if log ( x+y/3) =1/2[log x + log y],then x/y + y/x =?
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log[(x+y)/3] = 1/2[log x +log y]
⇒2 log[(x+y)/3] = log xy
⇒log[( x+y)/3]²= log xy
⇒[(x+y)/3]² = xy
⇒x²+ y² + 2xy = 9xy
⇒x²+y² =7xy
⇒(x²+y²)/xy =7
⇒x/y +y/x =7
⇒2 log[(x+y)/3] = log xy
⇒log[( x+y)/3]²= log xy
⇒[(x+y)/3]² = xy
⇒x²+ y² + 2xy = 9xy
⇒x²+y² =7xy
⇒(x²+y²)/xy =7
⇒x/y +y/x =7
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