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If a2 + b2 = 7ab, show thatlog(\(\frac{a+b}3\)) = 1/2 log a + 1/2 log b |
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Answer» a2 + b2 = 7ab a2 + 2ab + b2 = 7ab + 2ab (a + b)2 = 9ab \(\frac{(a+b)^2}9\) = ab \((\frac{a+b}3)^2 = ab\) Taking log on both sides, we get log\((\frac{a+b}3)^2=log (ab)\) 2 log\((\frac{a+b}3)\) = log a + log b Dividing throughout by 2, we get log\((\frac{a+b}3)=\frac12log\,a+\frac12 log\,b\) |
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