prove
Share
prove
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
a² + b² = 7ab
=> a² + b² + 2ab = 7ab + 2ab
=> (a + b)² = 9ab
=> (a + b) = 3√(ab)
=> (a + b)/3 = √ab
=> log{(a + b)/3} = log (√ab)
=> log{(a + b)/3} = (1/2)log(ab)
=> log{(a + b)/3} = (1/2)(loga+ logb)
=> log{(a + b)/3} = (loga)/2+ (logb)/2
=> log{(a + b)/3} - (loga)/2- (logb)/2 = 0 (Ans.)
i) By algebra identity, (a + b)² = a² + b² + 2ab
ii) Substituting for a² + b² = 7ab in the above, (a + b)² = 9ab
Taking square root on both sides, (a + b) = 3√(ab)
==> (a + b)/3 = √ab
Taking log on both sides, log{(a + b)/3} = (1/2)[log(a) + log(b)]
==> log{(a + b)/3} - {log(a)/2} - {Log(b)/2} = 0
[Kindly verify your requirement]
= Log((a+b)/2) - (1/2)Log(ab)
= (1/2)Log((a+b)^2 / (4ab))
= (1/2)Log((a^2 + b^2 + 2ab)/(4ab))
= (1/2)Log((7ab + 2ab)/(4ab))
= (1/2)Log(9/4)