SOLUTION: If {{{x^2 + y^2 = 7xy}}} then {{{ log ((x + y)) + log (1/3) }}} = ?

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Question 974318: If then = ?
Found 2 solutions by JoelSchwartz, MathTherapy:
Answer by JoelSchwartz(130) (Show Source):
You can put this solution on YOUR website!
x^2+y^2=7xy
(x^2+y^2)/yx=7xy/xy
x/y+y/x=7
x/y=(y/x)^-1
z=x/y
1/z=y/x
z+1/z=7
z^2+1=7z
z^2-7z+1=0
(z-3.5)^2=z^2-7z+12.25
(z-3.5)^2=11.25
z-3.5=+-3.3541019662496845446137605030969
z=6.8541019662496845446137605030969,0.14589803375031545538623949690309
x/y=6.8541019662496845446137605030969 You can also try to use 0.14589803375031545538623949690309
y=0.97915742374995493494482292901384
x=6.7112448233925416874709033602397
log(7.6904022471424966224157262892536)+log(1/3)=?
log(15.333185446665945625783833530888)=?
?=1.1856323882658255372240209090029

Answer by MathTherapy(10704) (Show Source):
You can put this solution on YOUR website!

If then = ? ==================================================================== I don't know what convoluted stuff that other person who responded wrote, but all that MUMBO JUMBO seems, in my opinion, to have absolutley nothing to do with the expression to be evaluated! If then = ? ---- Substituting 7xy for ----- Taking the square root of each side = ? --- Substituting ONLY the POSITIVE value (), for x + y ----- Applying = = , or