ifa3+b3=ab(8-3a-3b),then shoe that log (a=b/2)=1/3(loga+logb)
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a³ + b³ = a b ( 8 - 3 a - 3 b)
=> a³ + b³ + 3 a² b + 3 ab² = 8 a b
=> Log (a + b)³ = Log 8 a b = Log 8 + Log a + Log b
=> 3 Log (a+b) = 3 Log 2 + Log a + Log b
=> 3 (Log (a+b) - Log 2] = Log a + log b
=> Log [(a + b) / 2] = (Log a + Log b) / 3
=> a³ + b³ + 3 a² b + 3 ab² = 8 a b
=> Log (a + b)³ = Log 8 a b = Log 8 + Log a + Log b
=> 3 Log (a+b) = 3 Log 2 + Log a + Log b
=> 3 (Log (a+b) - Log 2] = Log a + log b
=> Log [(a + b) / 2] = (Log a + Log b) / 3
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