Evaluate : log a 2 b ÷ log a ( b ) 2

We have log a 2 b log a b 2 = log a 2 b 2 log a b = log a 2 b 4 log a b = 1 4 log b 2 b ∵ log m n = log n log m = 1 4 log b 2 + log b b = 1 4 log b 2 + 1

Evaluate : log a 2 b ÷ log a ( b ) 2

We have log a 2 b log a b 2 = log a 2 b 2 log a b = log a 2 b 4 log a b = 1 4 log b 2 b ∵ log m n = log n log m = 1 4 log b 2 + log b b = 1 4 log b 2 + 1

Evaluate: log 5 0 . 008

Let A = log 5 0 . 008 ⇒ A = log 5 8 1000 ⇒ A = log 5 1 125 ⇒ A = log 5 5 - 3 ⇒ A = - 3 log 5 5 ⇒ A = - 6

Prove that: log 3 log 2 log 3 81 = 1 .

LHS = log 3 log 2 log 3 81 ⇒ LHS = log 3 log 2 log 3 3 4 ⇒ LHS = log 3 log 2 8 ⇒ LHS = log 3 log 2 2 3 ⇒ LHS = log 3 3 ⇒ LHS = 1 = RHS

Prove that log a x × log b y = log b x × log a y .

We need to prove, log a x × log b y = log b x × log a y We know that log a m = log b m × log a b ∴ log a x = log b x × log a b and log b y = log b y × log a b So, LHS = log a x × log b y = log b x × log a b × log b y × log a b Similarly, RHS = log b x × log a y = log b x × log a b × log b y × log a b Then, LHS = RHS Hence, proved.

Prove that log a x = log b x × log c b × . .. × log n m × log a n

R . H . S . = log b x × log c b × . .. × log n m × log a n = log x log b × log b log c × . . . . . × log m log n × log n log a = log x log a = log a x = L . H . S .

Prove that: log a ( log b a ) log b ( log a b ) = − log a b

L . H . S = log a ( log b a ) log b ( log a b ) = log log b a log a × log b log log a b = log log b a log log a b × log b log a = log log b a log 1 log b a × log a b = log log b a log log b a - 1 × log a b = log log b a - log log b a × log a b = - log a b