – (log 5 – 2 log 6) = 2 – log n
Let us simplify the given expression to find n,
3 log 5 – 3 log 3 – log 5 + 2 log 6 = 2 – log n
2 log 5 – 3 log 3 + 2 log 6 = 2 (1) – log n
log 52 – log 33 + log 62 = 2 log 10 – log n [Since, 1 = log 10]
log 25 – log 27 + log 36 – log 102 = – log n
log n = – log 25 + log 27 – log 36 + log 100
= (log 100 + log 27) – (log 25 + log 36)
= log (100×27) – log (25×36)
= log (100×27)/(25×36)
log n = log 3
n = 3
Question 12. Given that log10 y + 2 log10 x = 2, express y in terms of x.
Answer:
Given:
log10 y + 2 log10 x = 2
Let us simplify the given expression,
log10 y + log10 x2 = 2(1)
log10 y + log10 x2 = 2 log10 10
log10 (y×x2) = log10 102
yx2 = 100
y = 100/x2
Question 13. Express log10 2 + 1 in the form log10x.
Answer:
Given:
log10 2 + 1
Let us simplify the given expression,
log10 2 + 1 = log10 2 + log10 10 [Since, 1 = log10 10 ]
= log10 (2×10)
= log10 20
Question 14. If a2 = log10 x, b2 = log10 y and a2/2 – b2/3 = log10 z. Express z in terms of x and y.
Answer:
Given:
a2 = log10 x
b2 = log10 y
a2/2 – b2/3 = log10 z
Let us substitute the given values in the expression, we get
log10 x/2 – log10 y/3 = log10 z
log10 x1/2 – log10 y1/3 = log10 z
log10 √x – log10 ∛y = log10 z
log10 √x/∛y = log10 z
√x/∛y = z
z = √x/∛y
Question 15. Given that log m = x + y and log n = x – y, express the value of log m²n in terms of x and y.
Answer:
Given:
log m = x + y
log n = x – y
log m²n
Let us simplify the given expression,
log m²n = log m2 + log n
= 2 log m + log n
By substituting the given values, we get
= 2 (x + y) + (x – y)
= 2x + 2y + x – y
= 3x + y
Question 16. Given that log x = m + n and log y = m – n, express the value of log (10x/y2) in terms of m and n.
Answer:
Given:
log x = m + n
log y = m – n
log (10x/y2)
Let us simplify the given expression,
log (10x/y2) = log 10x – log y2
= log 10 + log x – 2 log y
= 1 + log x – 2 log y
= 1 + (m + n) – 2(m – n)
= 1 + m + n – 2m + 2n
= 1 – m + 3n
Logarithms Exe-9.2
ML Aggarwal Class 9 ICSE Maths Solutions
Page 177
Question 17. If log x/2 = log y/3, find the value of y4/x6.
Answer:
Given:
log x/2 = log y/3
Let us simplify the given expression,
By cross multiplying, we get
3 log x = 2 log y
log x3 = log y2
so, x3 = y2
now square on both sides, we get
(x3)2 = (y2)2
x6 = y4
y4/x6 = 1
Question 18. Solve for x:
(i) log x + log 5 = 2 log 3
(ii) log3 x – log3 2 = 1
(iii) x = log 125/log 25
(iv) (log 8/log 2) × (log 3/log√3) = 2 log x
Answer:
(i) log x + log 5 = 2 log 3
Let us solve for x,
Log x = 2 log 3 – log 5
= log 32 – log 5
= log 9 – log 5
= log (9/5)
∴ x = 9/5
(ii) log3 x – log3 2 = 1
Let us solve for x,
log3 x = 1 + log3 2
= log3 3 + log3 2 [Since, 1 can be written as log3 3 = 1]
= log3 (3×2)
= log3 6
∴ x = 6
(iii) x = log 125/log 25
x = log 53/log52
= 3 log 5/ 2 log 5
= 3/2 [Since, log 5/log 5 = 1]
∴ x = 3/2
(iv) (log 8/log 2) × (log 3/log√3) = 2 log x
(log 23/log 2) × (log 3/log31/2) = 2 log x
(3log 2/log 2) × (log 3/½ log 3) = 2 log x
3 × 1/(½) = 2 log x
3 × 2 = 2 log x
6 = 2 log x
log x = 6/2
log x = 3
x = (10)3
= 1000
∴ x = 1000
Question 19. Given 2 log10 x + 1= log10 250, find
(i) x
(ii) log102x
Answer:
Given:
2 log10 x + 1= log10 250
(i) let us simplify the above expression,
log10 x2 + log10 10 = log10 250 [Since, 1 can be written as log10 10]
log10 (x2 × 10) = log10 250
(x2 × 10) = 250
x2 = 250/10
x2 = 25
x = √25
= 5
∴ x = 5
(ii) log10 2x
We know that, x = 5
So, log10 2x = log10 2×5
= log10 10
= 1
Question 20. If log x/log 5 = log y2/log 2 = log 9/log (1/3), find x and y.
Answer:
Given:
log x/log 5 = log y2/log 2 = log 9/log (1/3)
let us consider,
log x/log 5 = log 9/log (1/3)
log x = (log 9×log 5)/log (1/3)
= (log 32 × log 5) / (log 1 – log 3)
= (2 log 3 × log 5) / (-log 3) [log 1 = 0]
= -2 × log 5
= log 5-2
x = 5-2
= 1/52
= 1/25
Now,
log y2/log 2 = log 9/log (1/3)
log y2 = (log 9×log2)/log (1/3)
= (log 32 × log 2) / (log 1 – log 3)
= (2 log 3 × log 2) / (-log 3) [log 1 = 0]
= -2 × log 2
= log 2-2
y2 = 2-2
= 1/22
= ¼
= √(1/4)
= ½
Question 21. Prove the following:
(i) 3log 4 = 4log 3
(ii) 27log 2 = 8log 3
Answer:
(i) 3log 4 = 4log 3
Let us take log on both sides,
If log 3log 4 = log 4log 3
log 4 . log 3 = log 3 . log 4
log 22 . log 3 = log 3 . log 22
2 log 2 . log 3 = log 3 . 2 log 2
Which is true.
Hence proved.
(ii) 27log 2 = 8log 3
Let us take log on both sides,
If log 27log 2 = log 8log 3
log 2 . log 27 = log 3 . log 8
log 2 . log 33 = log 3 . log 23
log 2 . 3 log 3 = log 3 . 3 log 2
3 log2 . log 3 = 3 log2 . log 3
Which is true.
Hence proved.
Question 22. Solve the following equations:
(i) log (2x + 3) = log 7
(ii) log (x +1) + log (x – 1) = log 24
(iii) log (10x + 5) – log (x – 4) = 2
(iv) log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
(v) log (4y – 3) = log (2y + 1) – log 3
(vi) log10 (x + 2) + log10 (x – 2) = log103 + 3 log10 4
(vii) log (3x + 2) + log (3x – 2) = 5 log 2
Answer:
(i) log (2x + 3) = log 7
Let us simplify the expression,
2x + 3 = 7
2x = 7 – 3
2x = 4
x = 4/2
= 2
(ii) log (x +1) + log (x – 1) = log 24
Let us simplify the expression,
log [(x +1) (x – 1)] = log 24
log (x2 – 1) = log 24
(x2 – 1) = 24
x2 = 24 + 1
= 25
x = √25
= 5
(iii) log (10x + 5) – log (x – 4) = 2
Let us simplify the expression,
log (10x + 5) / (x – 4) = 2 log 10
log (10x + 5) / (x – 4) = log 102
(10x + 5) / (x – 4) = 100
10x + 5 = 100 (x – 4)
10x + 5 = 100x – 400
5 + 400 = 100x – 10x
90x = 405
x = 405/90
= 81/18
= 9/2
= 4.5
(iv) log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
Let us simplify the expression,
log10 [5× (5x + 1)] = log10 (x + 5) + log10 10
log10 [5× (5x + 1)] = log10 [(x + 5) × 10] [5× (5x + 1)] = [(x + 5) × 10]
25x + 5 = 10x + 50
25x – 10x = 50 – 5
15x = 45
x = 45/15
= 3
(v) log (4y – 3) = log (2y + 1) – log 3
Let us simplify the expression,
log (4y – 3) = log (2y + 1) / 3
(4y – 3) = (2y + 1) / 3
By cross multiplying, we get
3(4y – 3) = 2y + 1
12y – 9 = 2y + 1
12y – 2y = 9 + 1
10y = 10
y = 10/10
= 1
(vi) log10 (x + 2) + log10 (x – 2) = log103 + 3 log10 4
Let us simplify the expression,
log10 [(x + 2) × (x – 2)] = log10 3 + log10 43
log10 [(x + 2) × (x – 2)] = log10 (3×43)
[(x + 2) × (x – 2)] = (3×43)
(x2 – 4) = (3×4×4×4)
(x2 – 4) = 192
x2 = 192 + 4
= 196
x = √196
= 14
(vii) log (3x + 2) + log (3x – 2) = 5 log 2
Let us simplify the expression,
log (3x + 2) + log (3x – 2) = log 25
log [(3x + 2) × (3x – 2)] = log 32
log (9x2 – 4) = log 32
(9x2 – 4) = 32
9x2 = 32 + 4
9x2 = 36
x2 = 36/9
x2 = 4
x = √4
= 2
Question 23. Solve for x:
log3 (x + 1) – 1 = 3 + log3 (x – 1)
Answer:
Given:
log3 (x + 1) – 1 = 3 + log3 (x – 1)
Let us simplify the expression,
log3 (x + 1) – log3 (x – 1) = 3 + 1
log3 (x + 1) / (x – 1) = 4 log3 3 [Since, log3 3 = 1]
log3 (x + 1) / (x – 1) = log3 34
(x + 1) / (x – 1) = 34
By cross multiplying, we get
(x + 1) = 81 (x – 1)
x + 1 = 81x – 81
81x – x = 1 + 81
80x = 82
x = 82/80
= 41/40
= 1 1/40
Question 24. Solve for x:
5log x + 3log x = 3log x+1 – 5log x – 1
Answer:
Given:
5log x + 3log x = 3log x+1 – 5log x – 1
Let us simplify the expression,
5log x + 3log x = 3log x . 31 – 5log x . 5-1
5log x + 3log x = 3.3log x – 1/5 . 5log x
5log x + 1/5 . 5log x = 3.3log x – 3log x
(1 + 1/5) 5log x = (3 – 1) 3log x
(6/5) 5log x = 2(3log x)
5log x / 3log x = (2×5)/6
(5/3)log x = 10/6
(5/3)log x = 5/3
(5/3)log x = (5/3)1
So, by comparing the powers
log x = 1
log x = log 10
x = 10
Question 25. If log (x-y)/2 = ½ (log x + log y), prove that x2 + y2 = 6xy
Answer:
Given:
log (x-y)/2 = ½ (log x + log y)
Let us simplify,
log (x-y)/2 = ½ (log x×y)
log (x-y)/2 = ½ log xy
log (x-y)/2 = log (xy)1/2
(x-y)/2 = (xy)1/2
By squaring on both sides, we get
[(x-y)/2]2 = [(xy)1/2]2
(x – y)2/4 = xy
By cross multiplying, we get
(x – y)2 = 4xy
x2 + y2 – 2xy = 4xy
x2 + y2 = 4xy + 2xy
x2 + y2 = 6xy
Hence proved.
Question 26. If x2 + y2 = 23xy, Prove that log (x + y)/5 = ½ (log x + log y)
Answer:
Given:
x2 + y2 = 23xy
So, the above equation can be written as
x2 + y2 = 25xy – 2xy
x2 + y2 + 2xy = 25xy
(x + y)2 = 25xy
(x + y)2 / 25 = xy
Now by taking log on both sides, we get
log [(x + y)2 / 25] = log xy
log [(x + y)/5]2 = log xy
2 log (x+y)/5 = log x + log y
log (x+y)/5 = ½ log x + log y
Hence proved.
Question 27. If p = log10 20 and q = log10 25, find the value of x if 2 log10 (x + 1) = 2p – q
Answer:
Given:
p = log10 20
q = log10 25
Then,
2 log10 (x + 1) = 2p – q
Now substitute the values of p and q, we get
2 log10 (x + 1) = 2 log10 20 – log10 25
= 2 log10 20 – log10 52
= 2 log10 20 – 2 log10 5
2 log10 (x + 1) = 2 (log10 20 – log10 5)
log10 (x + 1) = (log10 20 – log10 5)
= log10 (20/5)
log10 (x + 1) = log10 4
(x + 1) = 4
x = 4 – 1
= 3
Question 28. Show that:
(i) 1/log2 42 + 1/log3 42 + 1/log7 42 = 1
(ii) 1/log8 36 + 1/log9 36 + 1/log18 36 = 2
Answer:
(i) 1/log2 42 + 1/log3 42 + 1/log7 42 = 1
Let us consider LHS:
1/log2 42 + 1/log3 42 + 1/log7 42
By using the formula, logn m = logm / logn
1/log2 42 + 1/log3 42 + 1/log7 42 = 1/(log 42/log2) + 1/(log 42/log3) + 1/(log 42/log7)
= log2/log 42 + log3/log 42 + log7/log 42
= (log2 + log3 + log7)/log 42
= (log 2×3×7)/log 42
= log 42 / log 42
= log 42/log 42
= 1
= RHS
(ii) 1/log8 36 + 1/log9 36 + 1/log18 36 = 2
Let us consider LHS:
1/log8 36 + 1/log9 36 + 1/log18 36
By using the formula, logn m = logm / logn
1/log8 36 + 1/log9 36 + 1/log18 36 = 1/(log 36/log8) + 1/(log 36/log9) + 1/(log 36/log18)
= log8/log 36 + log9/log 36 + log18/log 36
= (log8 + log9 + log18)/log 36
= (log 8×9×18)/log 36
= log 362/log 36
= 2 log 36/log 36
= 2
= RHS
Question 29. Prove the following identities:
(i) 1/loga abc + 1/logb abc + 1/logc abc = 1
(ii) logb a. logc b. logd c = logd a
Answer:
(i) 1/loga abc + 1/logb abc + 1/logc abc = 1
Let us consider LHS:
1/loga abc + 1/logb abc + 1/logc abc
By using the formula, logn m = logm / logn
1/loga abc + 1/logb abc + 1/logc abc = 1/(log abc/loga) + 1/(log abc/logb) + 1/(log abc/logc)
= loga/ log abc + logb/ log abc + logc/ log abc
= (loga + logb + logc)/ log abc
= (log a×b×c)/ log abc
= log abc/log abc
= 1
= RHS
(ii) logb a. logc b. logd c = logd a
Let us consider LHS:
logb a. logc b. logd c = (log a/ log b) × (log b/ log c) × (log c/ log d)
= log a/log d
= logd a
= RHS
Question 30. Given that loga x = 1/ α, logb x = 1/β, logc x = 1/γ, find logabc x.
Answer:
It is given that:
loga x = 1/ α, logb x = 1/β, logc x = 1/γ
So,
loga x = 1/ α => log x/loga = 1/ α => loga = α log x
logb x = 1/β => log x/logb = 1/ β => logb = β log x
logc x = 1/γ => log x/logc = 1/ γ => logc = γ log x
Now,
logabc x = log x/log abc
= log x/(log a + log b + log c)
= log x/(α log x + β log x + γ log x)
= log x/log x(α+ β+ γ)
= 1/(α+ β+ γ)
Question 31. Solve for x:
(i) log3 x + log9 x + log81 x = 7/4
(ii) log2 x + log8 x + log32 x = 23/15
Answer:
(i) log3 x + log9 x + log81 x = 7/4
let us simplify the expression,
1/logx 3 + 1/logx 9 + 1/logx 81 = 7/4
1/logx 31 + 1/logx 32 + 1/logx 34 = 7/4
1/logx 3 + 1/2logx 3 + 1/4logx 3 = 7/4
1/logx 3 [1 + ½ + ¼] = 7/4
1/logx 3 [(4+2+1)/4] = 7/4
log3 x [7/4] = 7/4
log3 x = (7/4) × (4/7)
log3 x = 1
log3 x = log3 3 [Since, 1= loga a]
On comparing, we get
x = 3
(ii) log2 x + log8 x + log32 x = 23/15
let us simplify the expression,
1/logx 2 + 1/logx 8 + 1/logx 32 = 23/15
1/logx 21 + 1/logx 23 + 1/logx 25 = 23/15
1/logx 2 + 1/3logx 2 + 1/5logx 2 = 23/15
1/logx 2 [1 + 1/3 + 1/5] = 23/15
log2 x [(15 + 5 + 3)/15] = 23/15
log2 x [23/15] = 23/15
log2 x = (23/15) × (15/23)
log2 x = 1
log2 x = log2 2 [Since, 1= loga a]
On comparing, we get
x = 2
— : End of ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions :–
Return to :- ML Aggarawal Maths Solutions for ICSE Class-9
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