Given, 2 log 10 x + 1 2 log 10 y = 1 , express y in terms of x .

Given, 2 log 10 x + 1 2 log 10 y = 1 ⇒ log 10 x 2 + log 10 y 1 2 = 1 ∵ n log a m = log a m n ⇒ log 10 x 2 × y 1 2 = log 10 10 ∵ log 10 10 = 1 ; log a m + log a n = log m n ⇒ x 2 × y 1 2 = 10 ⇒ y 1 2 = 10 x 2 ⇒ y 1 2 × 2 = 10 x - 2 2 ⇒ y = 100 x - 4 Hence, y = 100 x - 4

If log p + q 3 = 1 2 ( log p + log q ) , prove that p 2 + q 2 = 7 p q .

Given: log p + q 3 = 1 2 ( log p + log q ) Now, 2 log p + q 3 = log p + log q log p + q 3 2 = log p + log q log p 2 + q 2 + 2 p q 9 = log p q Now campare the index both side it must be equal, ⇒ p 2 + q 2 + 2 p q 9 = p q ⇒ p 2 + q 2 + 2 p q = 9 p q ⇒ p 2 + q 2 = 7 p q , hence proved.

If x 2 + y 2 = 51 x y , prove that, log x - y 7 = 1 2 ( log x + log y )

The given expression is: x 2 + y 2 = 51 x y ⇒ x 2 + y 2 - 2 x y = 49 x y ⇒ x - y 2 = 7 2 x y ⇒ x - y 7 2 = x y Now taking log on both sides, we get ⇒ log x - y 7 2 = log x y ⇒ 2 log x - y 7 = log x + log y log x - y 7 = 1 2 ( log x + log y ) Hence proved.

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ICSE Mathematics>Algebra>Chapter 7 - Logarithms>Exercise

O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 3: Exercise

Author:O P Malhotra, S K Gupta & Anubhuti Gangal

O P Malhotra Mathematics Solutions for Exercise - O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 3: Exercise

Attempt the practice questions from Exercise 3: Exercise with hints and solutions to strengthen your understanding. ICSE Mathematics solutions are prepared by Experienced Embibe Experts.