Is this a trick question? My buddies little brother brought me this question he was assigned for homework. Either I'm crazy, the teacher messed up, or it's a trick question. 
I'm assuming $x\in\mathbb R$.

Solve for $x$

$$\log_2(x-2) - \log_2(x+2) = 2$$
Since $\log_2(x-2)$ is only defined for $x >2 $, and since $\log_2(x+2) > \log_2(x-2)$ over [2, $\infty)$, there exists no $x$ that satisfies the equation? 
Right??
 A: Good intuitive approach! Alternately, we can use log rules to see that, if $x$ is a solution to the equation, then $$\log_2\frac{x-2}{x+2}=2,$$ meaning that $$\frac{x-2}{x+2}=2^2=4,$$ so that $$x-2=4(x+2)=4x+8,$$ which we can solve to determine that $x=-\frac{10}3.$ But then neither $x+2$ nor $x-2$ is positive, so this is certainly not allowed in the original equation. Thus, there are no real solutions.
A: Actually, $\log_2(x-2)$ is defined only for $x>2$, but otherwise what you say is correct. Perhaps there’s a typo, and it was supposed to be $$\log_2(x+2)-\log_2(x-2)=2\;.$$ That leads fairly straightforwardly to the solution $x=\frac{10}3$, using methods appropriate to the level.
A: If you just want to solve it
$\log_2(x-2)-\log_2(x+2)=2$ or $\log_2(\frac{x-2}{x+2})=2$ 
which is equivalent to $\frac{x-2}{x+2}=4$ which gives $x=\frac{-10}{3}$.
A: $\log_2{\frac{x-2}{x+2}}=\log_2{4}$
$x-2=4(x+2)$
$3x = -10$
$x = -\frac{10}{3}$
Hence equation has no solution in $\mathbb{R}$, since both $\log_2{(x+2)}$ and $\log_2{(x-2)}$ are in $\mathbb{C}$.
A: In some mathematics texts, all log(x) is automatically considered as log|x|.
Thus, if x was found to be -10/3, the original equation still makes sense. Because ....
$log_2(x-2) - log_2(x+2)$
$= log_2|x-2| - log_2|x+2|$
= …
$= log_2|-16/3| - log_2|-4/3|$
$= log_2(16/3) - log_2(4/3)$
$= log_2(4)$
= 2
= RHS
