Question

x^2+y^2=23xy probe log(x+y)/5=1/2( logx+logy)

Answer 1

The answer is given below :

Given,

${x}^{2} + {y}^{2} = 23xy \\ \\ or \: \: {(x + y)}^{2} - 2xy = 23xy \\ \\ or \: \: {(x + y)}^{2} = 25xy \\ \\ taking \: \: log \: \: we \: \: get \\ \\ log \: {(x + y)}^{2} = log \: (25xy) \\ \\ or \: \: 2 \: log(x + y) = log(25) + logx + logy \\ \\ or \: \: 2 \: log(x + y )= log ({5}^{2} ) + logx + logy \\ \\ or \: \: 2 \: log(x + y) - 2 \: log5 = logx + logy \\ \\ or \: \: 2 \: (log(x + y) - log5) = logx + logy \\ \\ or \: \: 2log (\frac{x + y}{5}) = logx + logy \\ \\ or \: \: log( \frac{x + y}{5} ) = \frac{1}{2} (logx + logy)$


[PROVED]

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