Question

if log x+y/3=1/2 logx+logy prove that x/y +y/x=7

Answer 1

it is given that,
log{(x + y)/3}= 1/2(logx + logy)

or, log{(x + y)/3} = 1/2 logxy [ we know, logm + logn = logmn , similarly logx + logy = logxy ]

or, 2log{(x + y)/3}= logxy

or, log{(x + y)/3}² = logxy [ as we know, nlogm = logmⁿ similarly , 2log(x + y/3) = log(x + y/3)² ]

or, (x + y)²/9 = xy

or, x² + y² + 2xy = 9xy

or, x² + y² = 7xy

or, x²/xy + y²/xy = 7xy/xy

or, x/y + y/x = 7

hence, x/y + y/x = 7

Answer 2

HELLO DEAR,



GIVEN:-
log{(x + y)/3}= 1/2(logx + logy)

=> log{(x + y)/3} = 1/2 logxy

[ as,logm + logn = logmn ]

=> 2log{(x + y)/3}= logxy

=> log{(x + y)/3}² = logxy

[ as , nlogm = logmⁿ ]

=> (x + y)²/9 = xy

=> x² + y² + 2xy = 9xy

=> x² + y² = 7xy

=> x²/xy + y²/xy = 7xy/xy

=> x/y + y/x = 7

hence, x/y + y/x = 7



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