Find the value of $\log_{xz}(xy^4z) \times \log_{xy}(xyz^4)$

Question

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

Answer 1

$$ (1+4\log_{xz}y)\times (1+4\log_{xy}z) =(1+4\,\frac{\log_{x}y}{\log_{x}(xz)})\times (1+4\,\frac{\log_{x}z}{\log_x(xy)})\\ =(1+4\,\frac{\log_{x}y}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{1+\log_x(y)})\\ =(1+4\,\frac{\log_{x}y}{1+\log_{x}(z)})\times (1+4\,\frac{\log_{x}z}{1+\log_x(y)}) $$

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}

Answer 2

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}