Do we have logarithm inequalities?

The logarithm can be simplified to obtain $x^2-4y^2=4 \tag{\$\star\$}$ Let $|x|-|y|=k$ ; thus, $(\star)$ becomes $3y^2-2k|y|+4-k^2=0$ or in general $3y^2 \pm 2ky+4-k^2=0$ Since $x,y \in \mathbb{R}$ , it follows that $\Delta=4k^2-12(4-k^2) \geq 0$ or simply $k^2-3 \geq 0$ Thus, $k \geq \sqrt{3}$ Which implies that $\boxed{a=3}$

First of all, $x$ certainly cannot be negative, as one of $x-2y$ or $x+2y$ will be negative, and negative numbers do not have real logarithms. Also, $y$ can either be positive or negative. It does not matter, as the absolute value is taken, and if y is negative, i.e. $y=-k$ for some $k$ , we get the equation $\log_4(x+2k)+\log_4(x-2]k)=1$ , which is basically the same with the original equation. So, we can assume without loss of generality that $x$ and $y$ are both positive, and we want to minimise $x-y$ .

So, we let $a= x+2y$ and $b=x-2y$ . Then we have that $log_4(ab) = 1$ , or $ab=4$ . We then solve for $x$ and $y$ in terms of $a$ and $b$ .

$a+b = (x+2y)+(x-2y) = 2x$ , giving us $x=\frac{a+b}{2}$

$a-b = (x+2y)-(x-2y) = 4y$ , giving us $y=\frac{a-b}{4}$

So, $x-y = \frac{a+b}{2}-\frac{a-b}{4} = \frac{a+3b}{4}$

So, we need to minimise $\frac{a+3b}{4}$ . By AM-GM, we get

$\frac{a+3b}{2} \geq \sqrt{3ab} = \sqrt{12}$ (since ab = 4)

So, $\frac{a+3b}{2} \geq 2\sqrt{3}$ , dividing both sides by 2 gives us

$\frac{a+3b}{4} \geq \sqrt{3}$

So, $\sqrt{a} = \sqrt{3}$ , $a = \boxed{3}$

It follows that x, 2 and 2y form sides of right angled triangle where x is the hypotenuse(if we consider only the positive values of x,y). Let @ be the angle between sides x and 2y. Using sine rule,x= 2 cosec@, y=cot@. Now minimum value of x-y= 2cosec@-cot@ can be found by calculus which is when @= 60° i.e. √3

We can also use the hyperbola approach and differentiate to get the min. Value

Firstly we have to solve the logarithmic equation to get the values of $x$ and $y$ .

$log_{4}(x+2y)+log_{4}(x-2y)=1$

$log_{4}(x+2y)(x-2y)=log_{4}4$

$log_{4}(x^{2}-4y^{2})=log_{4}4$

$log_{4}\frac{(x^{2}-4y^{2})}{4}=0$

So, $\frac{x^{2}-4y^{2}}{4}=0$

$\frac{x^{2}}{4}-y^{2}=0$

$\frac{x^{2}}{4}=y^{2}$

$x^{2}=4y^{2}$

$x=2y$

Now, $|x|-|y|=|2y|-(-y)$

$2y+y=3y$

So the difference between the values of $x$ and $y$ is the square root of $3$ where $3$ is an integer.

So, $\sqrt{a}=\sqrt{3}$

Therefore, $a=3$

So, the answer is: $a=\boxed{3}$