Find the value of $\log_{xz}(xy^4z) \times \log_{xy}(xyz^4)$ if $x,y,z \gt 1$ and $x^2=yz$
find the value of 
$$E=\log_{xz}(xy^4z) \times \log_{xy}(xyz^4)$$ what i did is 
$$E=\log_{xz}(xy^4z) \times \log_{xy}(xyz^4)=(1+4\log_{xz}y)\times (1+4\log_{xy}z)$$  after that i could not proceed
 A: $$
(1+4\log_{xz}y)\times (1+4\log_{xy}z)
=(1+4\,\frac{\log_{x}y}{\log_{x}(xz)})\times (1+4\,\frac{\log_{x}z}{\log_x(xy)})\\
=(1+4\,\frac{\log_{x}y}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{1+\log_x(y)})\\
=(1+4\,\frac{\log_{x}y}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{1+\log_x(y)})
$$
Now, since $yz=x^2$, we have $y=x^2/z$, and so 
$$
\log_xy=\log_xx^2-\log_xz=2-\log_xz.
$$
Then
\begin{align}
(1+4\log_{xz}y)\times (1+4\log_{xy}z)
&=(1+4\,\frac{\log_{x}y}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{1+\log_x(y)})\\
&=(1+4\,\frac{2-\log_{x}z}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{3-\log_x(z)})\\
&=\frac{(9-3\log_xz)(3+3\log_xz)}{(1+\log_xz)(3-\log_xz)}\\ \ \\
&=3\times3=9.
\end{align}
A: Late to the party, but it is maybe simpler to replace directly $x^2=yz$, then $E =$
$$\log_{xz}(x^3y^3) \times \log_{xy}(x^3z^3)$$
then using $\log_{b}(x) = \frac{\log(x)}{\log(b)}$
$$= \frac{\log(x^3y^3)}{\log(xz)}\times\frac{\log(x^3z^3)}{\log(xy)}$$
then since $\log(x^n) = n \log(x)$
\begin{align}
&= \frac{\log((xy)^3)}{\log(xz)}\times\frac{\log((xz)^3)}{\log(xy)}\\ \ \\
&= 3\,\frac{\log(xy)}{\log(xz)}\times 3\,\frac{\log(xz)}{\log(xy)}\\ \ \\
&= 9
\end{align}
