Solution for Simultaneous Logarithmic Equations Following are the Equations:
\begin{align}
\log x +\frac {\log(xy^8)}{((\log x)^2+(\log y)^2)} &= 2 \\
\log y + \frac{\log(x^8/y)}{((\log x)^2 +(\log y)^2)} &= 0
\end{align}
I tried substituting $\log x$ and $\log y$ with $a$ and $b$ but it results in a cubic equation with two variables .
 A: These are
$\log x +\dfrac{\log(x/y^8)}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{\log((x^8)y)}{(\log x)^2 +(\log y)^2)}
=0
$.
Expanding the logs,
$\log x +\dfrac{\log x-8\log y}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{8\log x+\log y}{(\log x)^2 +(\log y)^2)}
=0
$.
Letting
$\log x = a, \log y = b$,
$a +\dfrac{a-8b}{a^2+b^2}
= 2
$
and
$b+\dfrac{8a+b}{a^2 +b^2)}
=0
$.
The 8 is a mysterious constant,
so replace it by $c$.
$a +\dfrac{a-cb}{a^2+b^2}
= 2
$
and
$b+\dfrac{ca+b}{a^2 +b^2}
=0
$.
Now, solve.
Looks like we will get a cubic,
like you wrote.
$2(a^2+b^2)
=a(a^2+b^2) +a-cb
$
and
$0=
b(a^2+b^2)+ca+b
$.
From the first,
$a^2+b^2
=\dfrac{cb-a}{a-2}
$.
Putting this in the second,
$\begin{array}\\
0
&=b\dfrac{cb-a}{a-2}+ca+b\\
&=\dfrac{b(cb-a)+(ca+b)(a-2)}{a-2}\\
&=\dfrac{cb^2-ab+ca^2-2ca+ab-2b}{a-2}\\
&=\dfrac{cb^2+ca^2-2ca-2b}{a-2}\\
\end{array}
$
so,
if $a \ne 2$,
$0
=cb^2+ca^2-2ca-2b
$.
If $a=2$,
$0
=b(a^2+b^2)+ca+b
=b(4+b^2)+2c+b
$
so
$0
=b^3+5b+2c
$.
For $c=8$,
this has a negative real
(about -1.8771)
and two complex roots.
This looks like a mess,
so I am probably 
doing something wrong,
so I'll stop here.
