## 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 ⇒ x^{2} + y^{2} + 2xy = 4xy

⇒ x^{2} + y^{2} + 2xy – 4xy = 0 ⇒ x^{2} + y^{2} – 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 a^{2} + b^{2} = 27 ab, prove that

Answer:

a^{2} + b^{2} = 27ab

⇒ a^{2} + b^{2} – 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_{x}32 = x – 4

(v) log_{7} (2x^{2} – 1) = 2

(vi) log (x^{2} – 21) = 2

(vii) log_{6} (x – 2) (x + 3) = 1

(viii) log_{6} (x – 2) + log_{6} (x + 3) = 1

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

Answer:

⇒ x = +5, -5

(vi) log (x^{2} – 21) = 2

(10)^{2} = x^{2} – 21 ⇒ 100 = x^{2} – 21

⇒ x^{2} – 21 = 100 ⇒ x^{2} = 100 + 21

⇒ x^{2} = 121 ⇒ x = ±\(\sqrt{121}\) ⇒ x = ±11

∴ x = 11, -11

(vii) log_{6} (x – 2) (x + 3) = 1 {∵ log_{a} a= 1}

Comparing,

(x – 2)(x + 3) = 6

⇒ x^{2} + 3x – 2x – 6 = 6

⇒ x^{2} + x – 6 – 6 = 0

⇒ x^{2} + x- 12 = 0

⇒ x^{2} + 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) log_{6} (x – 2) + log_{6} (x + 3) = 1

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

Comparing,

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

⇒ x^{2} + x – 6 – 6 = 0 ⇒ x^{2} + x – 12 = 0

⇒ x^{2} + 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 (x^{2} – 1) = log 11 + log 9 [∵ a^{2} – b^{2} = (a + b)(a – b)]

⇒ log(x^{2} – 1) = log(11 × 9) ⇒ x^{2} – 1 = 11 × 9

⇒ x^{2} – 1 = 99 × x^{2} = 99 + 1 ⇒ x^{2} = 100

⇒ x^{2} = (10)2^{2} ⇒ x = 10

Question 8.

Solve for x and y:

Answer:

Question 9.

If a = 1 + log_{x}yz, b = 1 + log_{y} 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