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
 A: You have
$$(18\log_32\log_43-3\log_32\log_43)(2\log_23\log_32-2\log_32\log_83)\;,$$
which is correct. Clearly
$$18\log_32\log_43-3\log_32\log_43=15\log_32\log_43\;,$$
so we can immediately simplify it to
$$15\log_32\log_43(2\log_23\log_32-2\log_32\log_83)\;.\tag{1}$$
Now if $4^x=3$, then $2^{2x}=3$, so $\log_23=2\log_43$, or $\log_43=\frac12\log_23$. Similarly, you can verify that $\log_83=\frac13\log_23$. Thus, we can further simplify $(1)$ to
$$\frac{15}2\log_32\log_23\left(2\log_23\log_32-\frac23\log_32\log_23\right)\;.\tag{2}$$
This is extremely easy to evaluate if you know something about products of the form $\log_ab\log_ba$. If not, use the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. I’ll complete the calculation below but leave it spoiler-protected; mouse-over to see it.

 A useful general fact is that $\log_ab\log_ba=1$; this follows easily from the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. Thus, $\log_23\log_32=1$, and $(2)$ is simply $\dfrac{15}2\left(2-\dfrac23\right)=\dfrac{15}2\cdot\dfrac43=10$.

A: 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 $\log_ba=\frac{\log a}{\log b}$, $$\begin{vmatrix}
       \log_3512 & \log_43  \\
        \log_38 & \log_49 
 \end{vmatrix} =\begin{vmatrix}
       9\frac{\log 3}{\log 2} & \frac{\log3}{2\log 2}  \\
        3\frac{\log2}{\log 3} & \frac{2\log3}{2\log 2}
 \end{vmatrix} =9-\frac32$$
A: Hint: use the property of logarithm that $\log_ab=\frac{\log b}{\log a}$.
A: Recall that for $a > 1$, $\log_a : (0, \infty) \to \mathbb{R}$ and $$\log_a x = \frac{\ln x}{\ln a}.$$ As such, whenever we have a product of logarithms, we can swap their bases. That is $$\log_a x\log_b y = \frac{\ln x}{\ln a}\frac{\ln y}{\ln b} = \frac{\ln x}{\ln b}\frac{\ln y}{\ln a} = \log_b x\log_a y.$$
