$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $ Find all x's 
It is given that $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $$ Find all possible values of $x$.

What I did: $$\frac{1}{\log_2(x-2)}+\frac{1}{\log_2(x+2)}=\frac{5}{6} $$ $$\log_{(x-2)}2 +\log_{(x+2)}2 =\frac{5}{6}$$
What is the next ? I stuck in here..
I am looking for algebraic approach. The answer is $+6,-6$

Let $a = \log_2(x-2)$ and $b = \log_2(x+2)$.  Then your equation becomes:
$$\frac{1}{a} + \frac{1}{b} = \frac{5}{6}$$
$$\frac{a + b}{ab} = \frac{5}{6}$$
$$6(a + b) = 5ab$$
$$b = \frac{6a}{5a - 6}$$
$$\log_2 (2^a + 4) = \frac{6a}{5a - 6}$$
This has the integer solution $a = 2$.  Unfortunately, while this solution is easy to verify, it's hard to see how to explicitly solve for $a$.
Anyhow, with $\log_2(x-2) = 2$, we get $x - 2 = 4$, or $x = 6$.
