iflog[x+y/3]=1/2[logx+logy] thenx/y+y/x is
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Log [ (x+y)/3 ] = 1/2 * [ Log x + Log y ]
= 1/2 * Log xy
= Log √(xy)
(x+ y )/3 = √xy
x + y = √(xy) / 3
x² + y² = xy/9
x/y + y/x = 1/9
= 1/2 * Log xy
= Log √(xy)
(x+ y )/3 = √xy
x + y = √(xy) / 3
x² + y² = xy/9
x/y + y/x = 1/9
kvnmurty:
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