The number of solutions $$(x,y)$$ of the simultaneous equations $$\log_4 x - \log_2 y = 0$$, $$x^2 = 8+2y^2$$ is ------.

The first equation gives $$\log_2 x = 2\log_2 y$$, i.e. $$x = y^2$$ 

Now, both $$x,y>0$$ as they are the arguments of the log function in the given equations

Substituting into the second equation gives $$y^4-2y^2-8=0$$,

Which can be factored as:

$$(y^2-4)(y^2+2)=0$$

So $$y^2=4$$, 

$$y=2$$ (taking the positive root) 

$$x=4$$. 

This gives exactly $$1$$ valid solution pair which is $$(4,2)$$