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

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: If {{{x^2 + y^2 = 7xy}}} then {{{ log ((x + y)) + log (1/3) }}} = ?      Log On


   

Question 974318: If x%5E2+%2B+y%5E2+=+7xy then +log+%28%28x+%2B+y%29%29+%2B+log+%281%2F3%29+ = ?
Answer by JoelSchwartz(130) About Me  (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