<s>[INST] Lograthmic equation $ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}$ solutions $$ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}.$$
I made the graph using wolfram alpha it is giving answer as 
6. But how to solve it algebraically?
base of logarithm is 2.
Tried using taking Lcm but then two different log terms 
are getting formed.
 [/INST] $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2} = \frac{5}{12}$$
Rewrite the logs using $$\log_a b^c = c\log_a b$$
and factor. Then, simplify both sides.
$$\frac{1}{2}\cdot\frac{1}{\log_2\vert x-2\vert}+\frac{1}{2}\cdot\frac{1}{\log_2\vert x+2\vert} = \frac{5}{12}$$
$$\frac{1}{2}\bigg(\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert}\bigg) = \frac{5}{12}$$
$$\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert} = \frac{5}{6}$$
$$\frac{\log_2\vert x-2\vert+\log_2\vert x+2\vert}{\log_2\vert x-2\vert\cdot\log_2\vert x+2\vert} = \frac{5}{6}$$
Set $\color{blue}{a = \log_2\vert x-2\vert}$ and $\color{purple}{b = \log_2\vert x+2\vert}$.
$$\frac{\color{blue}{a}+\color{purple}{b}}{\color{blue}{a}\color{purple}{b}} = \frac{5}{6}$$
Can you take it on from here? (Hint: Solve for possible values of $a$ and $b$. Then, plug in $\color{blue}{\log_2\vert x-2\vert = a}$ and $\color{purple}{\log_2\vert x+2\vert = b}$ and check for any extraneous solutions, in case there are any.)
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