[Math] Find the product of the following determinants (involving logarithms with different bases)

determinantlogarithms

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

Best Answer

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 $\dfrac{15}2\left(2-\dfrac23\right)=\dfrac{15}2\cdot\dfrac43=10$.

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