this is log{(a+b)/2} or {log(a+b)}/2

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this is log{(a+b)/2} or {log(a+b)}/2

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log (a+b)/2 = 1/2(log a + log b)

log (a+b)/2 = 1/2(log ab)

log (a+b)/2 = log (√ab)

⇒(a+b)/2 = √ab

Squaring on both sides we get

(a+b)²/4 = ab

(a+b)² = 4ab

a² + b² + 2ab - 4ab = 0

a² + b² - 2ab = 0

(a-b)² = 0

(a-b) = 0

⇒ a = b

Hence proved.