Friday, March 24, 2023

# ML Aggarwal Logarithms Chapter Test Class 9 ICSE Maths Solutions

ML Aggarwal Logarithms Chapter Test Class 9 ICSE Maths APC Understanding Solutions. Solutions ofĀ  Chapter Test. This post is the Solutions ofĀ  ML AggarwalĀ Ā Chapter 9- Logarithms forĀ ICSEĀ MathsĀ Class-9.Ā  APC UnderstandingĀ ML AggarwalĀ Solutions (APC) Avichal Publication Solutions of Chapter-9Ā Logarithms forĀ ICSEĀ BoardĀ Class-9.Ā Visit official websiteĀ CISCEĀ for detail information about ICSE Board Class-9.

## ML Aggarwal Logarithms Chapter Test 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 Chapter Test Edition 2021-2022

### Chapter Test Solutions of ML Aggarwal for ICSE Class-9 Ch-9, Logarithms

Note:- Before viewing Solutions of Chapter -9 LogarithmsĀ Ā Class-9 of MLĀ AggarwalĀ Solutions . Ā Read the Chapter Carefully. Then solve all example given in Exercise-9.1, Exercise-9.2, MCQs, Chapter Test.

### Logarithms Chapter Test

ML Aggarwal Class 9 ICSE Maths Solutions

Page 179

#### Question 3. Prove the following

= 2[log 11 ā log 13] + [log 130 ā log 77] ā [log 55 ā log 91]

= 2 [log ā log 13] + [log 13 Ć 10 ā log 11 Ć 7] ā [log 11 ā log 13 Ć 7]

= 2 [log 11 ā log 13] + [(log 13 + log 10) ā (log 11 + log 7)] ā [(log 11 + log 5) ā (log 13 + log 7)]

= 2 log 11 ā 2 log 13 + log 10 ā log 11 ā log 7 ā log 11 ā log 5 + log 13 + log 7

= (2 log 11 ā log 11 ā log 11) + (-2 log 13 + log 13 + log 13) + log 10 ā log 5 + (log 7 ā log 7)

= 0 + 0 + log 10 ā log 5 + 0 = log 10 ā log 5

= log(10/5)

log 2 = R.H.S

Hence, Result is proved.

#### Question 4. If log (m + n) = log m + log n, show that n = m/(m-1)

Given log (m + n) = log m + log n
ā log (m + n) = log mn ā m + n = mn
ā m = mn ā n ā m = n (m -1)
ā n (m ā 1) = m ā n = m/(m-1)
Hence, result is proved.

#### Question 5. If log (x+y)/2 = 1/2 (log x + log y) prove that x = y.

Squaring
ā (x + y)2Ā = 4xy ā x2Ā + y2Ā + 2xy = 4xy
ā x2Ā + y2Ā + 2xy ā 4xy = 0 ā x2Ā + y2Ā ā 2xy = 0
ā (x ā y)2Ā = 0 ā x ā y = 0
ā“ x = y Hence proved.

#### Question 6. If a, b are positive real numbers, a > b and a2Ā + b2Ā = 27 ab, prove that

a2Ā + b2Ā = 27ab
ā a2Ā + b2Ā ā 2ab = 25ab

#### Question 7. Solve the following equations for x:

(i) logx 1/49 = -2

(ii) logx 1/4ā2 = -5

(iii) logx 1/243 = 10

(iv) logx 32 = x-4

(v) log7 (2xĀ² – 7)Ā  = 2

(vi) log (xĀ² – 21)Ā  = 2

(vii) log6 (x – 2)(x + 3)Ā  = 1

(viii) log6 (x – 2) + log6 (x + 3)Ā  = 1

(ix) log (x + 1) + log (x – 1) = log 11 + 2 log 3

ā x = +5, -5

##### (vi) log (x2 ā 21) = 2

(10)2Ā = x2Ā ā 21 ā 100 = x2Ā ā 21
ā x2Ā ā 21 = 100 ā x2Ā = 100 + 21
ā x2Ā = 121 ā x = Ā±ā121 ā x = Ā±11
ā“ x = 11, -11

##### (vii) log6Ā (x ā 2) (x + 3) = 1 {āµ loga a= 1}

Comparing,
(x ā 2)(x + 3) = 6
ā x2Ā + 3x ā 2x ā 6 = 6
ā x2Ā + x ā 6 ā 6 = 0
ā x2Ā + x- 12 = 0
ā x2Ā + 4x ā 3x- 12 = 0
ā x (x + 4) ā 3 (x + 4) = 0
ā (x + 4)(x ā 3) = 0
Either x + 4 = 0, then x = -4
or x ā 3 = 0, then x = 3
Hence x = 3, -4

##### (viii) log6Ā (x ā 2) + log6 (x + 3) = 1

log6Ā (x ā 2) (x + 3) = 1 = log6Ā 6 {āµ logaĀ a = 1}
Comparing,
(x ā 2) (x + 3) = 6 => x2Ā + 3x ā 2x ā 6 = 6
ā x2Ā + x ā 6 ā 6 = 0 ā x2Ā + x ā 12 = 0
ā x2Ā + 4x ā 3x ā 12 = 0
ā x (x + 4) ā 3 (x + 4) = 0
ā (x + 4)(x ā 3) = 0
Either x + 4 = 0, then x = -4
or x ā 3 = 0, then x = 3
ā“ x = 3, -4

##### (ix) log (x +1) + log (x -1) = log 11 + 2 log 3

ā log [(x + 1) (x ā 1)] = log 11 + log (3)2
ā log (x2Ā ā 1) = log 11 + log 9 [āµ a2Ā ā b2Ā = (a + b)(a ā b)]
ā log(x2Ā ā 1) = log(11 Ć 9) ā x2Ā ā 1 = 11 Ć 9
ā x2Ā ā 1 = 99 Ć x2Ā = 99 + 1 ā x2Ā = 100
ā x2Ā = (10)22Ā ā x = 10

#### Question 9. If a = 1 + logxyz, b = 1 + logyĀ zx and c = 1 + logzxy, then show that ab + bc + ca = abc.

a = 1 + logxyz
b = 1 + logyzx
c = 1 + logzxy
a = 1 + logxyz = logxx + logxyz

āĀ  : End of ML Aggarwal Logarithms Chapter Test Class 9 ICSE Maths Solutions :ā

Thanks

RELATED ARTICLES