If x2 + y2 = 23xy, prove that log x+y/5 = 1/2(log x + log y).​

2 thoughts on “If x2 + y2 = 23xy, prove that log x+y/5 = 1/2(log x + log y).​”

1. Answer:

   [tex] {x}^{2} + {y}^{2} = 23xy \\ {x}^{2} + {y}^{2} + 2xy = 23xy + 2xy \\ {(x + y)}^{2} = 25xy \\ \frac{ {(x + y)}^{2} }{25} = xy \\ {( \frac{x + y}{5}) }^{2} = xy[/tex]

   Now applying log on both sides of this simplified equation,

   [tex] log( {( \frac{x + y}{5}) }^{2} ) = log(xy) [/tex]

   Now using the properties of logarithms,

   [tex]2 log( \frac{x + y}{5} ) = log(x) + log(y) [/tex]

   [tex] log( \frac{x + y}{5} ) = \frac{ log(x) + log(y) }{2} [/tex]

   Hence proved.

   Reply

2. Step-by-step explanation:

   (x+y)2 = x2 + 2xy + y2 = 25xy

   log(x+y)2 = 2log(x+y)

   But, log(x+y)2 = log(25xy)

   = log25 + logx + logy

   = log(52) + logx + logy

   = 2log5 + logx + logy

   So, 2log(x+y) = 2log5 + logx + logy

   2(log(x+y) – log5) = logx + logy

   log(x+y) – log5 = ½[logx + logy]

   log{(x+y)/5} = ½[logx + logy]

   Reply