Is it hard to prove this identity: $$2 \log (a) \log (b)=\log(a b)^2-\log(a)^2-\log(b)^2$$
for $a>1$ and $b>1$?
$$\begin{align}(\log ab)^2 &= (\log a + \log b)^2 \\ &= (\log a)^2 + (\log b)^2 + 2\log a \log b \end{align}$$
Now, just rearrange by leaving $2\log a \log b$ on the right hand side.
Here's a hint: $A^2 + 2AB +B^2 = (A+B)^2$. Here's another hint: $\log a + \log b = \log(ab)$.
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$$\begin{align}(\log ab)^2 &= (\log a + \log b)^2 \\ &= (\log a)^2 + (\log b)^2 + 2\log a \log b \end{align}$$
Now, just rearrange by leaving $2\log a \log b$ on the right hand side.