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Answer 1

Hey mate!

Here is yr answer....

Answer 2

Solution:

Given

log[(x+y)/3]=1/2[logx+logy]

=> 2log[(x+y)/3]=[logx+logy]

=> log[(x+y)/3]²=log(xy)

/* we know the logarithmic laws:

i) nloga = log aⁿ

ii) log m + log n = log (mn)

*)

Remove log both sides of the equation, we get

=> [(x+y)/3]² = xy

=> (x+y)²/9 = xy

=> (x+y)² = 9xy

=> x²+y² + 2xy = 9xy

=> x²+y² = 9xy - 2xy

=> x²+y² = 7xy

Divide each term by xy, we get

=> x²/xy + y²/xy = 7xy/xy

=> x/y + y/x = 7

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