ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms Chapter Test

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms Chapter Test

Question 1.

Expand (log _{a} sqrt[3]{x^{7} y^{8} div sqrt[4]{z}})

Answer:

Question 2.

Find the value of (log _{sqrt{3}} 3 sqrt{3}-log _{5}(0 cdot 04))

Answer:

Question 3.

Prove the following



Answer:



= 2[log 11 – log 13] + [log 130 – log 77] – [log 55 – log 91]
= 2 [log – log 13] + [log 13 × 10 – log 11 × 7] – [log 11 – log 13 × 7]
= 2 [log 11 – log 13] + [(log 13 + log 10) – (log 11 + log 7)] – [(log 11 + log 5) – (log 13 + log 7)]
= 2 log 11 – 2 log 13 + log 10 – log 11 – log 7 – log 11 – log 5 + log 13 + log 7

= (2 log 11 – log 11 – log 11) + (-2 log 13 + log 13 + log 13) + log 10 – log 5 + (log 7 – log 7)

= 0 + 0 + log 10 – log 5 + 0 = log 10 – log 5

= log(left(frac{10}{5}right)) = 1og 2 = R.H.S

Hence, Result is proved.

Question 4.

If log (m + n) = log m + log n, show that n = (frac{m}{m-1}).

Answer:

Given log (m + n) = log m + log n

⇒ log (m + n) = log mn ⇒ m + n = mn

⇒ m = mn – n ⇒ m = n (m -1)

⇒ n (m – 1) = m ⇒ n = (frac{m}{m-1})

Hence, result is proved.

Question 5.

If (log frac{x+y}{2}=frac{1}{2}(log x+log y)), prove that x = y.

Answer:



Squaring

⇒ (x + y)² = 4xy ⇒ x² + y² + 2xy = 4xy

⇒ x² + y² + 2xy – 4xy = 0 ⇒ x² + y² – 2xy = 0

⇒ (x – y)² = 0 ⇒ x – y = 0

∴ x = y Hence proved.

Question 6.

If a, b are positive real numbers, a > b and a² + b² = 27 ab, prove that



Answer:

a² + b² = 27ab

⇒ a² + b² – 2ab = 25ab





Hence proved.

Solve the following equations for x

Question 7.

Solve the following equations for x:

(i) log_x(frac{1}{49}) = -2

(ii) log_x(frac{1}{4 sqrt{2}}) = -5

(iii) log_x(frac{1}{243}) = 10

(iv) log_x32 = x – 4

(v) log₇ (2x² – 1) = 2

(vi) log (x² – 21) = 2

(vii) log₆ (x – 2) (x + 3) = 1

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

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

Answer:





⇒ x = +5, -5

(vi) log (x² – 21) = 2

(10)² = x² – 21 ⇒ 100 = x² – 21

⇒ x² – 21 = 100 ⇒ x² = 100 + 21

⇒ x² = 121 ⇒ x = ±(sqrt{121}) ⇒ x = ±11

∴ x = 11, -11

(vii) log₆ (x – 2) (x + 3) = 1 {∵ log_a a= 1}

Comparing,

(x – 2)(x + 3) = 6

⇒ x² + 3x – 2x – 6 = 6

⇒ x² + x – 6 – 6 = 0

⇒ x² + x- 12 = 0

⇒ x² + 4x – 3x- 12 = 0

⇒ x (x + 4) – 3 (x + 4) = 0

⇒ (x + 4)(x – 3) = 0

Either x + 4 = 0, then x = -4

or x – 3 = 0, then x = 3

Hence x = 3, -4

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

log₆ (x – 2) (x + 3) = 1 = log₆ 6 {∵ log_a a = 1}

Comparing,

(x – 2) (x + 3) = 6 => x² + 3x – 2x – 6 = 6

⇒ x² + x – 6 – 6 = 0 ⇒ x² + x – 12 = 0

⇒ x² + 4x – 3x – 12 = 0

⇒ x (x + 4) – 3 (x + 4) = 0

⇒ (x + 4)(x – 3) = 0

Either x + 4 = 0, then x = -4

or x – 3 = 0, then x = 3

∴ x = 3, -4

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

⇒ log [(x + 1) (x – 1)] = log 11 + log (3)²

⇒ log (x² – 1) = log 11 + log 9 [∵ a² – b² = (a + b)(a – b)]
⇒ log(x² – 1) = log(11 × 9) ⇒ x² – 1 = 11 × 9

⇒ x² – 1 = 99 × x² = 99 + 1 ⇒ x² = 100

⇒ x² = (10)2² ⇒ x = 10

Question 8.

Solve for x and y:



Answer:

Question 9.

If a = 1 + log_xyz, b = 1 + log_y zx and c = 1 + log_zxy, then show that ab + bc + ca = abc.

Answer:

a = 1 + log_xyz

b = 1 + log_yzx

c = 1 + log_zxy

a = 1 + log_xyz = log_xx + log_xyz

ML Aggarwal Class 9 Solutions for ICSE Maths