**ML Aggarwal Logarithms Chapter Test Class 9 ICSE Maths APC Understanding Solutions.** Solutions ofĀ Chapter Test. This post is the Solutions ofĀ **ML Aggarwal**Ā Ā Chapter 9- Logarithms forĀ **ICSE**Ā MathsĀ **Class-9.**Ā APC UnderstandingĀ **ML Aggarwal**Ā Solutions (APC) Avichal Publication Solutions of Chapter-9Ā Logarithms for**Ā ICSEĀ **Board**Ā Class-9****.Ā **Visit official websiteĀ **CISCE**Ā 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 Aggarwa**lĀ **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 1. loga …..**

**Answer:Ā **

**Question 2. Find the value of logā3 3****ā3 – log5 (0.04)**

**Answer:Ā **

**Question 3. Prove the following**

**Answer:Ā **

= 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)**

**Answer:Ā **

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.**

**Answer:Ā **

Squaring

ā (x + y)^{2}Ā = 4xy ā x^{2}Ā + y^{2}Ā + 2xy = 4xy

ā x^{2}Ā + y^{2}Ā + 2xy ā 4xy = 0 ā x^{2}Ā + y^{2}Ā ā 2xy = 0

ā (x ā y)^{2}Ā = 0 ā x ā y = 0

ā“ x = y Hence proved.

**Question 6. If a, b are positive real numbers, a > b and a**^{2}Ā + b^{2}Ā = 27 ab, prove that

^{2}Ā + b

^{2}Ā = 27 ab, prove that

**Answer:Ā **

a^{2}Ā + b^{2}Ā = 27ab

ā a^{2}Ā + b^{2}Ā ā 2ab = 25ab

** **

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

(i) logx 1/49 = -2

(ii) logx 1/4ā2 = -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

**Answer:Ā **

ā x = +5, -5

**(vi) log (x**^{2} ā 21) = 2

^{2}ā 21) = 2

(10)^{2}Ā = x^{2}Ā ā 21 ā 100 = x^{2}Ā ā 21

ā x^{2}Ā ā 21 = 100 ā x^{2}Ā = 100 + 21

ā x^{2}Ā = 121 ā x = Ā±ā121 ā x = Ā±11

ā“ x = 11, -11

**(vii) log**_{6}Ā (x ā 2) (x + 3) = 1 {āµ log_{a} a= 1}

_{6}Ā (x ā 2) (x + 3) = 1 {āµ log

_{a}a= 1}

Comparing,

(x ā 2)(x + 3) = 6

ā x^{2}Ā + 3x ā 2x ā 6 = 6

ā x^{2}Ā + x ā 6 ā 6 = 0

ā x^{2}Ā + x- 12 = 0

ā x^{2}Ā + 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) log**_{6}Ā (x ā 2) + log_{6} (x + 3) = 1

_{6}Ā (x ā 2) + log

_{6}(x + 3) = 1

log_{6}Ā (x ā 2) (x + 3) = 1 = log_{6}Ā 6 {āµ log_{a}Ā a = 1}

Comparing,

(x ā 2) (x + 3) = 6 => x^{2}Ā + 3x ā 2x ā 6 = 6

ā x^{2}Ā + x ā 6 ā 6 = 0 ā x^{2}Ā + x ā 12 = 0

ā x^{2}Ā + 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 (x^{2}Ā ā 1) = log 11 + log 9 [āµ a^{2}Ā ā b^{2}Ā = (a + b)(a ā b)]

ā log(x^{2}Ā ā 1) = log(11 Ć 9) ā x^{2}Ā ā 1 = 11 Ć 9

ā x^{2}Ā ā 1 = 99 Ć x^{2}Ā = 99 + 1 ā x^{2}Ā = 100

ā x^{2}Ā = (10)2^{2}Ā ā x = 10

**Question 8. Solve for x and y:**

**Question 9. If a = 1 + log**_{x}yz, b = 1 + log_{y}Ā zx and c = 1 + log_{z}xy, then show that ab + bc + ca = abc.

_{x}yz, b = 1 + log

_{y}Ā zx and c = 1 + log

_{z}xy, then show that ab + bc + ca = abc.

**Answer:**

a = 1 + log_{x}yz

b = 1 + log_{y}zx

c = 1 + log_{z}xy

a = 1 + log_{x}yz = log_{x}x + log_{x}yz

āĀ : End of ML Aggarwal Logarithms Chapter Test Class 9 ICSE Maths Solutions :ā

Return to :-Ā **Ā ML Aggarawal Maths Solutions for ICSEĀ Class-9**

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