Prove that log 75 16 - 2 log 5 9 + log 32 243 = log 2 .

Given log 75 16 - 2 log 5 9 + log 32 243 = log 75 - log 16 - 2 ( log 5 - log 9 ) + log 32 - log 243 = log 3 · 5 2 - log 2 4 - 2 ( log 5 - log 3 2 ) + log 2 5 - log 3 5 = 2 log 5 + log 3 - 4 log 2 - 2 log 5 + 4 log 3 + 5 log 2 - 5 log 3 = log 2 Hence proved.

Solve the equation a 2 x · b 3 x = c 5 .

Given equation is a 2 x · b 3 x = c 5 Taking log on both sides, we have log a 2 x · b 3 x = log c 5 ⇒ 2 x · log a + 3 x · log b = 5 log c ⇒ x ( 2 log a + 3 log b ) = 5 log c ∴ x = 5 log c 2 log a + 3 log b

Solve the equation a x + 1 b x - 1 = c 2 x .

Given equation is a x + 1 b x - 1 = c 2 x Taking logarithms on both sides, we have log a x + 1 b x - 1 = log c 2 x ⇒ x + 1 log a - x - 1 log b = 2 x log c ⇒ x ( log a - log b - 2 log c ) = - log a - log b ∴ x = log a + log b 2 log c + log b - log a

Solve the equations a 2 x b 3 y = m 5 a 3 x b 2 y = m 10 .

Given equations are a 2 x · b 3 y = m 5 ; a 3 x · b 2 y = m 10 Taking logarithms of both sides in both the equations, we get log a 2 x · b 3 y = log m 5 ⇒ 2 x log a + 3 y log b = 5 log m … 1 & log a 3 x · b 2 y = log m 10 ⇒ 3 x log a + 2 y log b = 10 log m … 2 Multiply Equation ( 1 ) by 3 and ( 2 ) by 2 and then subtract 5 y log b = - 5 log m ∴ y = - log m log b Similarly, multiply equation 1 by 2 and 2 by 3 and then subtract ∴ x = 4 log m log a .

If log x 2 y 3 = a and log x y = b , find log x and log y .

Given equations are log x 2 y 3 = a and log x y = b ∴ 2 log x + 3 log y = a … 1 and log x - log y = b … 2 Multiplying Eq. 2 by 2 and subtracting from Eq. 1 , we get 5 log y = a - 2 b ∴ log y = a - 2 b 5 Similarly, multiplying Eq. ( 2 ) by 3 and adding Eq. 1 , we get 5 log x = a + 3 b ∴ log x = a + 3 b 5

If a 3 - x b 5 x = a x + 5 · b 3 x , show that x log b a = log a .

Since a 3 - x · b 5 x = a x + 5 · b 3 x ⇒ a 3 - x a x + 5 = b 3 x b 5 x ⇒ a 3 - x - x + 5 = b 3 x - 5 x ⇒ a - 2 x + 2 = b - 2 x Taking log on both sides, we get - 2 x - 2 log a = - 2 x log b or x + 1 log a = x log b or x log a + log a = x log b or log a = x log b - log a or log a = x log b a Hence proved.

E M B I B E

Higher Algebra>Chapter 16 - Logarithms>EXAMPLES. XVI.a.

H S Hall and S R Knight Solutions for Chapter: Logarithms, Exercise 1: EXAMPLES. XVI.a.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Logarithms, Exercise 1: EXAMPLES. XVI.a.

Attempt the practice questions on Chapter 16: Logarithms, Exercise 1: EXAMPLES. XVI.a. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.