<s>[INST] $\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 $+6,-6$
 [/INST] Ah you meant $\log_2((x-2)^2)$ not $(\log_2(x-2))^2$? That wasn't clear to me. Are you sure that its +- 6? Wolframalpha gives completely different answers, no matter whether I use $\log_2((x-2)^2)$ or $(\log_2(x-2))^2$.
This might help you a bit, but don't expect really nice answers unless you've made a mistake in the phrasing of your question.
$$\frac{1}{\log_2((x-2)^2)} + \frac{1}{\log_2((x+2)^2)} = \frac{5}{12}$$
$$\frac{1}{2\log_2(x-2)} + \frac{1}{2\log_2(x+2)} = \frac{5}{12}$$
$$\frac{1}{\log_2(x-2)} + \frac{1}{\log_2(x+2)} = \frac{5}{6}$$
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