(Q76).If log(
x+y
3
) =
1
2
(logx + logy), then find the value of
x
y
+
y
x
log(
x+y
3
) =
1
2
(logx + logy)
log(
x+y
3
) = (logx
1
2
+
1
2
)
log(
x+y
3
) = (log √x + log √
)
log(
x+y
3
) = log √x√
(
x+y
3
) = √x√
(
x+y
3
) = √
x + y =
√xy
(x + y)
2
= (
√xy)
2
x
2
+ y
2
+2xy =
xy
x
2
+ y
2
=
xy - 2xy
x
2
+ y
2
=
xy
x
2
+ y
2
xy
=
x
2
xy
+
y
2
xy
=
x
y
+
y
x
=
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