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.
 A: $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2} = \frac{5}{12}$$
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.)
A: Define $f:\mathbb{R}\setminus\{\pm1,\pm3\}\to\mathbb{R}$ by $$f(x):=\frac{1}{\log_2\big((x-2)^2\big)}+\frac{1}{\log_2\big((x+2)^2\big)}$$ for all real numbers $x\neq \pm1,\pm2,\pm3$, and $f(\pm 2)$ is defined to be $\dfrac14$.  Note that $f$ is a continuous even function (i.e., $f(-x)=f(x)$).  Therefore, it suffices to solve $f(x)=y$ for $x\geq 0$.  
Observe that $f$ is strictly increasing on $[0,1)$ with local minimum $f(0)=1$ and without upper bound.  Also, $f$ is strictly increasing on $(1,2]$, with local maximum $f(2)=\dfrac14$ and without lower bound.  On $[2,3)$, $f$ is strictly decreasing with local maximum $f(2)=\dfrac14$ and without lower bound.  Then, $f$ is strictly decreasing on $(3,\infty)$ with local infimum $\lim\limits_{x\to\infty}\,f(x)=0$ and without upper bound.  I leave the proof of the above analysis to you.    
Hence, for any real number $y$, the number $n(y)$ of $x\in\mathbb{R}$ such that $f(x)=y$ is
$$n(y)=\left\{\begin{array}{ll}
4&\text{if }y>1\,,\\
3&\text{if }y=1\,,\\
2&\text{if }\frac14< y<1\,,\\
4&\text{if }y=\frac14\,,\\
6&\text{if }0<y<\frac14\,,\\
4&\text{if }y\leq 0\,.
\end{array}\right.$$
In particular, $n\left(\dfrac{5}{12}\right)=2$.  Since $x=\pm6$ are solutions to $f(x)=\dfrac{5}{12}$, they are the only solutions.
