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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 $b>1$?

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Use $$\log a^m=m\log a$$ to (dis)prove – lab bhattacharjee Mar 21 '14 at 13:17
Judging by Yiyuan Lee and Umberto P.'s answer,the correct answer should be "no". – rah4927 Mar 21 '14 at 17:39
up vote 21 down vote accepted

$$\begin{align}(\log ab)^2 &= (\log a + \log b)^2 \\ &= (\log a)^2 + (\log b)^2 + 2\log a \log b \end{align}$$

Now, just rearrange by leaving $2\log a \log b$ on the right hand side.

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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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