Find the product of the following determinants (involving logarithms with different bases) Find the product of the following determinants:

$$\begin{vmatrix}
       \log_3512 & \log_43  \\
        \log_38 & \log_49 
 \end{vmatrix} * \begin{vmatrix}
       \log_23 & \log_83  \\
        \log_34 & \log_34 
 \end{vmatrix}$$

I tried to solve it like this:
$$\begin{vmatrix}
       9\log_2 3 & \log_43  \\
        3\log_32 & 2\log_43 
 \end{vmatrix} * \begin{vmatrix}
       \log_23 & \log_83  \\
        2\log_32 & 2\log_32 
 \end{vmatrix}$$ $$(9\log_2 3*2\log_43-3\log_32*\log_43)(\log_23*2\log_32-2\log_32*\log_83)$$
How do I get the final answer?
Please offer your assistance

HINT:
Putting $\log_32=a,$
$\log_3512=9\log_32=9a, \log_38=3\log_32=3a,$
$\log_43=\frac{\log 3}{\log 4}=\frac1{\frac{\log 4}{\log 3}}=\frac1{\log_34}=\frac1{2\log_32}=\frac1{2a}$ and so on
$$\begin{vmatrix}
       \log_3512 & \log_43  \\
        \log_38 & \log_49 
 \end{vmatrix}  =\begin{vmatrix}
       9a & \frac1{2a}  \\
        3a & \frac1a 
 \end{vmatrix} =\frac{15}2\text{ as }a=\log_32\ne0$$  

Alternatively using $\log_ba=\frac{\log a}{\log b}$, $$\begin{vmatrix}
       \log_3512 & \log_43  \\
        \log_38 & \log_49 
 \end{vmatrix} =\begin{vmatrix}
       9\frac{\log 3}{\log 2} & \frac{\log3}{2\log 2}  \\
        3\frac{\log2}{\log 3} & \frac{2\log3}{2\log 2}
 \end{vmatrix} =9-\frac32$$
