if log x+y/3=1/2 logx+logy prove that x/y +y/x=7
Answers
Answered by
31
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
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
Answered by
28
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
I HOPE IT'S HELP YOU DEAR,
THANKS
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
I HOPE IT'S HELP YOU DEAR,
THANKS
Similar questions